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What is the difference in assumptions between the one-mean t-test and the one-mean z-test?

The one mean z-test - ϑ is known

The one mean t-test - ϑ unknown

The one mean t-test - ϑ unknown

What are the two ways to create a narrower confidence interval?

(1) decrease ϑ

(2) increase n (sample size)

(2) increase n (sample size)

What are the two ways to create a wider confidence interval?

(1) increase ϑ

(2) decrease n (sample size)

(2) decrease n (sample size)

Formula for confidence interval, sigma known

What does ∝ signify?

The area in both tails. Usually .10 or less

A sample of 36 soda cans was taken and the average number of ounces was found to be 12.4 oz. (assume ϑ=2). Find a 95% confidence interval for µ, the average number of oz in the can. Interpret your results.

(11.747, 13.053)

We are 95% confident that the average number of ounces in a soda can is between 11.8 ounces and 13.1 ounces.

We are 95% confident that the average number of ounces in a soda can is between 11.8 ounces and 13.1 ounces.

What determines "z" in the confidence interval formula?

Changing the confidence level.

What is the relationship between confidence (high or low) and the length of a confidence interval?

Higher confidence means a wider interval. Lower confidence means a narrower interval.

A sample of 25 typists has an average typing speed of 85 wpm. Assume ó=19. Find a 99% confidence interval for the average speed of typists. Interpret your results.

(75.2, 94.8)

We are 99% confident the average typing speed for typists is between 75.2 words per minute and 94.8 words per minute.

We are 99% confident the average typing speed for typists is between 75.2 words per minute and 94.8 words per minute.

What calculator function is used to find a confidence interval when ó is known?

STAT TESTS Z-interval

Formula to calculate E (margin of error)

What is the relationship between the margin of error and the length of the confidence interval?

The margin of error is half the length of the confidence interval.

Relate the precision with which x-bar estimates µ with the size of the margin of error.

The more precision with which x-bar estimates µ, the smaller the margin of error.

An AP poll found that 38% of parents said they were unlikely to give permission for their kids to be vaccinated at school (sample of 1003 adults). The margin of sampling error is +- 3.1 percentage points for all adults. What is the confidence interval? What is the length of the confidence interval?

(34.9, 41.1)

Length of confidence interval: 6.2

Length of confidence interval: 6.2

Formula to find the sample size required for a 1-∝ confidence interval.

The FAA estimated with 90% confidence that the mean flight time from Albuquerque NM to Dallas TX to be between 99 minutes and 107.8 minutes. Assume n = 9 and ϑ = 8. (1) Find the margin of error. (2) Find how many times larger must a sample size be to halve the margin of error.

(1) 4.4

(2) 4 (36/9)

(2) 4 (36/9)

What are the properties of the t-distribution?

a) centered at 0 (like the standard normal distribution)

b) symmetric/bell-shaped

c) wider than the normal distribution

d) area under curve = 1

e) uses degrees of freedom (n-1) to find t-values

b) symmetric/bell-shaped

c) wider than the normal distribution

d) area under curve = 1

e) uses degrees of freedom (n-1) to find t-values

Which would result in a wider confidence interval? 90% confidence level or 95% confidence level?

95% confidence level would result in a wider confidence interval. Increasing the confidence level increases the length of the confidence interval.

Which would result in a wider confidence interval?

n=100 or n=400

n=100 or n=400

n=100 would result in a wider confidence interval. Decreasing the sample size increases the length of the confidence interval.

What is the appropriate z-value for a 95% confidence level?

1.96

What is the appropriate z-value for a 90% confidence level?

1.645

What is the appropriate z-value for a 99% confidence level?

2.576

What is the appropriate z-value for an 80% confidence level?

1.282

What is the appropriate z-value for an 85% confidence level?

1.44

(Hint: invnorm (.075))

(Hint: invnorm (.075))

What is the appropriate z-value for a 50% confidence level?

.674

(Hint: invnorm (.25))

(Hint: invnorm (.25))

Suppose that a random sample of 50 bottles of a particular brand of cough medicine is selected and the alcohol content of each bottle is determined. Let µ denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the sample of 50 results in a 95% confidence interval for µ of (7.8, 9.4).

(1) Would a 90% confidence interval have been narrower or wider than the given interval?

(1) Would a 90% confidence interval have been narrower or wider than the given interval?

(1) The 90% confidence interval would have been narrower because decreasing the confidence level from 95% to 90% will decrease the confidence interval. The value of Z∝/2 for a 90% confidence level (1.28) is smaller than the Z∝/2 value for the 95% confidence level (1.64); therefore the confidence interval would be narrower.

Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding 95% confidence interval is repeated 100 times, 95 of the resulting interval will include µ. Is this statement correct? Why or why not?

** This statement is correct. In the long run, after computing the corresponding 95% confidence interval many times, 95 of the resulting confidence intervals will include µ.

Two intervals (114.4, 115.6) and (114.1, 115.9) are confidence intervals for µ=true average resonance frequency (in hertz) for all tennis rackets of a certain type. (1) What is the value of the sample mean resonance frequency? (2) The confidence level for one of these intervals is 90% and for the other it is 99%. Which is which, and how can you tell?

(1) x-bar = 115 hertz

(2) the 90% confidence interval is (114.4, 115.6) and the 99% confidence interval is (114.1, 115.9). The confidence interval is larger for the 99% confidence level than for the 90% confidence level.

(2) the 90% confidence interval is (114.4, 115.6) and the 99% confidence interval is (114.1, 115.9). The confidence interval is larger for the 99% confidence level than for the 90% confidence level.

Five hundred randomly selected working adults living in Calgary, Canada were asked how long, in minutes, their typical daily commute was. The resulting sample mean commute time was 28.5 minutes. Construct and interpret a 90% confidence interval for the mean commute time of working adult Calgary residents. (Assume the population standard deviation is 24.2 minutes).

(26.72, 30.28)

We are 90% confident that the average daily commute time for working adults living in Calgary, Canada, will be between 26.72 minutes and 30.28 minutes.

We are 90% confident that the average daily commute time for working adults living in Calgary, Canada, will be between 26.72 minutes and 30.28 minutes.

According to Bride's Magazine, getting married these days can be expensive when all costs are included. A simple random sample of 20 recent U.S. weddings yielded data on wedding costs in dollars (sum of data is $526,538). (1) use the data to obtain a point estimate for the population mean wedding cost, µ, of all recent U.S. weddings. (2) Is your point estimate in part (1) likely to equal µ exactly? Explain your answer.

(1) $26,326.9

(2) No. It is unlikely that a sample mean (x-bar) will exactly equal the population mean, µ. Some sampling error is to be anticipated.

(2) No. It is unlikely that a sample mean (x-bar) will exactly equal the population mean, µ. Some sampling error is to be anticipated.

Consumer Reports provides information on new automobile models - including price, mileage ratings, engine size, body size, and indicators of features. A simple random sample of 35 new models yields data on fuel tank capacity, in gallons (sum of data is 664.9 gallons). (1) Find a point estimate for the mean fuel tank capacity of all new automobile models. (2) Determine a 95.44% confidence interval for the mean fuel tank capacity of all new automobile models. Assume ϑ=3.50 gallons. (3) How would you decide whether fuel tank capacities for new automobile models are approximately normally distributed? (4) Must fuel tank capacities for new automobile models be exactly normally distributed for the confidence interval that you obtained in part (2) to be approximately correct? Explain your answer.

(1) x-bar = 18.997

(2) (17.81, 20.18) We can be 95.44% confident that the mean fuel tank capacity of all 2003 automobile models is somewhere between 17.81 and 20.18 gallons.

(3) obtain a normal probability plot of the data.

(4) No, because the sample size is large.

(2) (17.81, 20.18) We can be 95.44% confident that the mean fuel tank capacity of all 2003 automobile models is somewhere between 17.81 and 20.18 gallons.

(3) obtain a normal probability plot of the data.

(4) No, because the sample size is large.

Find the confidence level and ∝ for a 90% confidence interval.

Confidence level = .90

∝ = .10

∝ = .10

Find the confidence level and ∝ for a 99% confidence interval.

Confidence level = .99

∝ = .01

∝ = .01

What are the assumptions required for using the z-interval procedure?

(1) simple random sample

(2) normal population or large sample (≥30)

(3) sigma (ϑ) known

(2) normal population or large sample (≥30)

(3) sigma (ϑ) known

How important is the normality assumption for the z-interval procedure?

The z-interval procedure works well when the variable is normally distributed and reasonably well if the variable is not normally distributed and the sample size is small or moderate, provided the variable is not too far from being normally distributed.

What is meant by saying that a statistical procedure is "robust"?

A statistical procedure that works reasonably well even when one of its assumptions is violated (or moderately violated) is called a robust procedure relative to that assumption.

Assume that the population standard deviation is known. Is it reasonable to use the z-interval procedure to obtain a confidence interval for the population mean under each of the following circumstances: (1) the sample data contains no outliers, the variable under consideration is roughly normally distributed, and the sample size is 20, (2) the distribution of the variable under consideration is highly skewed and the sample size is 20, (3) the sample data contains no outliers, the sample size is 250, and the variable under consideration is far from being normally distributed.

(1) Reasonable, because of the roughly normal distribution, sample size need not be greater than 30 and outliers do not exist that might call into question the normality assumption.

(2) Not reasonable, because sample size is too small.

(3) Reasonable because of the large sample size.

(2) Not reasonable, because sample size is too small.

(3) Reasonable because of the large sample size.

Of 95% and 99% confidence levels, which will result in the confidence interval's giving a more precise estimate of µ?

The 95% confidence level because decreasing the confidence level improves the precision.

Use the one-mean z-interval procedure to find a confidence interval for the mean of the population from which the sample was drawn:

x-bar = 20

n = 36

ϑ = 3

confidence level = 95%

x-bar = 20

n = 36

ϑ = 3

confidence level = 95%

(19.0, 21.0)

Use the one-mean z-interval procedure to find a confidence interval for the mean of the population from which the sample was drawn:

x-bar = 30

n = 25

ϑ = 4

confidence level = 90%

x-bar = 30

n = 25

ϑ = 4

confidence level = 90%

(28.7, 31.3)

Use the one-mean z-interval procedure to find a confidence interval for the mean of the population from which the sample was drawn:

x-bar = 50

n = 16

ϑ = 5

confidence level = 99%

x-bar = 50

n = 16

ϑ = 5

confidence level = 99%

(46.8, 53.2)

A random sample of 18 venture-capital investments in the fiber optics business sector yielded the following data, in millions of dollars (sum of the data is $113.97 million). (1) determine a 95% confidence interval for the mean amount, µ, of all venture-capital investments in the fiber optics business sector. Assume that the population standard deviation is $2.04 million. (2) Interpret your answer from part (1).

(1) $5.389 million to $7.274 million

(2) We can be 95% confident that the mean amount of all venture-capital investments in the fiber optics business sector is somewhere between $5.389 million and $7.274 million.

(2) We can be 95% confident that the mean amount of all venture-capital investments in the fiber optics business sector is somewhere between $5.389 million and $7.274 million.

Given: safety limit set for cadmium in dry vegetables at 0.5 ppm. Random sample of 12 Bp mushrooms, data obtained sums to 6.31 ppm.

Find and interpret a 99% confidence interval for the mean cadmium level of Bp mushrooms. Assume a population standard deviation of cadmium levels in Bp mushrooms of 0.37 ppm.

Find and interpret a 99% confidence interval for the mean cadmium level of Bp mushrooms. Assume a population standard deviation of cadmium levels in Bp mushrooms of 0.37 ppm.

0.251 ppm to 0.801 ppm

We can be 99% confident that the mean cadmium level of all Bp mushrooms is somewhere between 0.251 ppm and 0.801 ppm.

We can be 99% confident that the mean cadmium level of all Bp mushrooms is somewhere between 0.251 ppm and 0.801 ppm.

According to an article, the mean duration of imprisonment for 32 patients with chronic PTSD was 33.4 months. Assuming that ϑ = 42 months, determine a 95% confidence interval for the mean duration of imprisonment, µ, of all East German political prisoners with chronic PTSD. Interpret your answer in words.

18.8 to 48.0 months

We can be 95% confident that the mean duration of imprisonment, µ, of all East German political prisoners with chronic PTSD is somewhere between 18.8 and 48.0 months.

We can be 95% confident that the mean duration of imprisonment, µ, of all East German political prisoners with chronic PTSD is somewhere between 18.8 and 48.0 months.

For the same value of ∝, are t-values greater than or less than z-values?

For the same value of ∝, t-values will be greater than z-values.

Find the area to the right of t=1.771 with df=13.

.05

Find the area to the right of t=2 with df = 30.

Between 0.05 and 0.025

Compare t-curves to the standard normal curve as the number of degrees of freedom becomes larger.

As the number of degrees of freedom becomes larger, t-curves look increasingly like the standard normal curve.

Formula for confidence interval for one population mean when ϑ is unknown.

Standardized version of x-bar

...

Studentized version of x-bar

...

How does the distributions of the standard and studentized versions of x-bar differ?

The studentized version has more spread (wider).

Assumptions for the One-Mean t-Interval Procedure

(1) simple random sample

(2) normal population or large sample

(3) ϑ unknown

(2) normal population or large sample

(3) ϑ unknown

Formula for the confidence interval for µ (t-interval procedure)

...

When is a confidence interval exact? When is a confidence interval approximately correct?

The confidence interval is exact for normal populations and is approximately correct for large samples from non-normal populations.

The publication Amusement Business provides figures on the cost for a family of four to spend the day at one of America's Amusement parks. A random sample of 25 families of four that attended amusement parks yielded the following costs, rounded to the nearest dollar (see 8.94 p 391). Obtain and interpret a 95% confidence interval for the mean cost for a family of four to spend the day at an American amusement park. (Note: x-bar=$193.32, s=$26.73).

(182.29, 204.35)

We are 95% confident that the average cost for a family of four to spend the day at an American amusement park will be between $182.29 and $204.35.

We are 95% confident that the average cost for a family of four to spend the day at an American amusement park will be between $182.29 and $204.35.

null hypothesis

The hypothesis to be tested

alternative hypothesis

The alternative to the null hypothesis

hypothesis test

The problem in a hypothesis test is to decide whether the null hypothesis should be rejected in favor of the alternative hypothesis.

two-tailed test

If the primary concern is deciding whether a population mean, µ, is different from a specified value µ₀, we express the alternative hypothesis as:

H₁: µ ≠ µ₀

H₁: µ ≠ µ₀

left-tailed test

If the primary concern is deciding whether a population mean, µ, is less than a specified value µ₀, we express the alternative hypothesis as:

H₁: µ < µ₀

H₁: µ < µ₀

right-tailed test

If the primary concern is deciding whether a population mean, µ, is greater than a specified value µ₀, we express the alternative hypothesis as:

H₁: µ > µ₀

H₁: µ > µ₀

one-tailed test

A hypothesis test that is either left-tailed or right-tailed.

A snack food company produces a 454 g bag of pretzels and insists that the mean net weight of the bags is 454 g. As part of its program, the quality assurance department periodically performs a hypothesis test to decide whether the packaging machine is working properly, that is, to decide whether the mean net weight of all bags packaged is 454 g. (1) Determine the null hypothesis for the hypothesis test. (2) Determine the alternative hypothesis for the hypothesis test. (3) Classify the hypothesis test as two tailed, left tailed, or right tailed.

(1) H₀: µ = 454 g

(2) H₁: µ ≠ 454 g

(3) two tailed

(2) H₁: µ ≠ 454 g

(3) two tailed

What mathematical signs are allowed in the null hypothesis?

=, ≤, ≥

What mathematical signs are allowed in the alternative hypothesis?

≠, <, >

The mean charitable contribution per household in the U.S. in 2000 is $1623. A researcher claims that the level of giving has changed since then. State the null and the alternative hypotheses.

H₀: µ = $1623

H₁: µ ≠ $1623

H₁: µ ≠ $1623

Federal law requires that a jar of peanut butter labeled 32 oz. must contain at least 32 oz. A consumer advocate feels that a certain manufacturer is shorting customers by underfilling jars so that the mean content is less than 32 oz. State the null and alternative hypotheses.

H₀: µ ≥ 32

H₁: µ < 32

left tailed test

H₁: µ < 32

left tailed test

A confidence interval for a population mean has a margin of error of 3.4 Determine the length of the confidence interval.

6.8

A confidence interval for a population mean has a margin of error of 3.4. If the sample mean is 52.8, obtain the confidence interval.

(49.4, 56.2)

True or False. The length of a confidence interval can be determined if you know only the margin of error.

True

True or False. The margin of error can be determined if you know only the length of the confidence interval.

True

True or False: The confidence interval can be obtained if you know only the margin of error.

False

True or False: The confidence interval can be obtained if you know only the margin of error and the sample mean.

True

The method for computing the sample size required to obtain a confidence interval with a specified confidence level and margin of error - the number resulting from the formula should be rounded up to the nearest whole number. (1) Why do you want a whole number? (2) Why do you round up instead of down?

(1) The sample size cannot be a fraction.

(2) The result (n) is the smallest value that will provide the required margin of error. If the number were rounded down, the sample size would not be sufficient to ensure the required margin of error.

(2) The result (n) is the smallest value that will provide the required margin of error. If the number were rounded down, the sample size would not be sufficient to ensure the required margin of error.

Infants treated for pulmonary hypertension, called the PH group, were compared with those not so treated, called the control group. One of the characteristics measured was head circumference. The mean head circumference of the 10 infants in the PH group was 34.2 cm. (1) Assuming that head circumferences for infants treated for PH are normally distributed with standard deviation 2.1 cm, determine a 90% confidence interval for the mean head circumference of all such infants. (2) Obtain the margin of error, E, for the confidence interval you found in part (1). (3) Explain the meaning of E in this context in terms of the accuracy of the estimate. (4) Determine the sample size required to have a margin of error of 0.5 cm with a 95% confidence level.

(1) (33.108, 35.292)

(2) E = 1.1 cm

(3) We can be 90% confident that the error made in estimating µ by x-bar is at most 1.1 cm.

(4) 68

(2) E = 1.1 cm

(3) We can be 90% confident that the error made in estimating µ by x-bar is at most 1.1 cm.

(4) 68

Explain the difference in the formulas for the standardized and the studentized version of x-bar.

The denominator of the standardized version of x-bar (z-score) uses the population standard deviation, ϑ, whereas the denominator of the studentized version of x-bar (t-score) uses the sample standard deviation, s.

A variable has a mean of 100 and a standard deviation of 16. 4 observations of this variable have a mean of 108 and a sample standard deviation of 12. Determine the observed value of the:

(1) standardized version of x-bar

(2) studentized value of x-bar

(1) standardized version of x-bar

(2) studentized value of x-bar

(1) 1

(2) 1.33

(2) 1.33

Two t-curves have degrees of freedom, 12 and 20, respectively. Which one more closely resembles the standard normal curve? Explain your answer.

The df=20 because as the number of degrees of freedom increases, t-curves look increasingly like the standard normal curve.

According to Scarborough Research, more than 85% of working adults commute by car. Of all U.S. cities, Washington, D.C. and New York City have the longest commute times. A sample of 30 commuters in the Washington, D.C. area yielded the following commute times, in minutes (data set, x-bar=27.97 minutes, s=10.04 minutes). (1) Find a 90% confidence interval for the mean commute time of all commuters in Washington, D.C. (2) Interpret your answer from part (1).

(1) (24.855, 31.085)

(2) We are 90% confident that the average commute time of all commuters in Washington, D.C. is between 24.855 minutes and 31.085 minutes.

(2) We are 90% confident that the average commute time of all commuters in Washington, D.C. is between 24.855 minutes and 31.085 minutes.

A data set gives the additional sleep in hours obtained by a sample of 10 patients using laevohysocyamine hydrobromide (with xbar=2.33 hr, s=2.002 hr). (1) Obtain and interpret a 95% confidence interval for the additional sleep that would be obtained on average for all people using laevohysocyamine hydrobromide. (2) Was the drug effective in increasing sleep? Explain your answer.

(1) 0.90 hr to 3.76 hr

We can be 95% confident that the additional sleep that would be obtained on average for all people using the drug is somewhere between 0.90 hr and 3.76 hr.

(2) It appears so, because, based on the confidence interval, we can be 95% confident that the mean additional sleep is somewhere between 0.90 hr and 3.76 hr and that, in particular, the mean is positive.

We can be 95% confident that the additional sleep that would be obtained on average for all people using the drug is somewhere between 0.90 hr and 3.76 hr.

(2) It appears so, because, based on the confidence interval, we can be 95% confident that the mean additional sleep is somewhere between 0.90 hr and 3.76 hr and that, in particular, the mean is positive.

Explain the meaning of the term hypothesis as used in inferential statistics.

A hypothesis is a statement that something is true.

Given: safety limit set for cadmium in dry vegetables at 0.5 ppm. A hypothesis test is to be performed to decide whether the mean cadmium level in Bp mushrooms is greater than the government's recommended limit. (1) Determine the null hypotheses, (2) Determine the alternative hypothesis, (3) classify the hypothesis test as two tailed, left tailed, or right tailed.

(1) null hypothesis H₀: µ = .5 ppm

(2) alternative hypothesis H₁: µ > .5 ppm

(3) right tailed test

(2) alternative hypothesis H₁: µ > .5 ppm

(3) right tailed test

The recommended dietary allowance (RDA) of iron for adult females under the age of 51 is 18 milligrams (mg) per day. A hypothesis test is to be performed to decide whether adult females under the age of 51 are, on average, getting less than the RDA of 18 mg of iron. (1) Determine the null hypothesis, (2) Determine the alternative hypothesis, (3) classify the hypothesis test as two tailed, left tailed, or right tailed.

(1) null hypothesis H₀: µ = 18 mg

(2) alternative hypothesis H₁: µ < 18 mg

(3) left tailed test

(2) alternative hypothesis H₁: µ < 18 mg

(3) left tailed test

According to the Bureau of Crime Statistics and Research of Australia, the mean length of imprisonment for motor-vehicle theft offenders in Australia is 16.7 months. You want to perform a hypothesis test to decide whether the mean length of imprisonment for motor-vehicle theft offenders in Sydney differs from the national mean in Australia. (1) Determine the null hypothesis, (2) Determine the alternative hypothesis, (3) classify the hypothesis test as two tailed, left tailed, or right tailed.

(1) H₀: µ = 16.7 months

(2) H₁: µ ≠ 16.7 months

(3) two tailed test

(2) H₁: µ ≠ 16.7 months

(3) two tailed test

test statistic

the z-score (or t-score) that determines if an average is unusual or not

Calculation formula for the test statistic (z-score or t-score)

...

What does the z or t test statistic tell us?

The z or t test statistic tells us how far x-bar is from µ in standard deviations (i.e. the number of standard deviations from the mean). It is the statistic used as a basis for deciding whether the null hypothesis should be rejected.

rejection region

The set of values for the test statistic that lead us to reject H₀ (tail or tails of the distribution)

non-rejection region

The set of values for the test statistic that lead us not to reject H₀

critical values

the boundaries for the rejection/non-rejection regions

Type I Error

Rejecting the null hypothesis when it is in fact true

Type II Error

Not rejecting the null hypothesis when it is in fact false.

Type I Error probability

the probability of a Type I error, denoted ∝, also called the significance level of the hypothesis test

significance level

the probability of making a Type I error, that is, of rejecting a true null hypothesis (denoted ∝)

Type II Error probability

the probability of a Type II error, denoted β - a Type II error occurs if the test statistic falls in the non-rejection region when in fact the null hypothesis is false.

What is the relationship between Type I and Type II Error probabilities?

For a fixed sample size, the smaller we specify the significance level, ∝, the larger will be the probability, β, of not rejecting a false null hypothesis.

Possible conclusions for a hypothesis test

If the null hypotheses is rejected, we conclude that the alternative hypothesis is true. If the null hypothesis is not rejected, we conclude that the data do not provide sufficient evidence to support the alternative hypothesis.

Steps for Hypothesis Tests for One Population Mean when ϑ is known (6)

(1) State the null and alternative hypotheses

(2) Decide on a value for ∝ (significance level)

(3) Compute the test statistic Z

(4) Find the critical values

(5) Conclusion

(6) Interpretation

(2) Decide on a value for ∝ (significance level)

(3) Compute the test statistic Z

(4) Find the critical values

(5) Conclusion

(6) Interpretation

Calculation of critical values (for hypothesis test for one population when ϑ is known)

+/- Z (∝/2) - two tailed test

- Z ∝ - left tailed test

Z∝ - right tailed test

- Z ∝ - left tailed test

Z∝ - right tailed test

Suppose a CEO of a company wants to determine whether the average amount of wasted time during an 8-hour day for employees at the company is less than 120 minutes. A random sample of 10 employees gave these results:

108, 131, 112, 113, 117, 113, 130, 105, 111, 128

Assume ϑ = 9.

Do these data provide evidence that the mean wasted time for this company is less than 120 minutes?

108, 131, 112, 113, 117, 113, 130, 105, 111, 128

Assume ϑ = 9.

Do these data provide evidence that the mean wasted time for this company is less than 120 minutes?

Conclusion: Do not reject H₀

Interpretation: There is not enough evidence to conclude that the average amount of wasted time at the company is less than 120 minutes.

Interpretation: There is not enough evidence to conclude that the average amount of wasted time at the company is less than 120 minutes.

True or False: If it is important not to reject a true null hypothesis, the hypothesis test should be performed at a small significance level.

True. The significance level ∝ of a hypothesis test is the probability of making a Type I error (rejecting a true null hypothesis). If this is important, the lower the probability of making such an error the better; thus you should use a small significance level.

True or False: For a fixed sample size, decreasing the significance level of a hypothesis test results in an increase in the probability of making a Type II error.

True. For a fixed sample size, the smaller you specify the significance level ∝, the larger will be the probability β, of not rejecting a false null hypothesis.

Define test statistic

The statistic used as a basis for deciding whether the null hypothesis should be rejected

Define rejection region

The set of values for the test statistic that leads to rejection of the null hypothesis.

Define non-rejection region

The set of values for the test statistic that leads to non-rejection of the null hypothesis.

Define critical values

The values of the test statistic that separate the rejection and non-rejection regions. A critical value is considered part of the rejection region.

Define significance level

The probability of making a Type I error, that is, of rejecting a true null hypothesis.

Identify the two types of incorrect decisions in a hypothesis test. For each incorrect decision, what symbol is used to represent the probability of making that type of error?

(1) Type I - rejecting a true null hypothesis (symbol ∝)

(2) Type II - not rejecting a false null hypothesis (symbol β)

(2) Type II - not rejecting a false null hypothesis (symbol β)

Would it be appropriate to use a t-interval for a sample size of 15? Explain.

It would not be appropriate because the assumption of a normal population or a large sample is not met. We know nothing of the population and the sample is small.

What is the appropriate t-value for the following confidence level and sample size: confidence level 95%, n=17

2.120

What is the appropriate t-value for the following confidence level and sample size: confidence level 90%, n=12

1.796

What is the appropriate t-value for the following confidence level and sample size: confidence level 99%, n=24

2.807

The following data are airborne times for United Airlines flight 448 from Albuquerque to Denver on 10 randomly selected days: 57, 54, 55, 51, 56, 48, 52, 51, 59, 59 . (1) Compute and interpret a 90% confidence interval for the mean airborne time for flight 448. (2) Based on your interval in part (1), if flight 448 is scheduled to depart at 10 a.m., what would you recommend for the published arrival time? Explain.

(1) (52.07, 56.33)

(2) Recommend an arrival time of 10:57 a.m., so that 0% of the flights would be late.

(2) Recommend an arrival time of 10:57 a.m., so that 0% of the flights would be late.

A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding machine. Strength can be measured by recording the force required to pull the pages from the binding. If this force is measured in pounds, how many books should be tested to estimate with 95% confidence to within 0.1 lb, the average force required to break the binding? Assume ϑ is known to be 0.8 lb.

246

To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose that the specifications state that the mean strength of welds should exceed 100 lb/ in². The inspection team decides to test H₀: µ = 100 versus H₁: µ > 100. Explain why this alternative hypothesis was chosen rather than µ < 100.

The alternative hypothesis was chosen because the mean strength of welds should be greater than 100, thus the alternative hypothesis H₁: µ > 100. The primary concern of the research is to decide whether the population mean is greater than the specified value (and meets specifications).

Does this pair comply with the rules for setting up hypotheses? If not, explain why.

H₀: µ = 15; H₁: µ = 15

H₀: µ = 15; H₁: µ = 15

Does not comply. H₁ must be stated as ≠ 15, <15, or > 15.

Does this pair comply with the rules for setting up hypotheses? If not explain why.

H₀: µ = 10; H₁: µ > 12

H₀: µ = 10; H₁: µ > 12

Does not comply. H₁ must use the same number as H₀, the null hypothesis.

Does this pair comply with the rules for setting up hypotheses? If not explain why.

H₀: µ = 123; H₁: µ < 123

H₀: µ = 123; H₁: µ < 123

Complies

Does this pair comply with the rules for setting up hypotheses? If not explain why.

H₀: µ = 123; H₁: µ = 125

H₀: µ = 123; H₁: µ = 125

Does not comply. H₁ must use the same number as H₀ and cannot contain the equal sign.

Does this pair comply with the rules for setting up hypotheses? If not explain why.

H₀: µ = 50; H₁: µ ≠ 50

H₀: µ = 50; H₁: µ ≠ 50

Complies

Researchers have postulated that because of differences in diet, Japanese children have a lower mean blood cholesterol level than U.S. children do. Suppose that the mean level for U.S. children is known to be 170. Let µ represent the true mean blood cholesterol level for Japanese children. What hypothesis should the researchers test? Give the null and alternative hypotheses.

H₀: µ = 170 (the null hypothesis is the hypothesis to be tested)

H₁: µ < 170

H₁: µ < 170

Using the test statistic formula for the z-score, determine the required critical value(s) for a right-tailed test with ∝ = 0.05.

1.645

Using the test statistic formula for the z-score, determine the required critical value(s) for a left-tailed test with ∝ = 0.05.

-1.645

Using the test statistic formula for the z-score, determine the required critical value(s) for a two-tailed test with ∝ = 0.05.

-1.96 and +1.96

Decide in the following situations whether the z-test is an appropriate method for conducting

the hypothesis test for a population mean: (a) no outliers, distribution highly skewed, sample size 24, (b) no outliers, mildly skewed, sample size 70

the hypothesis test for a population mean: (a) no outliers, distribution highly skewed, sample size 24, (b) no outliers, mildly skewed, sample size 70

(a) not appropriate

(b) appropriate

(b) appropriate

Cadmium, a heavy metal, is toxic to animals. Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. The Czech and Slovak governments have set a safety limit for cadmium in dry vegetables at 0.5 parts per million (ppm). A random sample of the edible mushroom Boletus pinicola with the resulting data: 0.24, 0.59, 0.62, 0.16, 0.77, 1.33, 0.92, 0.19, 0.33, 0.25, 0.59, 0.32 . At the 5% significance level, do the data provide sufficient evidence to conclude that the mean cadmium level in Boletus pinicola mushrooms is greater than the government's recommended limit of 0.5 ppm? Assume that the population standard deviation of cadmium levels in Boletus pinicola mushrooms is 0.37 ppm. (Note: The sum of the data is 6.31 ppm).

Given: significance level 0.05

ϑ = 0.37

H₀: µ = 0.5 ppm

H₁: µ > 0.5 ppm

Test statistic: z=0.24

Critical value: 1.645

Conclusion: 0.24 is in the non-rejection region

Interpretation: There is not enough evidence to conclude that the mean level of cadmium in Boletus pinicola mushrooms is greater than 0.5 ppm.

ϑ = 0.37

H₀: µ = 0.5 ppm

H₁: µ > 0.5 ppm

Test statistic: z=0.24

Critical value: 1.645

Conclusion: 0.24 is in the non-rejection region

Interpretation: There is not enough evidence to conclude that the mean level of cadmium in Boletus pinicola mushrooms is greater than 0.5 ppm.

Given:

n=45

x-bar = 14.68

ϑ = 4.2

∝ = .01

H₀: µ = 18

H₁: µ < 18

Reject or do not reject null hypothesis?

n=45

x-bar = 14.68

ϑ = 4.2

∝ = .01

H₀: µ = 18

H₁: µ < 18

Reject or do not reject null hypothesis?

Test statistic: z = -5.3

Critical value: -2.326

Conclusion: -5.3 is in the rejection region

Interpretation: There is sufficient evidence to conclude that ... the alternative hypothesis is true.

Critical value: -2.326

Conclusion: -5.3 is in the rejection region

Interpretation: There is sufficient evidence to conclude that ... the alternative hypothesis is true.

Describe the meaning of P-value of a hypothesis test

To obtain the P-value of a hypothesis test, we assume that the null hypothesis is true and compute the probability of observing a value of the test statistic as extreme as or more extreme than that observed. By extreme we mean "far from what we would expect to observe if the null hypothesis is true." We use the letter P to denote the P-value. The p-value is the area beyond the test statistic in either direction.

If you have a p-value of 0.0168 and a z-score of +/- 2.39, interpret the meaning of these values in context.

The probability of getting a z-score more extreme than +/- 2.39 is 0.0168.

On the calculator (TI-84), how do you find the area to the left of a particular z-score?

NORMALCDF (-1000, Z-score, 0, 1)

On the calculator (TI-84), how do you find the area to the right of a particular z-score?

NORMALCDF (Z-score, 1000, 0, 1)

What p-value(s) describe weak or no evidence against the null hypothesis?

P-value > .10

What p-value(s) describe moderate evidence against the null hypothesis?

.05 < P ≤ .10

What p-value(s) describe strong evidence against the null hypothesis?

.01 < P ≤ .05

What p-value(s) describe very strong evidence against the null hypothesis?

P ≤ .01

How does a large test statistic relate to the area in the tail?

A large test statistic means that there is a smaller area in the tail.

How does a small test statistic relate to the area in the tail?

A small test statistic means that there is a larger area in the tail.

2 methods for determining whether to reject or not reject the null hypothesis

(1) compare the test statistic to the critical values; where the test statistic falls (rejection region or non-rejection region)

(2) compare the p-value to ∝

If the p-value is low, H₀ must go!

If the p-value ≤ ∝, reject H₀

If the p-value > ∝, do not reject H₀

(2) compare the p-value to ∝

If the p-value is low, H₀ must go!

If the p-value ≤ ∝, reject H₀

If the p-value > ∝, do not reject H₀

A hot tub manufacturer advertises that with its heating equipment a temperature of 100 degrees F can be achieved in at most 15 minutes. A random sample of 20 tubs is selected and the time needed to reach 100 degrees is determined for each tub. The sample mean is 16 minutes with a standard deviation of 1 minute. Does this information cast doubt on the company's claim? Assume ∝ = 0.01.

(1) H₀: µ ≤ 15

H₁: µ > 15

(2) ∝ = 0.01

(3) Test statistic: t = 4.47

(4) P-value = .000013 (using t-test on calculator)

(5) Compare p-value to ∝

.000013 ≤ 0.01

Reject H₀

(6) There is enough evidence to suggest the average time for a hot tub to reach 100 degrees F is more than 15 minutes (p=.000013, ∝ = 0.01).

H₁: µ > 15

(2) ∝ = 0.01

(3) Test statistic: t = 4.47

(4) P-value = .000013 (using t-test on calculator)

(5) Compare p-value to ∝

.000013 ≤ 0.01

Reject H₀

(6) There is enough evidence to suggest the average time for a hot tub to reach 100 degrees F is more than 15 minutes (p=.000013, ∝ = 0.01).

An automobile manufacturer who wishes to advertise that one of its models achieves 30 mpg decides to carry out a fuel efficiency test. Six non-professional drivers are selected and each one drives a car from Phoenix to Los Angeles. The resulting fuel efficiencies in mpg are: 27.2, 29.3, 31.2, 28.4, 30.3, 29.6. Assuming that the fuel efficiency is normally distributed, do the data contradict the claim that the true average fuel efficiency is at least 30 mpg? Assume ∝ = 0.05.

(1) H₀: µ ≥ 30 mpg

H₁: µ < 30 mpg

(2) ∝ = 0.05

(3) test statistic t = -1.16

(4) p-value = .1493 (t-test on calculator)

(5) compare p-value to ∝

.1493 ≤ 0.05 ? NO

Do not reject H₀

(6) There is not enough evidence to conclude that the average fuel efficiency in mpg is less than 30 mpg.

H₁: µ < 30 mpg

(2) ∝ = 0.05

(3) test statistic t = -1.16

(4) p-value = .1493 (t-test on calculator)

(5) compare p-value to ∝

.1493 ≤ 0.05 ? NO

Do not reject H₀

(6) There is not enough evidence to conclude that the average fuel efficiency in mpg is less than 30 mpg.

For which of the following p-values would the null hypothesis be rejected at a level of ∝ = 0.05:

(a) .001

(b) .021

(c) .078

(d) .047

(e) .148

(a) .001

(b) .021

(c) .078

(d) .047

(e) .148

(a) reject

(b) reject

(c) do not reject

(d) reject

(e) do not reject

(b) reject

(c) do not reject

(d) reject

(e) do not reject

State two reasons why including the p-value is prudent when you are reporting the results of a hypothesis test.

(1) it allows you to assess significance at any desired level

(2) it permits you to evaluate the strength of the evidence against the null hypothesis

(2) it permits you to evaluate the strength of the evidence against the null hypothesis

What is the p-value of a hypothesis test?

The probability of observing a value of the test statistic as extreme or more extreme than that observed. By extreme we mean "far from what we would expect to observe if the null hypothesis is true."

When does the p-value provide evidence against the null hypothesis?

When the p-value is less than or equal to the significance level, ∝

True or False: The p-value is the smallest significance level for which the observed sample data result in rejection of the null hypothesis.

True

The p-value for a hypothesis test is 0.083. For each of the following significance levels, decide whether the null hypothesis should be rejected:

(1) ∝ = 0.05

(2) ∝ = 0.10

(3) ∝ = 0.06

(1) ∝ = 0.05

(2) ∝ = 0.10

(3) ∝ = 0.06

(1) Do not reject (0.083 > 0.05)

(2) Reject (0.083 ≤ 0.10)

(3) Do not reject (0.083 > 0.06)

(2) Reject (0.083 ≤ 0.10)

(3) Do not reject (0.083 > 0.06)

Determine the strength of the evidence against the null hypothesis:

(a) p = 0.06

(b) p = 0.35

(c) p = 0.027

(d) p = 0.004

(a) p = 0.06

(b) p = 0.35

(c) p = 0.027

(d) p = 0.004

(1) moderate

(2) weak or none

(3) strong

(4) very strong

(2) weak or none

(3) strong

(4) very strong

Determine the p-value.

Z = 2.03 right-tailed test

Z = 2.03 right-tailed test

0.0212

Determine the p-value.

Z = -0.31 right-tailed test

Z = -0.31 right-tailed test

0.6217

Determine the p-value.

Z = -0.74 left-tailed test

Z = -0.74 left-tailed test

0.2296

Determine the p-value.

Z = 1.16 left-tailed test

Z = 1.16 left-tailed test

0.8770

Determine the p-value.

Z = -1.66 two-tailed test

Z = -1.66 two-tailed test

0.0970

Determine the p-value.

Z = 0.52 two-tailed test

Z = 0.52 two-tailed test

0.6030

Cadmium, a heavy metal, is toxic to animals. Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. The Czech and Slovak governments have set a safety limit for cadmium in dry vegetables at 0.5 parts per million (ppm). A random sample of the edible mushroom Boletus pinicola with the resulting data: 0.24, 0.59, 0.62, 0.16, 0.77, 1.33, 0.92, 0.19, 0.33, 0.25, 0.59, 0.32 . At the 5% significance level, do the data provide sufficient evidence to conclude that the mean cadmium level in Boletus pinicola mushrooms is greater than the government's recommended limit of 0.5 ppm? Assume that the population standard deviation of cadmium levels in Boletus pinicola mushrooms is 0.37 ppm. (Note: The sum of the data is 6.31 ppm). USE THE P-VALUE APPROACH.

(1) H₀: µ = 0.5 ppm

H₁: µ > 0.5 ppm

(2) ∝ = 0.05

(3) test statistic: Z = 0.24

(4) p-value =0.4044 (calculator t-test)

(5) conclusion: compare p-value to ∝:

Is 0.4044 ≤ 0.05? No. Do not reject.

(6) There is not sufficient evidence to conclude that the average cadmium level in the mushrooms is greater than .5 ppm.

H₁: µ > 0.5 ppm

(2) ∝ = 0.05

(3) test statistic: Z = 0.24

(4) p-value =0.4044 (calculator t-test)

(5) conclusion: compare p-value to ∝:

Is 0.4044 ≤ 0.05? No. Do not reject.

(6) There is not sufficient evidence to conclude that the average cadmium level in the mushrooms is greater than .5 ppm.

According to Communications Industry Forecast & Report, the average person watched 4.66 hours of television per day in 2002. A random sample of 20 people gave the number of hours of television watched per day for last year (x-bar = 4.835 hours, s = 2.291 hours). At the 10% significance level, do the data provide sufficient evidence to conclude that the amount of television watched per day last year by the average person differed from that in 2002?

USE THE P-VALUE APPROACH.

USE THE P-VALUE APPROACH.

(1) H₀: µ = 4.66 hours

H₁: µ ≠ 4.66 hours

(2) ∝ = 0.10

(3) Test statistic t = 3.416

(4) p-value: p=0.7364 (from calculator t-test)

(5) Conclusion: compare p-value to ∝

Is 0.7364 ≤ 0.10? No. Do not reject.

(6) There is not sufficient evidence that the average number of daily t.v. viewing hours has changed from that in 2002 (average of 4.66 hours daily).

H₁: µ ≠ 4.66 hours

(2) ∝ = 0.10

(3) Test statistic t = 3.416

(4) p-value: p=0.7364 (from calculator t-test)

(5) Conclusion: compare p-value to ∝

Is 0.7364 ≤ 0.10? No. Do not reject.

(6) There is not sufficient evidence that the average number of daily t.v. viewing hours has changed from that in 2002 (average of 4.66 hours daily).

In the following problem, decide whether applying the t-test to perform a hypothesis test for the population mean appears reasonable.

The Florida State Center for Health Statistics reported that, for cardiovascular hospitalizations, the mean age of women is 71.9 years. At one hospital, a random sample of 20 of its female cardiovascular patients had the following ages, in years: 75.9, 78,2, 88.2, 58.9, 83.7, 76.1, 78.9, 97.6, 87.3, 52.8, 81.7, 65.8, 74.5, 56.4, 54.4, 86.4, 82.5, 53.8, 52.7, 72.4.

The Florida State Center for Health Statistics reported that, for cardiovascular hospitalizations, the mean age of women is 71.9 years. At one hospital, a random sample of 20 of its female cardiovascular patients had the following ages, in years: 75.9, 78,2, 88.2, 58.9, 83.7, 76.1, 78.9, 97.6, 87.3, 52.8, 81.7, 65.8, 74.5, 56.4, 54.4, 86.4, 82.5, 53.8, 52.7, 72.4.

The t-test is not reasonable because the sample is small (20) and a histogram of the data shows that the data is not normally distributed.

For the following:

Right tailed test, n=20, t=2.235,

(1) Estimate the p-value from the Table

(2) Based on your estimate from part (1), state at which significance levels the null hypothesis can be rejected, at which significance levels it cannot be rejected, and which significance levels it is not possible to decide.

Right tailed test, n=20, t=2.235,

(1) Estimate the p-value from the Table

(2) Based on your estimate from part (1), state at which significance levels the null hypothesis can be rejected, at which significance levels it cannot be rejected, and which significance levels it is not possible to decide.

(1) 0.01 < P < 0.025

(2) We can reject H₀ at any significance level of 0.025 or larger, and we cannot reject H₀ at any significance level of 0.01 or smaller. For significance levels between 0.01 and 0.025, the table is not sufficiently detailed to help us to decide whether to reject H₀.

(2) We can reject H₀ at any significance level of 0.025 or larger, and we cannot reject H₀ at any significance level of 0.01 or smaller. For significance levels between 0.01 and 0.025, the table is not sufficiently detailed to help us to decide whether to reject H₀.

For the following:

Left-tailed test, n=10, t = -3.381,

(1) Estimate the p-value from the Table

(2) Based on your estimate from part (1), state at which significance levels the null hypothesis can be rejected, at which significance levels it cannot be rejected, and which significance levels it is not possible to decide.

Left-tailed test, n=10, t = -3.381,

(1) Estimate the p-value from the Table

(2) Based on your estimate from part (1), state at which significance levels the null hypothesis can be rejected, at which significance levels it cannot be rejected, and which significance levels it is not possible to decide.

(1) P < 0.005

(2) We can reject H₀ at any significance level of 0.005 or larger. For significance levels smaller than 0.005, the table is not sufficiently detailed to help us to decide whether to reject H₀.

(2) We can reject H₀ at any significance level of 0.005 or larger. For significance levels smaller than 0.005, the table is not sufficiently detailed to help us to decide whether to reject H₀.

For the following:

Two-tailed test, n = 17, and t = -2.733,

(1) Estimate the p-value from the Table

(2) Based on your estimate from part (1), state at which significance levels the null hypothesis can be rejected, at which significance levels it cannot be rejected, and which significance levels it is not possible to decide.

Two-tailed test, n = 17, and t = -2.733,

(1) Estimate the p-value from the Table

(2) Based on your estimate from part (1), state at which significance levels the null hypothesis can be rejected, at which significance levels it cannot be rejected, and which significance levels it is not possible to decide.

(1) 0.01 < P < 0.02

(2) We can reject H₀ at any significance level of 0.02 or larger, and we cannot reject H₀ at any significance level of 0.01 or smaller. For significance levels between 0.01 and 0.02, the table is not sufficiently detailed to help us to decide whether to reject H₀.

(2) We can reject H₀ at any significance level of 0.02 or larger, and we cannot reject H₀ at any significance level of 0.01 or smaller. For significance levels between 0.01 and 0.02, the table is not sufficiently detailed to help us to decide whether to reject H₀.

Given:

x-bar = 20

s = 4

n = 32

H₀: µ = 22, H₁: µ < 22

∝ = 0.05

(1) Use the one-mean t-test to perform the required hypothesis test about the mean, µ, of the population from which the sample was drawn.

(2) Find (or estimate) the P-value and determine the strength of the evidence against the null hypothesis.

x-bar = 20

s = 4

n = 32

H₀: µ = 22, H₁: µ < 22

∝ = 0.05

(1) Use the one-mean t-test to perform the required hypothesis test about the mean, µ, of the population from which the sample was drawn.

(2) Find (or estimate) the P-value and determine the strength of the evidence against the null hypothesis.

(1) t = -2.82; critical value = -1.696

P-value = 0.004 (P < 0.05) Reject H₀.

(2) Very strong (P < 0.01)

P-value = 0.004 (P < 0.05) Reject H₀.

(2) Very strong (P < 0.01)

Given:

x-bar = 24

s = 4

n = 15

H₀: µ = 22, H₁: µ > 22

∝ = 0.05

(1) Use the one-mean t-test to perform the required hypothesis test about the mean, µ, of the population from which the sample was drawn.

(2) Find (or estimate) the P-value and determine the strength of the evidence against the null hypothesis.

x-bar = 24

s = 4

n = 15

H₀: µ = 22, H₁: µ > 22

∝ = 0.05

(1) Use the one-mean t-test to perform the required hypothesis test about the mean, µ, of the population from which the sample was drawn.

(2) Find (or estimate) the P-value and determine the strength of the evidence against the null hypothesis.

(1) t = 1.94; critical value = 1.761

p-value = 0.037 (0.025 < P < 0.05) Reject H₀.

(2) Strong (0.01 < P < 0.05)

p-value = 0.037 (0.025 < P < 0.05) Reject H₀.

(2) Strong (0.01 < P < 0.05)

Given:

x-bar = 23

s = 4

n = 24

H₀: µ = 22, H₁: µ ≠ 22

∝ = 0.05

(1) Use the one-mean t-test to perform the required hypothesis test about the mean, µ, of the population from which the sample was drawn.

(2) Find (or estimate) the P-value and determine the strength of the evidence against the null hypothesis.

x-bar = 23

s = 4

n = 24

H₀: µ = 22, H₁: µ ≠ 22

∝ = 0.05

(1) Use the one-mean t-test to perform the required hypothesis test about the mean, µ, of the population from which the sample was drawn.

(2) Find (or estimate) the P-value and determine the strength of the evidence against the null hypothesis.

(1) t = 1.22; critical values = +/- 2.069

p-value = .233 (P > 0.20) Do not reject H₀.

(2) Weak or none (P > 0.10)

p-value = .233 (P > 0.20) Do not reject H₀.

(2) Weak or none (P > 0.10)

What calculator function is sued to find a confidence interval when ϑ is unknown?

STATS - TESTS - T-Interval

What value does 1 - ∝ give you?

The confidence level

For a fixed confidence level, how does increasing the sample size affect precision?

Increasing the sample size improves the precision.

What factor determines the precision with which x-bar estimates µ?

The length of the confidence interval, determined by the margin of error.

point estimate

The value of a statistic used to estimate a parameter

A tarantula has two body parts. The anterior part of the body is covered above by a shell, or carapace. A simple random sample of 15 of adult male Brazilian giant tawny red tarantulas provided the following data on carapace length, in millimeters (mm): 15.7, 19.2, 16.4, 18.3, 19.8, 16.8, 19.7, 18.1, 18.9, 17.6, 18.0, 18.5, 19.0, 20.9, 19.5. Find and interpret a 95.44% confidence interval for the mean carapace length of all adult male Brazilian giant tawny red tarantulas. The population standard deviation is 1.76 mm.

(17.52, 19.34)

We can be 95.44% confident that the average carapace length of all adult male Brazilian giant tawny red tarantulas is between 17.52 mm and 19.34 mm.

We can be 95.44% confident that the average carapace length of all adult male Brazilian giant tawny red tarantulas is between 17.52 mm and 19.34 mm.

Explain why the margin of error determines the precision with which a sample mean estimates a population mean.

The length of a confidence interval, and thus the precision with which x-bar estimates µ, is determined by the margin of error.

A confidence interval for a population mean has length 20. (a) Determine the margin of error. (b) If the sample mean is 60, obtain the confidence interval.

(a) 10

(b) (50, 70)

(b) (50, 70)

For the confidence interval (5.289, 7.274), obtain the margin of error by:

(a) taking half the length of the confidence interval

(b) using the formula for E

(a) taking half the length of the confidence interval

(b) using the formula for E

margin of error = .94

For the confidence interval (18.8, 48): (a) Determine the margin of error E. (b) For a 95% confidence level, explain the meaning of E in this context in terms of the accuracy of the measurement.

(a) margin of error = 14.6

(b) We are 95% confident that the maximum error made in using x-bar to estimate µ is 14.6.

(b) We are 95% confident that the maximum error made in using x-bar to estimate µ is 14.6.

Professor Thomas Stanley of Georgia State University has surveyed millionaires since 1973. Among other information, Professor Stanley obtains estimates for the mean age, µ, of all U.S. millionaires. Suppose that one year's study involved a simple random sample of 36 U.S. millionaires whose mean age was 58.53 years with a sample standard deviation of 13.36 years. (a) If, for next year's study, a confidence interval for µ is to have a margin of error of 2 years and a confidence level of 95%, determine the required sample size. (b) Why did you use the sample standard deviation, s=13.36, in place of ϑ in your solution to part (a)? Why is it permissible to do so?

(a) 172

(b) We used s in place of ϑ because ϑ was unknown. We can do this because the sample of size 36 is large enough to provide an estimate of ϑ and the variation is not likely to change much from one year to the next.

(b) We used s in place of ϑ because ϑ was unknown. We can do this because the sample of size 36 is large enough to provide an estimate of ϑ and the variation is not likely to change much from one year to the next.

Suppose that a simple random sample is taken from a normal population having a standard deviation of 10 for the purpose of obtaining a 95% confidence interval for the mean of the population. (a) If the sample size is 4, obtain the margin of error. (b) Repeat part (a) for a sample size of 16. (c) Can you guess the margin of error for a sample size of 64? Explain your reasoning.

(a) 9.80

(b) 4.90

(c) It appears that quadrupling the sample size will halve the margin of error. Therefore, increasing n from 16 to 64 will decrease the margin of error from 4.90 to 2.45.

(b) 4.90

(c) It appears that quadrupling the sample size will halve the margin of error. Therefore, increasing n from 16 to 64 will decrease the margin of error from 4.90 to 2.45.

Explain why there is more variation in the possible values of the studentized version of x-bar than in the possible values of the standardized version of x-bar.

The variation in the possible values of the standardized version of x-bar is due only to the variation in x-bar, while the variation in the studentized version results not only from the variation in x-bar, but also from the variation in the sample standard deviation.

For a t-curve with df=6, use Table IV to find each t-value: (a) t (sub 0.10), (b) t (sub 0.025), (c) t (sub 0.001)

(a) 1.440

(b) 2.447

(c) 3.143

(b) 2.447

(c) 3.143

For a t-curve with df=21, find each t-value: (a) The t-value having area 0.10 to its right (b) t (sub 0.01), (c) The t-value having area 0.025 to its left (Hint: a t-curve is symmetric about 0), (d) The two t-values that divide the area under the curve into a middle 0.90 area and two outside areas of 0.05.

(a) 1.323

(b) 2.518

(c) -2.080

(d) +/- 1.721

(b) 2.518

(c) -2.080

(d) +/- 1.721

A simple random sample of size 100 is taken from a population with unknown standard deviation. A normal probability plot of the data displays significant curvature but no outliers. Can you reasonably apply the t-interval procedure? Explain your answer.

It is reasonable to use the t-interval procedure since the sample size is large and for large degrees of freedom (99), the t-distribution is very similar to the standard normal distribution. Another way of expressing this is that the sampling distribution of x-bar is approximately normal when n is large, so the standardized and studentized versions of x-bar are essentially the same.

With the following information, use the one-mean t-interval procedure to find a confidence interval for the mean of the population from with the sample was drawn:

x-bar = 20

n = 36

s = 3

confidence level = 95%

x-bar = 20

n = 36

s = 3

confidence level = 95%

(18.98, 21.02)

With the following information, use the one-mean t-interval procedure to find a confidence interval for the mean of the population from with the sample was drawn:

x-bar = 30

n = 25

s = 4

confidence level = 90%

x-bar = 30

n = 25

s = 4

confidence level = 90%

(28.63, 31.37)

With the following information, use the one-mean t-interval procedure to find a confidence interval for the mean of the population from with the sample was drawn:

x-bar =50

n = 16

s = 5

confidence level = 99%

x-bar =50

n = 16

s = 5

confidence level = 99%

(46.32, 53.68)

A variable of a population has a mean of 266 and a standard deviation of 16. Ten observations of this variable have a mean of 262.1 and a sample standard deviation of 20.4. Obtain the observed value of the (a) standardized version of x-bar, (b) studentized version of x-bar.

(a) -0.77

(b) -0.605

(b) -0.605

Explain the difference between a point estimate of a parameter and a confidence-interval estimate of a parameter.

A point estimate of a parameter consists of a single value with no indication of the accuracy of the estimate. A confidence interval consists of an interval of numbers obtained from a point estimate of the parameter together with a percentage that specifies how confident we are that the parameter lies in the interval.

Must the variable under consideration be normally distributed for you to use the z-interval procedure or t-interval procedure? Explain your answer.

No. The z-interval procedure can be used almost anytime with large samples because the sampling distribution of x-bar is approximately normal for large n. The same is true for the t-interval procedure because when n is large, the t-distribution is very similar to the normal distribution. However, when n is small, especially when n is 15 or less, the z-interval and t-interval procedures will not provide reliable estimates if the distribution of the underlying variable is not normal. For sample sizes in the range of 15 to 30, both procedures can be used if the data is roughly normal and has no outliers.

Suppose that you have obtained a sample with the intent of performing a particular statistical-inference procedure. What should you do before applying the procedure to the sample data? Why?

Before applying a particular statistical inference procedure, we should look at graphical displays of the sample data to see if there appear to be any violations of the conditions required for the use of the procedure.

A confidence interval for a population mean has a margin of error of 10.7. (a) Obtain the length of the confidence interval. (b) If the mean of the sample is 75.2, determine the confidence interval.

(a) 21.4

(b) (64.5, 85.9)

(b) (64.5, 85.9)

Dr. Thomas Stanley of Georgia State University has surveyed millionaires since 1973. Among other information, Stanley obtains estimates for the mean age, µ, of all U.S. millionaires. Suppose that 36 U.S. millionaires are randomly selected (mean = 58.53). Determine a 95% confidence interval for the mean age, µ, of all U.S. millionaires. Assume that the standard deviation of ages of all U.S. millionaires is 13.0 years.

(54.3, 62.8)

Abigail Camp Dimon found the mean shell length of 461 randomly selected specimens of N. trivittata to be 11.9 mm. (a) assuming that ϑ = 2.5 mm, obtain a 90% confidence interval for the mean length, µ, of all N. trivittata. (b) interpret your answer from part (a), (c) What properties should a normal probability plot of the data have for it to be permissible to apply the procedure that you used in part (a)? (d) find the margin of error E. (e) Explain the meaning of E as far as the accuracy of the estimate is concerned. (f) Determine the sample size required to have a margin of error of 0.1 mm and a 90% confidence level. (g) Find a 90% confidence interval for µ if a sample of the size determined in part (f) yields a mean of 12.0 mm.

(a) (11.71, 12.09)

(b) We can be 90% confident that the mean length of N. trivittata is somewhere between 11.71 and 12.09 mm.

(c) Since the sample size is very large, the distribution of sample means will be approximately normal regardless of the shape of the original distribution.

(d) 0.19

(e) We can be 90% confident that the maximum error made in using x-bar to estimate µ is 0.19 mm.

(f)1692

(g) (11.90, 12.10)

(b) We can be 90% confident that the mean length of N. trivittata is somewhere between 11.71 and 12.09 mm.

(c) Since the sample size is very large, the distribution of sample means will be approximately normal regardless of the shape of the original distribution.

(d) 0.19

(e) We can be 90% confident that the maximum error made in using x-bar to estimate µ is 0.19 mm.

(f)1692

(g) (11.90, 12.10)

For a t-curve with df = 18, obtain the t-values of the following:

(a) The t-value having area 0.025 to its right

(b) t (sub 0.05)

(c) The t-value having area 0.10 to its left

(d) The two t-values that divide the area under the curve into a middle 0.99 area and two outside 0.005 areas.

(a) The t-value having area 0.025 to its right

(b) t (sub 0.05)

(c) The t-value having area 0.10 to its left

(d) The two t-values that divide the area under the curve into a middle 0.99 area and two outside 0.005 areas.

(a) 2.101

(b) 1.734

(c) -1.330

(d) +/- 2.878

(b) 1.734

(c) -1.330

(d) +/- 2.878

In a Singapore edition of Business Times, diamond pricing was explored. The price of a diamond is based on the diamond's weight, color, and clarity. A simple random sample of 18 one-half carat diamonds had the following prices, in dollars: 1676, 1995, 1442, 1876, 1995, 2032, 1718, 1988, 1826, 2071, 2071, 2234, 1947, 2108, 1983, 1941, 2146, 2316. (a) Apply the t-interval procedure to these data to find a 90% confidence interval for the mean price of all one-half carat diamonds. Interpret your result. (note: x-bar = $1964.7 and s = $206.5.)

($1880.07, $2049.37)

We can be 90% confident that the mean diamond price for one-half carat diamonds is between $1880.07 and $2049.37.

We can be 90% confident that the mean diamond price for one-half carat diamonds is between $1880.07 and $2049.37.

True or False (and give a reason for your answer): If a 95% confidence interval for a population mean, µ, is from 33.8 to 39.0, the mean of the population must lie somewhere between 33.8 and 39.0

False. We are 95% confident that the mean lies in the interval from 33.8 to 39.0, but about 5% of the time, the procedure will produce an interval that does not contain the population mean. Therefore, we cannot say that the mean must lie in the interval.

If you obtained one thousand 95% confidence intervals for a population mean, µ, roughly how many of the intervals would actually contain µ?

950

Suppose that you intend to find a 95% confidence interval for a population mean by applying the one-mean z-interval procedure to a sample of size 100. (a) What would happen to the precision of the estimate if you used a sample of size 50 instead but kept the same confidence level of 0.95? (b) What would happen to the precision of the estimate if you changed the confidence level to 0.90 but kept the same sample size of 100?

(a) Reducing the sample size from 100 to 50 will reduce the precision of the estimate (result in a longer confidence interval).

(b) Reducing the confidence level from .95 to .90 while maintaining the sample size will increase the precision of the estimate (result in a shorter confidence interval).

(b) Reducing the confidence level from .95 to .90 while maintaining the sample size will increase the precision of the estimate (result in a shorter confidence interval).

Suppose that you plan to apply the one-mean z-interval procedure to obtain a 90% confidence interval for a population mean, µ. You know that ϑ = 12 and that you are going to use a sample of size 9. (a) What will be your margin of error? (b) What else do you need to know in order to obtain the confidence interval?

(a) 6.58

(b) To obtain the confidence interval, you also need to know x-bar.

(b) To obtain the confidence interval, you also need to know x-bar.

Decide whether the appropriate method for obtaining the confidence interval is the z-interval procedure, the t-interval procedure, or neither:

A random sample of size 17 is taken from a population. A normal probability plot of the sample data is found to be very close to linear (straight line). The population standard deviation is unknown.

A random sample of size 17 is taken from a population. A normal probability plot of the sample data is found to be very close to linear (straight line). The population standard deviation is unknown.

t-interval procedure

Decide whether the appropriate method for obtaining the confidence interval is the z-interval procedure, the t-interval procedure, or neither:

A random sample of size 50 is taken from a population. A normal probability plot of the sample data is found to be roughly linear. The population standard deviation is known.

A random sample of size 50 is taken from a population. A normal probability plot of the sample data is found to be roughly linear. The population standard deviation is known.

z-interval procedure

Decide whether the appropriate method for obtaining the confidence interval is the z-interval procedure, the t-interval procedure, or neither:

A random sample of size 25 is taken from a population. A normal probability plot of the sample data shows three outliers but is otherwise roughly linear. Checking reveals that the outliers are due to recording errors. The population standard deviation is known.

A random sample of size 25 is taken from a population. A normal probability plot of the sample data shows three outliers but is otherwise roughly linear. Checking reveals that the outliers are due to recording errors. The population standard deviation is known.

z-interval procedure

Decide whether the appropriate method for obtaining the confidence interval is the z-interval procedure, the t-interval procedure, or neither:

A random sample of size 20 is taken from a population. A normal probability plot of the sample data shows three outliers but is otherwise roughly linear. Removal of the outliers is questionable. The population standard deviation is unknown.

A random sample of size 20 is taken from a population. A normal probability plot of the sample data shows three outliers but is otherwise roughly linear. Removal of the outliers is questionable. The population standard deviation is unknown.

Neither procedure should be used.

Decide whether the appropriate method for obtaining the confidence interval is the z-interval procedure, the t-interval procedure, or neither:

A random sample of size 128 is taken from a population. A normal probability plot of the sample data shows no outliers but has significant curvature. The population standard deviation is known.

A random sample of size 128 is taken from a population. A normal probability plot of the sample data shows no outliers but has significant curvature. The population standard deviation is known.

z-interval procedure

Decide whether the appropriate method for obtaining the confidence interval is the z-interval procedure, the t-interval procedure, or neither:

A random sample of size 13 is taken from a population. A normal probability plot of the sample data shows no outliers but has significant curvature. The population standard deviation is unknown.

A random sample of size 13 is taken from a population. A normal probability plot of the sample data shows no outliers but has significant curvature. The population standard deviation is unknown.

Neither procedure should be used.

Following are the arterial blood pressures, in millimeters of mercury (mm Hg), for a random sample of 16 children of diabetic mothers:

(x-bar = 85.99 mm Hg, s = 8.08 mm Hg)

(a) Apply the t-interval procedure to find a 95% confidence interval for the mean arterial blood pressure of all children of diabetic mothers. Interpret your result.

(x-bar = 85.99 mm Hg, s = 8.08 mm Hg)

(a) Apply the t-interval procedure to find a 95% confidence interval for the mean arterial blood pressure of all children of diabetic mothers. Interpret your result.

(81.69, 90.29)

We can be 95% confident that the mean arterial blood pressure of all children of diabetic mothers is somewhere between 81.69 and 90.29 mm Hg.

We can be 95% confident that the mean arterial blood pressure of all children of diabetic mothers is somewhere between 81.69 and 90.29 mm Hg.

Researchers at the University of Washington and Harvard University analyzed records of breast cancer screening and diagnostic evaluations. Discussing the benefits and downsides of the screening process, the article states that, although the rate of false-positives is higher than previously thought, if radiologists were less aggressive in following up on suspicious tests, the rate of false-positives would fall but the rate of missed cancers would rise. Suppose that such a screening test is used to decide between a null hypothesis of H₀: no cancer is present, and an alternative hypothesis of H₁: Cancer is present.

(a) Would a false-positive (thinking cancer is present when in fact it is not) be a Type I or Type II error? (b) Describe a Type I error in the context of this problem and discuss the (real-life) consequences of making a Type I error. (c) Describe a Type II error in the context of this problem and discuss the (real-life) consequences of making a Type II error.

(a) Would a false-positive (thinking cancer is present when in fact it is not) be a Type I or Type II error? (b) Describe a Type I error in the context of this problem and discuss the (real-life) consequences of making a Type I error. (c) Describe a Type II error in the context of this problem and discuss the (real-life) consequences of making a Type II error.

(a) Type I

(b) The Type I error would be a false positive (thinking cancer is present when there is none) and would likely result in following up by the radiologists. There would be perhaps unnecessary additional tests which would increase costs.

(c) A Type II error would be thinking that cancer is not present when in fact it is (a false negative). Thinking that no cancer is present, there would likely be no follow-up. This is the most serious type of error in the scenario presented because there could be cancers that are missed and not treated.

(b) The Type I error would be a false positive (thinking cancer is present when there is none) and would likely result in following up by the radiologists. There would be perhaps unnecessary additional tests which would increase costs.

(c) A Type II error would be thinking that cancer is not present when in fact it is (a false negative). Thinking that no cancer is present, there would likely be no follow-up. This is the most serious type of error in the scenario presented because there could be cancers that are missed and not treated.

Find the critical value(s) for the following tests:

(a) Right tail test with ∝ = 0.10

(b) Left tail test with ∝ = 0.01

(c) Two tail test with ∝ = 0.05

(a) Right tail test with ∝ = 0.10

(b) Left tail test with ∝ = 0.01

(c) Two tail test with ∝ = 0.05

(a) 1.282

(b) -2.326

(c) +/- 1.96

(b) -2.326

(c) +/- 1.96

In a study of computer use, 1000 randomly selected Canadian internet users were asked how much time they spend using the Internet in a typical week. The mean of the 1000 resulting observations was 12.7 hours. Assume the population standard deviation is 5 hours. Carry out a hypothesis test with a significance level of 0.05 to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than 12.5 hours.

1. H₀: µ = 12.5 hours, H₁: µ > 12.5 hours

2. ∝ = 0.05

3. test statistic: z = 1.26

4. critical value: 1.645

5. conclusion: Do not reject the null hypothesis.

6. interpretation: There is not enough evidence to conclude that the average time spent using the Internet by Canadians is greater than 12.5 hours.

2. ∝ = 0.05

3. test statistic: z = 1.26

4. critical value: 1.645

5. conclusion: Do not reject the null hypothesis.

6. interpretation: There is not enough evidence to conclude that the average time spent using the Internet by Canadians is greater than 12.5 hours.

A credit bureau analysis of undergraduate student credit records found that the average number of credit cards in an undergraduate's wallet was 4.09. It was also reported that in a random sample of 132 undergraduates, the sample mean number of credit cards carried was 2.6. Assume the population standard deviation as 1.2. Is there enough evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau figure of 4.09?

1. H₀: µ = 4.08, H₁: µ < 4.09

2. ∝ = 0.05 (not given)

3. Test statistic: z = -14.26

4. Critical value: -1.645

5. Conclusion: Reject the null hypothesis.

6. There is sufficient evidence to conclude that the average number of credit cards that undergraduates report is less than the credit bureau's figure of 4.09.

2. ∝ = 0.05 (not given)

3. Test statistic: z = -14.26

4. Critical value: -1.645

5. Conclusion: Reject the null hypothesis.

6. There is sufficient evidence to conclude that the average number of credit cards that undergraduates report is less than the credit bureau's figure of 4.09.

Consider the null and alternative hypotheses:

H₀: µ = 30 lb (mean has not increased)

H₁: µ > 30 lb (mean has increased)d

where µ is last year's mean cheese consumption for all Americans. Explain what each of the following would mean:

(a) Type I error

(b) Type II error

(c) correct decision

Now suppose that the results of carrying out the hypothesis test lead to nonrejection of the null hypothesis. Classify that decision by error type or as a correct decision if in fact last year's mean cheese consumption

(d) has not increased from the 2001 mean of 30.0 lb

(e) has increased from the 2001 mean of 30 lb.

H₀: µ = 30 lb (mean has not increased)

H₁: µ > 30 lb (mean has increased)d

where µ is last year's mean cheese consumption for all Americans. Explain what each of the following would mean:

(a) Type I error

(b) Type II error

(c) correct decision

Now suppose that the results of carrying out the hypothesis test lead to nonrejection of the null hypothesis. Classify that decision by error type or as a correct decision if in fact last year's mean cheese consumption

(d) has not increased from the 2001 mean of 30.0 lb

(e) has increased from the 2001 mean of 30 lb.

(a) a Type I error would occur if, in fact, µ = 30 lb, but the results of the sampling lead to the conclusion that µ > 30 lb.

(b) a Type II error would occur if, in fact, µ > 30 lb, but the results of the sampling fail to lead to that conclusion

(c) A correct decision would occur if, in fact, µ = 30 lb and the results of the sampling do not lead to the rejection of that fact; if, in fact µ > 30 lb and the results of the sampling lead to that conclusion

(d) If, in fact, last year's mean consumption of cheese for all Americans has not increased over the 2001 mean of 30.0 lb, and we do not reject the null hypothesis that µ = 30 lb, we made a correct decision

(e) If, in fact, last year's mean consumption of cheese for all Americans has increased over the 2001 mean of 30.0 lb, and we fail to reject the null hypothesis that µ = 30 lb, we made a Type II error.

(b) a Type II error would occur if, in fact, µ > 30 lb, but the results of the sampling fail to lead to that conclusion

(c) A correct decision would occur if, in fact, µ = 30 lb and the results of the sampling do not lead to the rejection of that fact; if, in fact µ > 30 lb and the results of the sampling lead to that conclusion

(d) If, in fact, last year's mean consumption of cheese for all Americans has not increased over the 2001 mean of 30.0 lb, and we do not reject the null hypothesis that µ = 30 lb, we made a correct decision

(e) If, in fact, last year's mean consumption of cheese for all Americans has increased over the 2001 mean of 30.0 lb, and we fail to reject the null hypothesis that µ = 30 lb, we made a Type II error.

According to an FBI document, the mean value lost to purse snatching was $332 in 2002. For last year, 12 randomly selected purse-snatching offenses yielded the following values lost, to the nearest dollar: 207, 237, 422, 226, 272, 205, 362, 348, 165, 266, 269, 430. (a) use a t-test with either the critical-value approach or the P-value approach to decide, at the 5% significance level, whether last year's mean value lost to purse snatching has decreased from the 2002 mean. The mean and standard deviation of the data are $284.1 and $86.9, respectively. (b) Perform the required hypothesis test, using the Wilcoxon signed-rank test.

(a) H₀: µ = 332, H₁: µ < 332

∝ = 0.05

test statistic: t = -1.909

Critical value = -1.782

Conclusion: reject H₀

Interpretation: At the 5% significance level, the data provides sufficient evidence to conclude that the mean value lost because of purse snatching has decreased from the 2002 mean of $332.00

(b) Critical value: 12(13)/2 - 61 = 17

W = 17

Since the W-value is less than or equal to the critical value, we reject the null hypothesis.

∝ = 0.05

test statistic: t = -1.909

Critical value = -1.782

Conclusion: reject H₀

Interpretation: At the 5% significance level, the data provides sufficient evidence to conclude that the mean value lost because of purse snatching has decreased from the 2002 mean of $332.00

(b) Critical value: 12(13)/2 - 61 = 17

W = 17

Since the W-value is less than or equal to the critical value, we reject the null hypothesis.

In May, 2002, the average cost of a private room in a nursing home was $168 per day. For August 2003, a random sample of 11 nursing homes yielded the following daily costs, in dollars, for a private room in a nursing home: 73, 159, 199, 182, 192, 208, 181, 129, 182, 282, 250. Use the Wilcoxon signed-rank test to decide at the 10% significance level whether the average cost for a private room in a nursing home in August 2003 exceeded that in May 2002.

Test Statistic: W = 48

Reject the null hypothesis.

Reject the null hypothesis.

Assumptions for the Wilcoxon Signed Rank Test

(1) simple random sample

(2) symmetric distribution (triangular, uniform, symmetric bimodal)

(2) symmetric distribution (triangular, uniform, symmetric bimodal)

Steps for the Wilcoxon Signed Rank Test

(1) Determine null and alternative hypotheses

(2) Determine value for ∝

(3) Construct the table to find W, the test statistic

(4) Find critical value(s)

(5) Number line; plot test statistic and critical value(s)

(6) Conclusion: reject H₀ or do not reject H₀ based on where test statistic falls

(7) Interpret results

(2) Determine value for ∝

(3) Construct the table to find W, the test statistic

(4) Find critical value(s)

(5) Number line; plot test statistic and critical value(s)

(6) Conclusion: reject H₀ or do not reject H₀ based on where test statistic falls

(7) Interpret results

In the Wilcoxon signed rank test, how do you assign ranks if there is a tie (1) between 2 values of |D|, (2) between 3 values of |D|

(1) If there is a tie between two values of |D|, then average the ranks and assign that average to the two values

(2) If there is a tie between three values of |D|, then assign the middle rank to all three values

(2) If there is a tie between three values of |D|, then assign the middle rank to all three values

In the Wilcoxon signed rank test, if an observation is equal to µ₀, what should happen to that observation?

The observation is to be tossed out and the sample size reduced accordingly.

Test statistic for z-test

Test statistic for t-test

The National Center for Test Statistics reports that the median birth weight of U.S. babies was 7.4 lb in 2002. A random sample of this year's births provided the following weights, in pounds: 8.6, 8.8, 7.4, 8.2, 5.3, 9.2, 13.8, 5.6, 7.8, 6.0, 5.7, 11.6, 9.2, 7.2. Con we conclude that this year's median birth weight differs from that in 2002? Use a significance level of 0.05.

(1) H₀: µ = 7.4 lb

H₁: µ ≠ 7.4 lb

(2) ∝ = 0.05

(3) Test statistic (from created table): W = 57.5

(4) Critical values - Right side 74, Left side 17

(5) Conclusion: Do not reject H₀

(6) Interpretation: There is not enough evidence to conclude that the median birth weight of U.S. babies is different than 7.4 lbs in 2002.

H₁: µ ≠ 7.4 lb

(2) ∝ = 0.05

(3) Test statistic (from created table): W = 57.5

(4) Critical values - Right side 74, Left side 17

(5) Conclusion: Do not reject H₀

(6) Interpretation: There is not enough evidence to conclude that the median birth weight of U.S. babies is different than 7.4 lbs in 2002.

For the 2002 baseball season, the median baseball salary was determined to be $800,000. A random sample of 14 salaries was conducted with the following salaries in 2005 (in thousands) as follows: 316, 326, 331, 332, 335, 550, 750, 950, 1300, 3000, 4000, 6500, 8250, 13100.At the significance level of 0.05, is there enough evidence to conclude that the median baseball salary for the 2005 baseball season is more than $800,000? Use the Wilcoxon signed rank test.

(1) H₀: µ = 800

H₁: µ > 800

(2) ∝ = 0.05

(3) Test Statistic (Construct Table): W = 71

(4) Critical Value = 79

(5) Conclusion: Do not reject H₀

(6) Interpretation: At the 5% significance level, there is not sufficient evidence to conclude that the median baseball salary is greater than that in 2002 ($800,000).

H₁: µ > 800

(2) ∝ = 0.05

(3) Test Statistic (Construct Table): W = 71

(4) Critical Value = 79

(5) Conclusion: Do not reject H₀

(6) Interpretation: At the 5% significance level, there is not sufficient evidence to conclude that the median baseball salary is greater than that in 2002 ($800,000).

Standardized version of the difference of two means (x-bar₁ - x-bar₂)

Sampling Distribution for x-bar₁ - x-bar₂

Mean µ₁ - µ₂

Std Deviation: See above

Shape: normal distribution

Std Deviation: See above

Shape: normal distribution

Technically, what is a nonparametric method? In current statistical practice, how is that term used?

A nonparametric method is an inferential method not concerned with parameters (such as µ and ϑ). Common statistical practice is to refer to most methods that can be applied without assuming normality as nonparametric.

What assumption must be met in order to use the Wilcoxon signed rank test?

a symmetric distribution

In a Wilcoxon signed rank test, an observation equals µ₀ (the value given for the mean in the null hypothesis), that observation should be removed and the sample size reduced by 1. Why does that need to be done?

Because the D-value for such an observation equals 0, a sign cannot be attached to the rank of |D|

The Wilcoxon signed rank test can be used to perform a hypothesis test for a population median, η, as well as for the population mean. Why is that so?

For a symmetric distribution, the mean and median are equal.

During the late 1800s, Lake Wingra in Madison, Wisconsin, was frozen over an average of 124.9 days per year. A random sample of eight recent years provided the following data on numbers of days that the lake was frozen over: 103, 80, 79, 135, 134, 77, 80, 111. At the 5% significance level, do the data provide sufficient evidence to conclude that the average number of ice days is less now than in the late 1800s? Use the Wilcoxon signed rank test.

(1) H₀: µ = 124.9

H₁: µ < 124.9

(2) ∝ = 0.05

(3) Test Statistic (from constructed table): W = 3

(4) Critical value = 6

(5) Conclusion: Reject H₀

(6) Interpretation: At the 5% significance level, there is sufficient evidence to conclude that the average number of ice days is less than that in the late 1880s.

H₁: µ < 124.9

(2) ∝ = 0.05

(3) Test Statistic (from constructed table): W = 3

(4) Critical value = 6

(5) Conclusion: Reject H₀

(6) Interpretation: At the 5% significance level, there is sufficient evidence to conclude that the average number of ice days is less than that in the late 1880s.

In 2002, the median age of U.S. residents was 35.7 years. A random sample of 10 U.S. residents taken this year yielded the following data, in years: 42, 45, 62, 49, 14, 39, 57, 11, 36, 26. At the 1% significance level, do the data provide sufficient evidence to conclude that the median age of today's U.S. residents has increased from the 2002 median age of 35.7 years? Use the Wilcoxon signed rank test.

(1) H₀: µ = 35.7

H₁: µ > 35.7

(2) ∝ = .01

(3) Test Statistic (from constructed table): W = 33

(4) Critical value: 50

(5) Conclusion: Do not reject H₀

(6) Interpretation: At the 1% significance level, there is not sufficient evidence to conclude that the median age of today's U.S. residents has increased from the 2002 median age of 35.7 years.

H₁: µ > 35.7

(2) ∝ = .01

(3) Test Statistic (from constructed table): W = 33

(4) Critical value: 50

(5) Conclusion: Do not reject H₀

(6) Interpretation: At the 1% significance level, there is not sufficient evidence to conclude that the median age of today's U.S. residents has increased from the 2002 median age of 35.7 years.

Consider the following quantities: µ, ϑ, x-bar, s.

Which are parameters and which are statistics? Which are fixed numbers and which are variables?

Which are parameters and which are statistics? Which are fixed numbers and which are variables?

Parameters: µ and ϑ

Statistics: x-bar and s (sample standard deviation)

Parameters are fixed numbers and statistics are variables.

Statistics: x-bar and s (sample standard deviation)

Parameters are fixed numbers and statistics are variables.

Why do you need to know the sampling distribution of the difference between two sample means in order to perform a hypothesis test to compare two population means?

So that you can determine whether the observed difference between the two sample means can be reasonably attributed to sampling error or whether that difference suggests that the null hypothesis of equal population means is false and the alternative hypothesis is true.

Identify the assumption for using the two-means z-test and the two-means z-interval procedure that renders those procedures generally impractical.

The assumption that ϑ is known (because population standard deviations are usually unknown).

How do you calculate the test statistic for the w-test?

From a constructed chart using (1) observations, (2) difference between observation and µ₀, (3) absolute values of values calculated in step 2, (4) ranking of the absolute values, (5) signing ranks according to sign in step 2, (6) w = sum of positive ranks.

With the following hypothesis test: (a) identify the variable, (b) identify the two populations, (c) determine the null and alternative hypotheses, (d) classify the hypothesis test as two tailed, left tailed, or right tailed:

Samples of adolescent offspring of diabetic mothers (ODM) and non-diabetic mothers (ONM) were taken and evaluated for potential differences in vital measurements, including blood pressure and glucose tolerance. A hypothesis test is to be performed to decide whether the mean systolic blood pressure of ODM adolescents exceeds that of ONM adolescents.

Samples of adolescent offspring of diabetic mothers (ODM) and non-diabetic mothers (ONM) were taken and evaluated for potential differences in vital measurements, including blood pressure and glucose tolerance. A hypothesis test is to be performed to decide whether the mean systolic blood pressure of ODM adolescents exceeds that of ONM adolescents.

(a) systolic blood pressure

(b) ODM adolescents and ONM adolescents

(c) H₀: µ₁ = µ₂

H₁: µ₁ > µ₂

* µ₁ = mean systolic bp of ODM adolescents

* µ₂ = mean systolic bp of ONM adolescents

(d) right-tailed

(b) ODM adolescents and ONM adolescents

(c) H₀: µ₁ = µ₂

H₁: µ₁ > µ₂

* µ₁ = mean systolic bp of ODM adolescents

* µ₂ = mean systolic bp of ONM adolescents

(d) right-tailed

A variable of two populations has a mean of 40 and a standard deviation of 12 for one of the populations and a mean of 40 and a standard deviation of 6 for the other population. For independent samples of sizes 9 and 4 respectively, find the mean and standard deviation of x-bar₁ - x-bar₂.

Mean: 0

Standard Deviation: 5

Standard Deviation: 5

The Kelley Blue Book provides information on retail and trad-in values for used card and trucks. The retail value represents the price a dealer might charge after preparing the vehicle for sale. A 2003 Ford Mustang coupe has a 2006 Kelley Blue Book retail value of $12,850. We obtained the following asking prices, in dollars, for a sample of 2003 Ford Mustang coupes for sale in Phoenix, AZ: 13480, 12499, 12992, 11500, 12988, 10400, 12800, 12500, 12599, 12600. At the 10% significance level, do the data provide sufficient evidence to conclude that the mean asking price for 2003 Ford Mustang coupes in Phoenix is less than the 2006 Kelley Blue Book retail value? Use the Wilcoxon signed rank test.

(1) H₀ = 12,850

H₁ < 12,850

(2) ∝ = 0.10

(3) Test statistic: W₀ = 13

(4) Critical value = (10)(11)/2 - 41

(5) Since W < 14, reject H₀

(6) At the 10% significance level, there is sufficient evidence to conclude that the mean asking price for a 2003 Ford Mustang is less than the 2006 Kelly Blue Book value.

H₁ < 12,850

(2) ∝ = 0.10

(3) Test statistic: W₀ = 13

(4) Critical value = (10)(11)/2 - 41

(5) Since W < 14, reject H₀

(6) At the 10% significance level, there is sufficient evidence to conclude that the mean asking price for a 2003 Ford Mustang is less than the 2006 Kelly Blue Book value.

A certain pen has been designed to that true average writing lifetime under controlled conditions is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal probability plot of the resulting data support the use of a one-sample z-test. The relevant hypotheses are H₀: µ = 10, and H₁: µ < 10. Use the p-value method.

(a) If z = -1.8 and ∝ = 0.05 is selected, what conclusion is appropriate?

(b) If Z = -2.4 and ∝ = 0.10 is selected, what conclusion is appropriate?

(c) If z = - 0.89, what conclusion is appropriate?

(a) If z = -1.8 and ∝ = 0.05 is selected, what conclusion is appropriate?

(b) If Z = -2.4 and ∝ = 0.10 is selected, what conclusion is appropriate?

(c) If z = - 0.89, what conclusion is appropriate?

(a) It is appropriate to reject H₀.

(b) It is appropriate to reject H₀.

(c) Do not reject H₀ (weak or no evidence - P-value > 0.10)

(b) It is appropriate to reject H₀.

(c) Do not reject H₀ (weak or no evidence - P-value > 0.10)

A credit bureau analysis of undergraduate student credit records found that the average number of credit cards in an undergraduate's wallet was 4.09. It was also reported that in a random sample of 132 undergraduates, the sample mean number of credit cards carried was 2.6. Assume the population standard deviation was 1.2. Is there enough evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of 4.09?

(a) Hypotheses:

(b) ∝:

(c) Test statistic:

(d) P-value:

(e) Conclusion:

(f) Summary:

(a) Hypotheses:

(b) ∝:

(c) Test statistic:

(d) P-value:

(e) Conclusion:

(f) Summary:

(a) H₀: µ = 4.09

H₁: µ < 4.09

(b) ∝ = .05 (not given)

(c) Test statistic: z₀ = -14.27

(d) P-value: ≈ 0 (less than 0.0001)

(e) Conclusion: Reject H₀

(f) Interpretation: There is enough evidence to suggest that the average number of credit cards in an undergraduate's wallet is less than 4.09.

H₁: µ < 4.09

(b) ∝ = .05 (not given)

(c) Test statistic: z₀ = -14.27

(d) P-value: ≈ 0 (less than 0.0001)

(e) Conclusion: Reject H₀

(f) Interpretation: There is enough evidence to suggest that the average number of credit cards in an undergraduate's wallet is less than 4.09.

Fill in the blanks:

(a) For a two-tail test, the ____(a)_____ is the probability of observing a value of the test statistic t that is at least as large in magnitude as the value actually observed, which is the area under the t-curve that lies outside the interval from _(b)_______ to ________(c)______.

(a) For a two-tail test, the ____(a)_____ is the probability of observing a value of the test statistic t that is at least as large in magnitude as the value actually observed, which is the area under the t-curve that lies outside the interval from _(b)_______ to ________(c)______.

(a) p-value

(b) - |t₀|

(c) |t₀|

(b) - |t₀|

(c) |t₀|

Fill in the blanks:

(Use the Book tables) For a right-tailed test with n=15, ∝ = 0.01 and a value of the test statistic of t = 3.458, the test statistic lies to the right of ____(a)____. This means the true p-value is (choose one: larger/smaller) ____(b)____ than ____(c)____.

(Use the Book tables) For a right-tailed test with n=15, ∝ = 0.01 and a value of the test statistic of t = 3.458, the test statistic lies to the right of ____(a)____. This means the true p-value is (choose one: larger/smaller) ____(b)____ than ____(c)____.

(a) 2.977

(b) smaller

(c) 0.005

(b) smaller

(c) 0.005

What are the 3 assumptions of the t-test?

(1) simple random sample

(2) normal population or a large sample

(3) ϑ unknown

(2) normal population or a large sample

(3) ϑ unknown

A hot tub manufacturer advertises that with its heating equipment, a temperature of 100° F can be achieved in at most 15 minutes. A random sample of 25 tubs is selected, and the time necessary to achieve a 100° F temperature is determined for each tub. The sample average time and sample standard deviation are 17.5 min and 2.2 min, respectively. Does this information cast doubt on the company's claim? Carry out a hypothesis test with a significance level of 0.05.

(a) Hypotheses:

(b) ∝:

(c) Test statistic:

(d) p-value:

(e) Conclusion:

(f) Summary:

(a) Hypotheses:

(b) ∝:

(c) Test statistic:

(d) p-value:

(e) Conclusion:

(f) Summary:

(a) H₀: µ ≤ 15 minutes

H₁: µ > 15 minutes

(b) ∝ = 0.05

(c) Test statistic: t₀ = 5.68

(d) p-value: < .0001 (≈.00000374)

(e) Conclusion: Reject H₀

(f) Summary: There is enough evidence to suggest the average time for a hot tub to reach 100° F is greater than 15 minutes.

H₁: µ > 15 minutes

(b) ∝ = 0.05

(c) Test statistic: t₀ = 5.68

(d) p-value: < .0001 (≈.00000374)

(e) Conclusion: Reject H₀

(f) Summary: There is enough evidence to suggest the average time for a hot tub to reach 100° F is greater than 15 minutes.

Test statistic formula for independent samples (2 population means, equal standard deviations)

(Y-bars should show as x-bars)

Critical value(s) for independent samples (2 population means, equal standard deviations)

Two tailed: +/- t (sub ∝/2), df = n₁ + n₂ - 2

Left tailed : - t (sub ∝), df = n₁ + n₂ - 2

Right tailed: t (sub ∝), df = n₁ + n₂ - 2

Left tailed : - t (sub ∝), df = n₁ + n₂ - 2

Right tailed: t (sub ∝), df = n₁ + n₂ - 2

Test statistic formula for independent samples (2 population means, standard deviations not equal)

Critical value(s) for independent samples (2 population means, standard deviations not equal)

Two tailed: +/- t (sub ∝/2), df = Δ

Left tailed : - t (sub ∝), df = Δ

Right tailed: t (sub ∝), df = Δ

Left tailed : - t (sub ∝), df = Δ

Right tailed: t (sub ∝), df = Δ

2-mean confidence interval formula for independent samples (standard deviations equal)

2-mean confidence interval formula for independent samples (standard deviations not equal)

* - refers to:

t (sub ∝/2), df = Δ

t (sub ∝/2), df = Δ

Pooled T-Test Assumptions

(1) simple random sample

(2) normal population or large sample

(3) independent samples

(4) equal population standard deviations

(2) normal population or large sample

(3) independent samples

(4) equal population standard deviations

Non-pooled T-Test Assumptions

(1) simple random sample

(2) normal population or large sample

(3) independent samples

(4) unequal population standard deviations

(2) normal population or large sample

(3) independent samples

(4) unequal population standard deviations

How do you know whether to use the pooled t-test or the non-pooled t-test for 2-mean independent samples?

Compare the standard deviations of the two samples. If one is nearly twice the other, use the non-pooled t-test. If they are fairly close, use the pooled t-test.

Of the 4 pooled t-test assumptions (simple random sample, independent samples, normal population or large samples, equal population standard deviations), how important is each of these?

Simple random samples and independent samples are essential assumptions. Moderate violations of the normality assumption are permissible even for small or moderate size samples. Moderate violations of the equal standard deviations requirement are not serious provided the two sample sizes are roughly equal.

The U.S. Bureau of Prisons publishes data in Prison Statistics on the times served by prisoners released from federal institutions for the first time. Independent random samples of released prisoners in the fraud and firearms offense categories yielded the following information on time served, in months:

Fraud: x-bar₁ = 10.12, s₁ = 4.90, n₁ = 10

Firearms: x-bar₂ = 18.78, s₂ = 4.64, n₂ = 10

At the 5% significance level, do the data provide sufficient evidence to conclude that the mean time served for fraud is less than that for firearms offenses?

Fraud: x-bar₁ = 10.12, s₁ = 4.90, n₁ = 10

Firearms: x-bar₂ = 18.78, s₂ = 4.64, n₂ = 10

At the 5% significance level, do the data provide sufficient evidence to conclude that the mean time served for fraud is less than that for firearms offenses?

(1) H₀: µ₁ = µ₂

H₁: µ₁ < µ₂

(2) ∝ = 0.05

(3) Test Statistic: t = -4.058

(4) Critical Value = -1.734, P-value: = .000369

(5) Conclusion: Reject H₀

(6) At the 5% significance level, there is sufficient evidence to conclude that the mean time served for fraud is less than that for firearms offenses.

H₁: µ₁ < µ₂

(2) ∝ = 0.05

(3) Test Statistic: t = -4.058

(4) Critical Value = -1.734, P-value: = .000369

(5) Conclusion: Reject H₀

(6) At the 5% significance level, there is sufficient evidence to conclude that the mean time served for fraud is less than that for firearms offenses.

L. Smith and D. Haukos examined the relationship of species richness and diversity to playa area and watershed disturbance. Independent random samples of 126 playa with cropland and 98 playa with grassland in Southern Great Plains yielded the following summary statistics for the number of native species:

Cropland: x-bar₁ = 14.06, s₁ = 4.83, n₁ = 126

Wetland: x-bar₂ = 15.36, s₂ = 4.95, n₂ = 98

At the 5% significance level, do the data provide sufficient evidence to conclude that a difference exists in the mean number of native species in the two regions?

Cropland: x-bar₁ = 14.06, s₁ = 4.83, n₁ = 126

Wetland: x-bar₂ = 15.36, s₂ = 4.95, n₂ = 98

At the 5% significance level, do the data provide sufficient evidence to conclude that a difference exists in the mean number of native species in the two regions?

(1) H₀: µ₁ = µ₂

H₁: µ₁ ≠ µ₂

(2) ∝ = 0.05

(3) Test Statistic: t = -1.98

(4) Critical Value = -1.971, P-value: = 0.0493

(5) Conclusion: Reject H₀

(6) At the 5% significance level, there is sufficient evidence to conclude that there is a difference in the mean number of native species in the two regions.

H₁: µ₁ ≠ µ₂

(2) ∝ = 0.05

(3) Test Statistic: t = -1.98

(4) Critical Value = -1.971, P-value: = 0.0493

(5) Conclusion: Reject H₀

(6) At the 5% significance level, there is sufficient evidence to conclude that there is a difference in the mean number of native species in the two regions.

Suppose that you know that a variable is normally distributed on each of two populations. Further suppose that you want to perform a hypothesis test based on independent random samples to compare the two population means. In each case, decide whether you would use the pooled or nonpooled t-test.

(1) You know the population standard deviations are equal.

(2) You know that the population standard deviations are not equal.

(3) The sample standard deviations are 23.6 and 25.2, and each sample size is 25.

(4) The sample standard deviations are 23.6 and 59.2.

(1) You know the population standard deviations are equal.

(2) You know that the population standard deviations are not equal.

(3) The sample standard deviations are 23.6 and 25.2, and each sample size is 25.

(4) The sample standard deviations are 23.6 and 59.2.

(1) pooled

(2) nonpooled

(3) pooled

(4) nonpooled

(2) nonpooled

(3) pooled

(4) nonpooled

Researchers randomly and independently selected 32 former prisoners diagnosed with chronic PTSD and 20 former prisoners that were diagnosed with PTSD after release from prison but had since recovered (remitted). The ages, in years, at arrest yielded the following summary statistics:

Chronic: x-bar₁ = 25.8, s₁ = 9.2, n₁ = 32

Remitted: x-bar₂ = 22.1, s₂ = 5.7, n₂ = 20

At the 10% significance level, is there sufficient evidence to conclude that a difference exists in the mean age at arrest of East German prisoners with chronic PTSD and remitted PTSD?

Chronic: x-bar₁ = 25.8, s₁ = 9.2, n₁ = 32

Remitted: x-bar₂ = 22.1, s₂ = 5.7, n₂ = 20

At the 10% significance level, is there sufficient evidence to conclude that a difference exists in the mean age at arrest of East German prisoners with chronic PTSD and remitted PTSD?

(1) H₀: µ₁ = µ₂

H₁: µ₁ ≠ µ₂

(2) ∝ = 0.10

(3) Test Statistic: t = 1.79

(4) Critical Value = +/- 1.677, P-value: = 0.0794

(5) Conclusion: Reject H₀

(6) At the 10% significance level, there is sufficient evidence to conclude that there is a difference in the mean age of arrest of East German prisoners with chronic PTSD and remitted PTSD.

H₁: µ₁ ≠ µ₂

(2) ∝ = 0.10

(3) Test Statistic: t = 1.79

(4) Critical Value = +/- 1.677, P-value: = 0.0794

(5) Conclusion: Reject H₀

(6) At the 10% significance level, there is sufficient evidence to conclude that there is a difference in the mean age of arrest of East German prisoners with chronic PTSD and remitted PTSD.

Use the nonpooled t-test and the nonpooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval:

x-bar₁ = 10 x-bar₂ = 12

s₁ = 2 s₂ = 5

n₁ = 15 n₂ = 15

(a) two tailed test, ∝ = 0.05

(b) 95% confidence interval

x-bar₁ = 10 x-bar₂ = 12

s₁ = 2 s₂ = 5

n₁ = 15 n₂ = 15

(a) two tailed test, ∝ = 0.05

(b) 95% confidence interval

(a) t = -1.44

critical values = +/- 2.101

do not reject H₀

(b) (-4.92, 0.92)

critical values = +/- 2.101

do not reject H₀

(b) (-4.92, 0.92)

Use the nonpooled t-test and the nonpooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval:

x-bar₁ = 20 x-bar₂ = 18

s₁ = 4 s₂ = 5

n₁ = 10 n₂ = 15

(a) right-tailed test, ∝ = 0.05

(b) 90% confidence interval

x-bar₁ = 20 x-bar₂ = 18

s₁ = 4 s₂ = 5

n₁ = 10 n₂ = 15

(a) right-tailed test, ∝ = 0.05

(b) 90% confidence interval

(a) t = 1.11

critical value = 1.717

do not reject H₀

(b) (-1.10, 5.10)

critical value = 1.717

do not reject H₀

(b) (-1.10, 5.10)

Use the nonpooled t-test and the nonpooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval:

x-bar₁ = 20 x-bar₂ = 24

s₁ = 6 s₂ = 2

n₁ = 20 n₂ = 15

(a) left-tailed test, ∝ = 0.05

(b) 90% confidence interval

x-bar₁ = 20 x-bar₂ = 24

s₁ = 6 s₂ = 2

n₁ = 20 n₂ = 15

(a) left-tailed test, ∝ = 0.05

(b) 90% confidence interval

(a) t = -2.78

critical value = -1.711

reject H₀

(b) (-6.46, -1.54)

critical value = -1.711

reject H₀

(b) (-6.46, -1.54)

Researchers obtained the following data on the number of acute postoperative days in the hospital using the dynamic and static systems:

dynamic:

Dynamic: x-bar₁ = 7.36, s₁ = 1.22, n₁ = 14

Static: x-bar₂ = 10.50, s₂ = 4.59, n₂ = 6

At the 5% significance level, do the data provide sufficient evidence to conclude that the mean number of acute postoperative days in the hospital is smaller with the dynamic system than with the static system?

dynamic:

Dynamic: x-bar₁ = 7.36, s₁ = 1.22, n₁ = 14

Static: x-bar₂ = 10.50, s₂ = 4.59, n₂ = 6

At the 5% significance level, do the data provide sufficient evidence to conclude that the mean number of acute postoperative days in the hospital is smaller with the dynamic system than with the static system?

(1) H₀: µ₁ = µ₂

H₁: µ₁ < µ₂

(2) ∝ = 0.05

(3) Test Statistic: t = -1.645

(4) Critical Value = -2.015, P-value: = 0.0781

(5) Conclusion: Do not reject H₀

(6) At the 5% significance level, there is not sufficient evidence to conclude that the mean number of acute postoperative days in the hospital is smaller with the dynamic system than with the static system.

H₁: µ₁ < µ₂

(2) ∝ = 0.05

(3) Test Statistic: t = -1.645

(4) Critical Value = -2.015, P-value: = 0.0781

(5) Conclusion: Do not reject H₀

(6) At the 5% significance level, there is not sufficient evidence to conclude that the mean number of acute postoperative days in the hospital is smaller with the dynamic system than with the static system.

Researchers randomly and independently selected 32 former prisoners diagnosed with chronic PTSD and 20 former prisoners that were diagnosed with PTSD after release from prison but had since recovered (remitted). The ages, in years, at arrest yielded the following summary statistics:

Chronic: x-bar₁ = 25.8, s₁ = 9.2, n₁ = 32

Remitted: x-bar₂ = 22.1, s₂ = 5.7, n₂ = 20

Obtain a 90% confidence interval for the difference µ₁ - µ₂, between the mean ages at arrest of East German prisoners with chronic PTSD and remitted PTSD. Interpret your result.

Chronic: x-bar₁ = 25.8, s₁ = 9.2, n₁ = 32

Remitted: x-bar₂ = 22.1, s₂ = 5.7, n₂ = 20

Obtain a 90% confidence interval for the difference µ₁ - µ₂, between the mean ages at arrest of East German prisoners with chronic PTSD and remitted PTSD. Interpret your result.

(.23712, 7.1629) or rounded (.2, 7.2); df = Δ = +/- 1

We can be 90% confident that the difference between the mean ages at arrest of East German prisoners with chronic PTSD and remitted PTSD is between 0.2 and 7.2 years.

We can be 90% confident that the difference between the mean ages at arrest of East German prisoners with chronic PTSD and remitted PTSD is between 0.2 and 7.2 years.

What does "pooling" refer to, in the context of a hypotheses test for two population means?

Since we cannot use the population standard deviation as a basis for calculating the test statistic in a hypothesis test for two population means (because ϑ is unknown), we use sample information to estimate ϑ, the unknown population standard deviation. We first have to estimate the unknown population variance, ϑ², and to do so, we take the two sample variances (s₁² and s₂²) and pool (combine into one standard deviation Sp) by weighting them according to sample size (actually degrees of freedom).

Do children diagnosed with ADHD have smaller brains than children without this condition? Brain scans were completed for 152 children with ADHD and 139 children of similar age without ADHD. Summary values for total cerebral volume (in milliliters) are given below:

With ADHD, n=152, x-bar = 1059.4, s = 117.5

Without ADHD, n=139, x-bar=1104.5, s = 111.3

Do these data provide evidence that the mean brain volume of children with ADHD is smaller than the mean for children without ADHD? Test the relevant hypotheses by using a 0.05 level of significance.

With ADHD, n=152, x-bar = 1059.4, s = 117.5

Without ADHD, n=139, x-bar=1104.5, s = 111.3

Do these data provide evidence that the mean brain volume of children with ADHD is smaller than the mean for children without ADHD? Test the relevant hypotheses by using a 0.05 level of significance.

(1) H₀: µ₁ = µ₂

H₁: µ₁ < µ₂

(2) ∝ = 0.05

(3) Test Statistic: t = -3.3527

(4) P-value: .0004516

(5) Conclusion: Reject H₀

(6) Interpretation: There is sufficient evidence to conclude that the mean brain volume of children with ADHD is smaller than the mean for children without ADHD.

H₁: µ₁ < µ₂

(2) ∝ = 0.05

(3) Test Statistic: t = -3.3527

(4) P-value: .0004516

(5) Conclusion: Reject H₀

(6) Interpretation: There is sufficient evidence to conclude that the mean brain volume of children with ADHD is smaller than the mean for children without ADHD.

An article investigated the driving behavior of teenagers by observing their vehicles as they left a high school parking lot and then again at a site approximately 1/2 mile from the school.

For this test, use a 0.01 significance level.

The following measurements represent the amount by which the speed limit was exceeded by male drivers and female drivers:

Male: 1.3, 1.3, 0.9, 2.1, 0.7, 1.3, 5.0, 1.3, 0.6, 2.1

Fem: -0.2, 0.5, 1.1, 0.7, 1.1, 1.2, 0.1, 0.9, 0.5, 0.5

Doe these data provide convincing evidence support for the claim that, on average, male teenage drivers exceed the speed limit by more than do female teenage drivers?

For this test, use a 0.01 significance level.

The following measurements represent the amount by which the speed limit was exceeded by male drivers and female drivers:

Male: 1.3, 1.3, 0.9, 2.1, 0.7, 1.3, 5.0, 1.3, 0.6, 2.1

Fem: -0.2, 0.5, 1.1, 0.7, 1.1, 1.2, 0.1, 0.9, 0.5, 0.5

Doe these data provide convincing evidence support for the claim that, on average, male teenage drivers exceed the speed limit by more than do female teenage drivers?

(1) H₀: µ₁ = µ₂

H₁: µ₁ > µ₂

(2) ∝ = 0.01

(3) Test Statistic: t = 2.374

(4) P-value: .0181

(5) Conclusion: Do not reject H₀

(6) Interpretation: There is not sufficient evidence to conclude that the male teenage drivers exceed the speed limit by more than do female drivers.

H₁: µ₁ > µ₂

(2) ∝ = 0.01

(3) Test Statistic: t = 2.374

(4) P-value: .0181

(5) Conclusion: Do not reject H₀

(6) Interpretation: There is not sufficient evidence to conclude that the male teenage drivers exceed the speed limit by more than do female drivers.

The logic behind the Wilcoxon Signed-Rank Test

The Wilcoxon signed-rank test is based on the assumption that the variable under consideration has a symmetric distribution - one that can be divided into two pieces that are mirror images of each other - but does not require that its distribution be normal or have any other specific shape.

two-tailed test

If the primary concern is deciding whether a population mean, µ, is different from a specified value ₀, we express the alternative hypothesis as:

H₁: µ ≠ µ₀

A hypothesis test whose alternative hypothesis has this form is called a two-tailed test.

H₁: µ ≠ µ₀

A hypothesis test whose alternative hypothesis has this form is called a two-tailed test.

left-tailed test

If the primary concern is deciding whether a population mean, µ, is less than a specified value µ₀, we express the alternative hypothesis as:

H₁: µ < µ₀

A hypothesis test whose alternative hypothesis has this form is called a left-tailed test.

H₁: µ < µ₀

A hypothesis test whose alternative hypothesis has this form is called a left-tailed test.

right-tailed test

If the primary concern is deciding whether a population mean, µ, is greater than a specified value µ₀, we express the alternative hypothesis as:

H₁: µ > µ₀

A hypothesis test whose alternative hypothesis has this form is called a right-tailed test.

H₁: µ > µ₀

A hypothesis test whose alternative hypothesis has this form is called a right-tailed test.

Basic logic of hypothesis testing

Take a random sample from the population. If the sample data are consistent with the null hypothesis, do not reject the null hypothesis; if the sample data are inconsistent with the null hypothesis (in the direction of the alternative hypothesis), reject the null hypothesis and conclude that the alternative hypothesis is true.

When to use the one-mean z-interval procedure

(a) For small samples - say, of size less than 15 - the z-interval procedure should be used only when the variable under consideration is normally distributed or very close to being so.

(b) For samples of moderate size - say, between 15 and 30 - the z-interval procedure can be used unless the data contain outliers or the variable under consideration is far from being normally distributed.

(c) For large samples - say, of size 30 or more - the z-interval procedure can be used essentially without restriction. However, if outliers are present and their removal is not justified, you should compare the confidence intervals obtained with and without the outliers to see what effect the outliers have. If the effect is substantial, use a different procedure or take another sample.

(d) If outliers are present but their removal is justified and results in a data set for which the z-interval procedure is appropriate (as previously stated), the procedure can be used.

(b) For samples of moderate size - say, between 15 and 30 - the z-interval procedure can be used unless the data contain outliers or the variable under consideration is far from being normally distributed.

(c) For large samples - say, of size 30 or more - the z-interval procedure can be used essentially without restriction. However, if outliers are present and their removal is not justified, you should compare the confidence intervals obtained with and without the outliers to see what effect the outliers have. If the effect is substantial, use a different procedure or take another sample.

(d) If outliers are present but their removal is justified and results in a data set for which the z-interval procedure is appropriate (as previously stated), the procedure can be used.

A fundamental principle of data analysis

Before performing a statistical-inference procedure, examine the sample data. If any of the conditions required for using the procedure appear to be violated, do not apply the procedure. Instead use a different, more appropriate procedure, or, if you are unsure of one, consult a statistician.

Relate confidence and precision for a fixed sample size

For a fixed sample size, decreasing the confidence level improves the precision, and vice versa.

Margin of error, precision, and sample size

The length of a confidence interval for a population mean, µ, and therefore the precision with which x-bar estimates µ, is determined by the margin of error, E. For a fixed confidence level, increasing the sample size improves the precision, and vice versa.

Basic properties of t-curves

(1) The total area under a t-curve equals 1.

(2) A t-curve extends indefinitely in both directions, approaching but never touching, the horizontal axis as it does so.

(3) A t-curve is symmetric about 0.

(4) As the number of degrees of freedom becomes larger, t-curves look increasingly like the standard normal curve.

(2) A t-curve extends indefinitely in both directions, approaching but never touching, the horizontal axis as it does so.

(3) A t-curve is symmetric about 0.

(4) As the number of degrees of freedom becomes larger, t-curves look increasingly like the standard normal curve.

Relation between Type I and Type II error probabilities

For a fixed sample size, the smaller we specify the significance level, ∝, the larger will be the probability, β, of not rejecting a false null hypothesis.

Obtaining critical values

Suppose that a hypothesis test is to be performed at the significance level, ∝. Then the critical value(s) must be chosen so that, if the null hypothesis is true, the probability is ∝ that the test statistic will fall in the rejection region.

When to use the one-mean z-test

(1) For small samples - say of size less than 15 - the z-test should be used only when the variable under consideration is normally distributed or very close to being so.

(2) For samples of moderate size - say, between 15 and 30 - the z-test can be used unless the data contain outliers or the variable under consideration is far from being normally distributed.

(3) For large samples - say, of size 30 or more - the z-test can be used essentially without restriction. However, if outliers are present and their removal is not justified, you should perform the hypothesis test once with the outliers and once without them to see what effect the outliers have. If the conclusion is affected, use a different procedure or take another sample.

(4) If outliers are present but their removal is justified and results in a data set for which the z-test is appropriate (as previously stated), the procedure can be used.

(2) For samples of moderate size - say, between 15 and 30 - the z-test can be used unless the data contain outliers or the variable under consideration is far from being normally distributed.

(3) For large samples - say, of size 30 or more - the z-test can be used essentially without restriction. However, if outliers are present and their removal is not justified, you should perform the hypothesis test once with the outliers and once without them to see what effect the outliers have. If the conclusion is affected, use a different procedure or take another sample.

(4) If outliers are present but their removal is justified and results in a data set for which the z-test is appropriate (as previously stated), the procedure can be used.

Decision criterion for a hypothesis test using the p-value

If the P-value is less than or equal to the specified significance level, reject the null hypothesis; otherwise, do not reject the null hypothesis.

Wilcoxon Signed-Rank Test versus the t-test

Suppose that you want to perform a hypothesis test for a population mean. When deciding between the t-test and the Wilcoxon signed-rank test, follow these guidelines:

(1) If you are reasonably sure that the variable under consideration is normally distributed, use the t-test.

(2) If you are not reasonably sure that the variable under consideration is normally distributed but are reasonably sure that it has a symmetric distribution, use the Wilcoxon signed-rank test.

(1) If you are reasonably sure that the variable under consideration is normally distributed, use the t-test.

(2) If you are not reasonably sure that the variable under consideration is normally distributed but are reasonably sure that it has a symmetric distribution, use the Wilcoxon signed-rank test.

Choosing between a pooled and nonpooled t-procedure

Suppose you want to use independent simple random samples to compare the means of two populations. To decide between a pooled t-procedure and a nonpooled t-procedure, follow these guidelines: If you are reasonably sure that the populations have nearly equal standard deviations, use a pooled t-procedure; otherwise, use a nonpooled t-procedure.