Customer participation in store loyalty card programs. Customers who participate in a store's free loyalty card program save money on their purchases, but allow the store to keep track of the customer's shopping habits and potentially sell these data to third parties. A Pew Internet & American Life Project Survey (January 2016) revealed that 225 of a random sample of 250 U.S. adults would agree to participate in a store loyalty card program, despite the potential for information sharing. Let $p$ represent the true proportion of all customers who would participate in a store loyalty card program. Consider a store owner who claims that more than $80 \%$ of all customers would participate in a loyalty card program. Set up the null and alternative hypotheses for testing whether the true proportion of all customers who would participate in a store loyalty card program exceeds $.8$.

Customer participation in store loyalty card programs. Customers who participate in a store's free loyalty card program save money on their purchases, but allow the store to keep track of the customer's shopping habits and potentially sell these data to third parties. A Pew Internet & American Life Project Survey (January 2016) revealed that 225 of a random sample of 250 U.S. adults would agree to participate in a store loyalty card program, despite the potential for information sharing. Let $p$ represent the true proportion of all customers who would participate in a store loyalty card program. Compute a point estimate of $p$.

A statistics student used a computer program to test the null hypothesis $H_0: p=.5$ against the one-tailed alternative, $H_{\mathrm{a}}: p>$.5. A sample of 500 observations are input into SPSS, which returns the following results: $z=.44$, two-tailed $p$-value $=.33$. The student concludes, based on the $p$-value, that there is a $33 \%$ chance that the alternative hypothesis is true. Do you agree? If not, correct the interpretation.

Name four sources of information to use in collecting secondary data.

The yield of a chemical process is being studied. From previous experience with this process the standard deviation of yield is known to be 3. The past 5 days of plant operation have resulted in the following yields: 91.6, 88.75, 90.8, 89.95, and 91.3%. Use $\alpha = 0.05.$ (a) Is there evidence that the mean yield is not 90%? Use the P-value approach. (b) What sample size would be required to detect a true mean yield of 85% with probability 0.95? (c) What is the type II error probability if the true mean yield is 92%? (d) Find a 95% two-sided CI on the true mean yield. (e) Use the CI found in part (d) to test the hypothesis.

Paolo can walk three dogs or mow two lawns in two hours. Ashanti can walk six dogs or mow three lawns in two hours. Ashanti's opportunity cost for each additional dog walked is:

a. 2.5 lawns mowed.

b. 0.5 lawn mowed.

c. 1.5 lawns mowed.

d. 2.0 lawns mowed.

A random sample of n measurements was selected from a population with unknown mean $\mu$ and standard deviation $\sigma = 20$. Calculate a 95% confidence interval for $\mu$ for each of the following situations:

a. $n=75$, $\overline{x}=28$

b. $n=200$, $\overline{x}=102$

c. $n=100$, $\overline{x}=15$

d. $n=100$, $\overline{x}=4.05$

e. Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts a–d? Explain.

Consider the Minitab output shown on the right. (a) Fill in the missing values in the output, including a bound on the P-value. Can the null hypothesis be rejected at the 0.05 level? Why? (b) Is this a one-sided or a two-sided test? (c) Use the output and the t table to find a 99% two-sided CI on the difference in means. (d) How can you tell that there is insufficient evidence to reject the null hypothesis just by looking at the computer output and not making any additional computations?

$$\begin{matrix} \text{Paired T-Test and CI:X1, X2}\\ \text{Paired T for X1 - X2}\\ \quad & \text{N} & \text{Mean} & \text{StDev} & \text{SE Mean}\\ \text{X1} & \text{12} & \text{74.2430} & \text{1.8603} & \text{0.5370}\\ \text{X2} & \text{12} & \text{73.4797} & \text{1.9040} & \text{0.5496}\\ \text{Difference} & \text{12} & \text{?} & \text{2.905560} & \text{0.838763}\\ 95\% \text{CI for mean difference: (-1.082805, 2.609404)}\\ \text{T-Test of mean difference = 0 (vs not = 0):}\\ \text{T-Value = ?, P-Value = ?}\\ \end{matrix}$$

Consider the Minitab output shown below.

$$\begin{matrix} \text{\text{One-Sample T: X}}\\ \text{\text{ Test of mu = 91 vs not = 91}}\\ \text{Variable} & \text{N} & \text{Mean} & \text{StDev} & \text{SE Mean} & 95\% CI & \text{T} & \text{P}\\ \text{X} & \text{25} & \text{92.5805} & \text{?} & \text{0.4673} & \text{(91.6160, ?)} & \text{3.38} & \text{0.002}\\ \end{matrix}$$

(a) Fill in the missing values in the output. Can the null hypothesis be rejected at the 0.05 level? Why? (b) Is this a one-sided or a two-sided test? (c) If the hypotheses had been $H_{0} : \mu=90 \text { versus } H_{1} : \mu \neq 90$ would you reject the null hypothesis at the 0.05 level? (d) Use the output and the t-table to find a 99% two-sided CI on the mean. (e) What is the P-value if the alternative hypothesis is $H_{1} : \mu>91 ?$

A consumer products company is formulating a new shampoo and is interested in foam height (in millimeters). Foam height is approximately normally distributed and has a standard deviation of 20 millimeters. The company wishes to test H$_0$ : μ = 175 millimeters versus H$_1$: μ > 175 millimeters, using the results of n = 10 samples. Find the type I error probability α if the critical region is $\overline{x}$ > 185. b. What is the probability of type II error if the true mean foam height is 185 millimeters? c. Find ß for the true mean of 195 millimeters.

A random sample of n observations is selected from a normal population to test the null hypothesis that

$$\mu = 10$$

Specify the rejection region for each of the following combinations of

$$H _ { a } , \alpha , \text { and } n$$

$$H _ { a } : \mu \neq 10 ; \alpha = .05 ; n = 14$$

The Food and Drug Administration (FDA) produces a quarterly report called Total Diet Study . The FDA’s report covers more than 200 food items, each of which is analyzed for dangerous chemical compounds. One Total Diet Study reported that no pesticides at all were found in 65% of the domestically produced food samples. Consider a random sample of 800 food items analyzed for the presence of pesticides. a. Compute

$$\mu \text { and } \sigma$$

for the random variable x , the number of food items found without any trace of pesticide. b. On the basis of a sample of 800 food items, is it likely that you would observe fewer than half without any traces of pesticide? Explain.

The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of 1/8 inch. The width of a door is normally distributed with a mean of 23 7/8 inches and a standard deviation of 1/16 inch. Assume independence. a. Determine the mean and standard deviation of the difference between the width of the casing and the width of the door. b. What is the probability that the width of the casing minus the width of the door exceeds 1/4 inch? c. What is the probability that the door does not fit in the casing?

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. You wish to test H$_0$: μ = 100 versus H$_1$: μ ≠ 100 with a sample of n = 9 specimens.
a. If the acceptance region is defined as 98.5 ≤ $\overline{x}$ ≤ 101.5, find the type I error probability α.
b. Find β for the case in which the true mean heat evolved is 103.
c. Find β for the case where the true mean heat evolved is 105. This value of β is smaller than the one found in part (b). Why?