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Non-Credit Test 4 Review

Terms in this set (72)

Students in a discussion of gun control in a sociology class at Foothill Community College argue that Republicans are more likely to oppose gun control than Independents. They use data from an article titled "Gun Control Splits America," published March 23, 2010 in pewresarch.org by the Pew Research Center for the People and the Press. In this study 62% of Repubterm-69licans and 57% of Independents say that states should not be able to pass laws banning handguns.
For a claim that a larger proportion of Republicans oppose state laws banning handguns when compared to Independents, the null and alternative hypotheses are


The p-value is 0.06. If we conduct this test at a 5% level of significance, what would be an appropriate conclusion?
A. Reject H0 , and support Ha .
B. Support H0 , and reject Ha .
C. Fail to Reject H0 , do not support Ha .
C. Fail to Reject H0 , do not support Ha .
Watches and bacteria: A group of researchers investigated the contamination of medical personnel watches at a New York hospital, since there is a potential for patient exposure to potentially dangerous bacteria.
They sampled watches worn by physicians, physician assistants, and medical students at a teaching hospital in New York. Nearly half (47.3%) of the watches tested harbored microorganisms that can cause illness. By comparison, only one of the 10 watches worn by security guards tested positive for a disease-carrying microorganism. The researchers want to determine if the difference is statistically significant.
Which of the following is an appropriate statement of the null hypothesis, H0?

A. The proportion of contaminated wrist-watches from medical personnel is the same as the proportion of contaminated wrist-watches from security guards, i.e., H0: p = 47.3%.
B. The proportion of contaminated wrist-watches from medical personnel is not the same as the proportion of contaminated wrist-watches from security guards, i.e., H0: p not equal 47.3%.
C. The proportion of contaminated wrist-watches from medical personnel is greater than the proportion of contaminated wrist-watches from security guards, i.e., H0: p > 47.3%.
A. The proportion of contaminated wrist-watches from medical personnel is the same as the proportion of contaminated wrist-watches from security guards, i.e., H0: p = 47.3%.
The makers of Mini-Oats cereal have an automated packaging machine that is set to fill boxes with 24.1 ounces of cereal (as labeled on the box). At various times in the packaging process, we select a random sample of 100 boxes to see if the machine is (on average) filling the boxes as labeled.
On Tuesday morning, at 7:45 a.m., a random sample of 100 boxes produced an average amount of 23.8 ounces.
Which of the following is an appropriate statement of the null hypothesis?

A. The machine fills the boxes with the proper amount of cereal. The average is 24.1 ounces (H0: μ = 24.1)
B. The machine is not filling the boxes with the proper amount of cereal (H0: μ ≠ 24.1 ounces).
C. The machine is not putting enough cereal in the boxes. The average is less than 24.1 ounces (H0: μ < 24.1).
D. The machine fills the boxes with an average of 23.8 ounces (H0: μ = 23.8).
A. The machine fills the boxes with the proper amount of cereal. The average is 24.1 ounces (H0: μ = 24.1)
In 2010 polls indicated that 73% of Americans favored mandatory testing of students in public schools as a way to rate the school. This year in a poll of 1,000 Americans 71% favor mandatory testing for this purpose. Has public opinion changed since 2010?
We test the hypothesis that the percentage supporting mandatory testing is less than 73% this year. The p-value is 0.015.
Which of the following interpretation of this p-value is valid?

A. The probability that Americans have changed their opinion on this issue since 2010 is 0.015.
B. There is a 1.5% chance that the null hypothesis is true.
C. If 73% of Americans still favor mandatory testing this year, then there is a 1.5% chance that poll results will show 71% or fewer with this opinion.
C. If 73% of Americans still favor mandatory testing this year, then there is a 1.5% chance that poll results will show 71% or fewer with this opinion.
Genetically modified foods: According to a 2016 Pew Research survey, a majority of the American general public (57%) says that genetically modified (GM) foods are generally unsafe to eat. This month, in a survey of 500 randomly selected American adults, 59% says that GM foods are generally unsafe to eat. We test the hypothesis that the percentage who says that GM foods are generally unsafe to eat is greater than 57% this year. The p-value is 0.085.
Which of the following interpretations of this p-value is valid?

A. There is an 8.5% chance that 57% of Americans says that GM foods are generally unsafe to eat.
B. Assuming that more than 57% of Americans feel that GM foods are unsafe, the probability that 59% of 500 randomly selected Americans say that GM foods are generally unsafe to eat is 0.085.
C. If we assume that 57% of Americans says that GM foods are generally unsafe to eat, then there is an 8.5% chance that random sample results will show 59% or more who says that GM foods are generally unsafe to eat.
C. If we assume that 57% of Americans says that GM foods are generally unsafe to eat, then there is an 8.5% chance that random sample results will show 59% or more who says that GM foods are generally unsafe to eat.
Living with parents: The Pew Research Center reported that 36% of American Millennials (adults ages 18-31) still live at home with their parents. A group of students wants to conduct a study to determine whether this result is true for students at their campus. They survey 300 randomly selected students at their campus and determine that 43% of them live at home with their parents. With this data, they test the following hypotheses.
H0: Of Millennial students at their campus, 36% live at home with their parents.
Ha: More than 36% of Millennial students at their campus live at home with their parents.
In order to assess the evidence, which question best describes what we need to determine?

A. If we examine the proportion of students at their campus who still live at home with their parents, how likely is that proportion to be 36%?
B. If we examine the proportion of students at their campus who still live at home with their parents, how likely is that proportion to be more than 36%?
C. If we examine a sample of students at their campus and determine the proportion who still live at home with their parents, how likely is that proportion to be 43% or more?
C. If we examine a sample of students at their campus and determine the proportion who still live at home with their parents, how likely is that proportion to be 36%?
D. If we examine the proportion of students at their campus who still live at home with their parents, how likely is that proportion to be 43%?
C. If we examine a sample of students at their campus and determine the proportion who still live at home with their parents, how likely is that proportion to be 43% or more?
Food inspectors inspect samples of food products to see if they are safe. This can be thought of as a hypothesis test with the following hypotheses.
H0: The food is safe.
Ha: The food is not safe
Based on the hypotheses above, Is the following statement a Type I or Type II error?
The sample suggests that the food is safe, but it actually is not safe.

A. Type I
B. Type II
B. Type II
Food inspectors inspect samples of food products to see if they are safe. This can be thought of as a hypothesis test with the following hypotheses.
H0: The food is safe.
Ha: The food is not safe.
Based on the hypotheses above, is the following statement a Type I or Type II error?
The sample suggests that the food is not safe, but it actually is safe.

A. Type I
B. Type II
A. Type I
Pregnancy testing: A college student hasn't been feeling well and visits her campus health center. Based on her symptoms, the doctor suspects that she is pregnant and orders a pregnancy test. The results of this test could be considered a hypothesis test with the following hypotheses:

H0: The student is not pregnant
Ha: The student is pregnant.
Based on the hypotheses above, which of the following statements is considered aType I error?

A. The student is not pregnant, but the test result shows she is pregnant.
B. The student is pregnant, but the test result shows she is not pregnant.
C. The student is pregnant, and the test result shows she is pregnant.
D. The student is not pregnant, and the test result shows she is not pregnant.
A. The student is not pregnant, but the test result shows she is pregnant.
Pregnancy testing: A college student hasn't been feeling well and visits her campus health center. Based on her symptoms, the doctor suspects that she is pregnant and orders a pregnancy test. The results of this test could be considered a hypothesis test with the following hypotheses:
H0: The student is not pregnant
Ha: The student is pregnant.
Based on the hypotheses above, which of the following statements is considered a Type II error?

A. The student is not pregnant, but the test result shows she is pregnant.
B. The student is pregnant, but the test result shows she is not pregnant.
C. The student is not pregnant, and the test result shows she is not pregnant.
D. The student is pregnant, and the test result shows she is pregnant.
B. The student is pregnant, but the test result shows she is not pregnant.
A tire manufacturer has a 60,000 mile warranty for tread life. The manufacturer considers the overall tire quality to be acceptable if less than 5% are worn out at 60,000 miles. The manufacturer tests 250 tires that have been used for 60,000 miles. They find that 9 of them are worn out. With this data, we test the following hypotheses at the 5% significance level. The p-value is 0.15.
H0: The proportion of tires that are worn out after 60,000 miles is equal to 0.05.
Ha: The proportion of tires that are worn out after 60,000 miles is less than 0.05.
Which of the following conclusions is correct?

A. Accept H0
B. Fail to reject H0
C. Reject H0
B. Fail to reject H0
We conduct a study to determine whether the majority of community college students plan to vote in the next presidential election. We choose a significance level of 0.05. We survey 650 randomly selected community college students and find that 54% of them plan to vote. The p-value is 0.02.
H0: 50% of community college students plan to vote in the next presidential election.
Ha: More than 50% of community college students plan to vote in the next presidential election.
What can we conclude?

A. Nothing. The sample size is too small to represent all community college students.
B. The evidence suggests that the majority of community colleges students plan to vote in the next presidential election because 54% is greater than 50%.
C. The evidence suggests that the majority of community colleges students plan to vote in the next presidential election because the p-value is less than the significance level.
D. The evidence does not suggest that the majority of community colleges students plan to vote in the next presidential election because the difference between 50% and 54% is not statistically significant. A 4% difference could be due to random chance.
C. The evidence suggests that the majority of community colleges students plan to vote in the next presidential election because the p-value is less than the significance level.
Living with parents: A Pew Research analysis stated that in 2012, 36% of the nation's young adults ages 18 to 31—the so-called Millennial generation—were living in their parents' home. After reading the analysis, a statistics student wanted to design a study to determine if the percentage was higher for the Millennial students who attended his college. Before collecting data, he set the significance level for his test to be 0.05.
For which p-value would he reject the null hypothesis?

A. 0.07
B. 0.02
C. The magnitude of a p-value has no impact on whether he rejects or fails to reject the null hypothesis.
B. 0.02
Living with parents: The Pew Research Center reported that 36% of American Millennials (adults ages 18-31) still live at home with their parents. A group of students wants to conduct a study to determine whether this result is true for students at their campus. They survey 300 randomly selected students at their campus and determine that 43% of them still live at home with their parents. With this data, they test the following hypotheses at the 5% significance level. The p-value is 0.006.
H0: Of Millennial students at their campus, 36% live at home with their parents.
Ha: More than 36% of Millennial students at their campus live at home with their parents.
Which of the following conclusions is correct?

A. Accept H0
B. Fail to reject H0
C. Reject H0
C. Reject H0
Living with parents: The Pew Research Center reported that 36% of American Millennials (adults ages 18-31) still live at home with their parents. A group of students wants to conduct a study to determine whether this result is true for students at their campus. They survey 300 randomly selected students at their campus and determine that 43% of them still live at home with their parents. With this data, they test the following hypotheses at the 5% significance level. The p-value is 0.006.
H0: Of Millennial students at their campus, 36% live at home with their parents.
Ha: More than 36% of Millennial students at their campus live at home with their parents.
What can we conclude?

A. Nothing. The sample size is too small to represent students at their campus.
B. The evidence suggests that more than 36% of students at their campus live at home with their parents because 43% is greater than 36%.
C. The evidence suggests that more than 36% of students at their campus live at home with their parents because the p-value is less than the significance level.
D. The evidence does not suggest that more than 36% of students at their campus live at home with their parents because the difference between 43% and 36% is not statistically significant. A 7% difference could be due to random chance.
C. The evidence suggests that more than 36% of students at their campus live at home with their parents because the p-value is less than the significance level.
In 2016, the Centers for Disease Control (CDC) reported that the teen birth rate in Georgia was 24.3%. The Atlanta City Council orders a study to see if the teen birth rate within the city of Atlanta is lower.

Suppose that a study of 1000 female teenagers in Atlanta found that 234 of them gave birth in 2016.
Given that the population proportion is understood to be p = 0.243, what would be the null hypothesis, H0, for this study?

A. H0: p = 24.3%
B. H0: p < 24.3%
C. H0: p = 23.4%
D. Ha: p = 24.3%
D. Ha: p = 24.3%
According to the National Institute on Drug Abuse, a U.S. government agency, 17.3% of 8th graders in 2010 had used marijuana at some point in their lives. A school official hopes to show the percentage is lower in his district, testing H0: p = 0.173 versus Ha: p < 0.173.
The health department for the district uses anonymous random sampling and finds that 10% of 80 eighth graders surveyed had used marijuana.
Is the sample size condition for conducting a hypothesis test for a population proportion satisfied?

A. No, because 80 students is not enough to be representative of the students in the school district.
B. Yes, because (80)(.173) and (80)(1 ‑ 0.173) are both at least 10. This means we can use the normal distribution to model the distribution of sample proportions.
C. No, because (80)(.10) is less than 10. Because of this the sample size is too small to use the normal distribution to model the distribution of sample proportions.
D. Yes, because the sample is random. It represents the 8th graders in the district.
B. Yes, because (80)(.173) and (80)(1 ‑ 0.173) are both at least 10. This means we can use the normal distribution to model the distribution of sample proportions.
College students and STDs: A recent report estimated that 25% of all college students in the United States have a sexually transmitted disease (STD). Due to the demographics of the community, the director of the campus health center believes that the proportion of students who have a STD is lower at his college.
He tests H0: p = 0.25 versus Ha: p < 0.25. The campus health center staff select a random sample of 50 students and determine that 18% have been diagnosed with a STD.
Is the sample size condition for conducting a hypothesis test for a population proportion satisfied?

A. Yes, because the sample is random. It represents the college students at this campus.
B. No, because (50)(0.18) is less than 10. Because of this the sample size is too small to use the normal distribution to model the distribution of sample proportions.
C. No, because 50 students is not enough to be representative of the students at the college
D. Yes, because (50)(.25) and (50)(1 ‑ 0.25) are both at least 10. This means we can use the normal distribution to model the distribution of sample proportions.
D. Yes, because (50)(.25) and (50)(1 ‑ 0.25) are both at least 10. This means we can use the normal distribution to model the distribution of sample proportions.
A 2014 study by the reputable Gallup organization estimates that 44% of U.S. adults are underemployed. Underemployed means the person wants to work full time but is employed part time or unemployed. We want to know if the proportion is smaller this year. We select a random sample of 100 U.S. adults this year and find that 40% are underemployed.
After carrying out the hypothesis test for p = 0.44 compared to p < 0.44, we obtain a P‑value of 0.21. Which of the following interpretations of the P‑value is correct?

A. There is a 21% chance that 40% or fewer U.S. adults are underemployed this year if 44% were underemployed in 2014.
B. There is a 21% chance that 44% of U.S. adults are underemployed this year
C. There is a 21% chance that 40% of U.S. adults are underemployed this year.
D. There is a 21% chance that a sample of 100 U.S. adults will have 40% or fewer underemployed if 44% of the population is underemployed this year.
D. There is a 21% chance that a sample of 100 U.S. adults will have 40% or fewer underemployed if 44% of the population is underemployed this year.
Coffee: A popular chain of cafes has been receiving online complaints about one store location. Regular customers complained that the staff at this location consistently underfill their cups of coffee. The owner of the chain will personally visit the location to meet with the staff to determine if more than 10% of the 16 ounce cups are underfilled by 1 fluid ounce. She arranges to have "mystery shoppers" visit the store. Mystery shoppers are people who are paid to pose as a regular customer and provide feedback to the owner about customer service.
After 100 mystery shop visits, she finds that 18 of the 16 ounce cups of coffee were underfilled. She plans to test the hypotheses H0:p=0.10 versus Ha:p>0.10.
What is the test statistic?

A. z = -2.08
B. z = 2.67
C. z = -2.67
D. z = 2.08
B. z = 2.67
A quality control engineer at a potato chip company tests the bag filling machine by weighing bags of potato chips. Not every bag contains exactly the same weight. But if more than 16% of bags are over-filled then they stop production to fix the machine. They define over-filled to be more than 1 ounce above the weight on the package.

The engineer weighs 139 bags and finds that 31 of them are over-filled. He plans to test the hypotheses H0: p = 0.16 versus Ha: p > 0.16. What is the test statistic?
Z= 2.03
P= 0.16
N= 139
^P=31/139=
Z=^P-P/sqrt p(1-P)/N
Does secondhand smoke increase the risk of a low weight birth? A baby is "low birth weight" if it weighs less than 5.5 pounds at birth. According to the National Center of Health Statistics, about 7.8% of all babies born in the U.S. are categorized as low birth weight. Researchers randomly select 1200 babies whose mothers had extensive exposure to secondhand smoke during pregnancy. 10.4% of the sample are categorized as low birth weight.
Which of the following are the appropriate null and alternative hypotheses for this research question.

A. H0: p = 0.078; Ha: p ≠ 0.078
B. H0: p = 0.078; Ha: p > 0.078
C. H0: p = 0.104; Ha: p ≠ 0.104
D..H0: μ = 0.104; Ha: μ > 0.104
B. H0: p = 0.078; Ha: p > 0.078
Short-term classes: Does taking a class in a short-term format (8 weeks instead of 16 weeks) increase a student's likelihood of passing the course? For a particular course, the pass rate for the 16-week format is 59%. A team of faculty examine student data from 40 randomly selected accelerated classes and determine that the pass rate is 78%.
Which of the following are the appropriate null and alternative hypotheses for this research question?
A. H0: p = 0.59; Ha: p ≠ 0.59
B. H0: p = 0.59; Ha: p > 0.59
C. H0: p = 0.78; Ha: p ≠ 0.78
D. H0: p = 0.78; Ha: p > 0.59
B. H0: p = 0.59; Ha: p > 0.59
One population proportion test: Which of the following situations involves testing a claim about a single population proportion?

A. A recent report estimated that 25% of all college students in the United States have a sexually transmitted disease (STD). The director of the campus health center believes that the proportion of students with STDs is higher at their campus.

B. According to the American Heart Association guidelines, the daily maximum amount of sugar a teenager should have is 4 teaspoons. The parents of a local high school have expressed concerns that the average daily sugar intake of their students is higher than these guidelines and have been pressuring school officials to limit unhealthy snack and beverage choices offered in the cafeteria and student store.

C. Duke University researchers discovered that not only do women need more sleep than men but also that women tend to suffer more than men, both mentally and physically, if they are sleep deprived. A statistics class wanted to investigate the claims of this study and collected data to determine if women at their college sleep more, on average, than men.

D. The National Center for Health Statistics issued a report that stated that women in every age group in the United States were more likely than men to have serious mental health problems. A group of researchers want to design a study to further examine these results.
A. A recent report estimated that 25% of all college students in the United States have a sexually transmitted disease (STD). The director of the campus health center believes that the proportion of students with STDs is higher at their campus.
According to a Pew Research Center study, in May 2011, 41% of all American adults had a smart phone (one which the user can use to read email and surf the Internet). A communications professor at a university believes this percentage is higher among community college students. She selects 470 community college students at random and finds that 203 of them have a smart phone. Then in testing the hypotheses:
H0: p = 0.41 versus
Ha: p > 0.41,
what is the test statistic?
Z= 0.97
P= 0.41
N= 470
X=203
^P= X/N=
Z=^P-P/sqrt p(1-P)/N
According to a Pew Research Center study, in May 2011, 33% of all American adults had a smart phone (one which the user can use to read email and surf the Internet). A communications professor at a university believes this percentage is higher among community college students. She selects 349 community college students at random and finds that 138 of them have a smart phone. In testing the hypotheses:
H0: p = 0.33 versus
Ha: p > 0.33,
she calculates the test statistic as z = 2.5990.

Then the p‑value =
Then the p‑value = 0.0047

=1-NORM.S.DIST(2.5990,1)
According to the Pew Research Center, the proportion of the American population who use only a cellular telephone (no landline) is 37%. Jason claims that the proportion of young American adults who do not have a landline is greater than 37%. He conducts a survey with a sample of randomly selected young American adults and finds that 38% do not have landlines.
If we set up our null and alternative hypotheses as follows:
H0:p=0.37
Ha:p>0.37
and find that: "p-value"=0.418. Does this provide enough evidence to support Jason's claim? Use an α=0.05 level of significance.
Choose the correct answer below.

A. Since the p-value < α, do not reject the null hypothesis.
B. Since the p-value > α, do not reject the null hypothesis.
C. Since the p-value < α, reject the null hypothesis.
D. Since the p-value > α, reject the null hypothesis.
B. Since the p-value > α, do not reject the null hypothesis.
In a fictional study, suppose that a psychologist is studying the effect of daily meditation on resting heart rate. The psychologist believes the patients who not meditate have a higher resting heart rate. For a random sample of 45 pairs of identical twins, the psychologist randomly assigns one twin to one of two treatments. One twin in each pair meditates daily for one week, while the other twin does not meditate. At the end of the week, the psychologist measures the resting heart rate of each twin. Assume the mean resting heart rate is 80 heart beats per minute.
The psychologist conducts a T-test for the mean of the differences in resting heart rate of patients who do not meditate minus resting heart rate of patients who do meditate.
Which of the following is the correct null and alternative hypothesis for the psychologist's study?

A. H0: µ = 80; Ha: µ > 80
B. H0: µ = 0; Ha: µ ≠ 0
C. H0: µ = 0; Ha: µ > 0
C. H0: µ = 0; Ha: µ > 0
According to a 2014 research study of national student engagement in the U.S., the average college student spends 17 hours per week studying. A professor believes that students at her college study less than 17 hours per week. The professor distributes a survey to a random sample of 80 students enrolled at the college.
From her survey data the professor calculates that the mean number of hours per week spent studying for her sample is: ¯x= 15.6 hours per week with a standard deviation of s = 4.5 hours per week.
The professor chooses a 5% level of significance. What can she conclude from her data?

A. The data supports the professor's claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.
B. The professor cannot conclude that the average number of hours per week spent studying for students at her college is less than 17 hours per week. The sample mean of 15.6 is not significantly less than 17.
C. Nothing. The conditions for use of a t-model are not met. The professor cannot trust that the p-value is accurate for this reason.
A. The data supports the professor's claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.
A teacher is experimenting with computer-based instruction. In which situation could the teacher use a hypothesis test for a population mean?

A. She gives each student a pretest. Then she teaches a lesson using a computer program. Afterwards, she gives each student a posttest. The teacher wants to see if the difference in scores will show an improvement.
B. She randomly divides the class into two groups. One group receives computer-based instruction. The other group receives traditional instruction without computers. After instruction, each student has to solve a single problem. The teachers wants to compare the proportion of each group who can solve the problem.
C. The teacher uses a combination of traditional methods and computer-based instruction. She asks students if they liked computer-based instruction. She wants to determine if the majority prefer the computer-based instruction.
A. She gives each student a pretest. Then she teaches a lesson using a computer program. Afterwards, she gives each student a posttest. The teacher wants to see if the difference in scores will show an improvement.
The Food and Drug Administration (FDA) is a U.S. government agency that regulates (you guessed it) food and drugs for consumer safety. One thing the FDA regulates is the allowable insect parts in various foods. You may be surprised to know that much of the processed food we eat contains insect parts. An example is flour. When wheat is ground into flour, insects that were in the wheat are ground up as well.
The mean number of insect parts allowed in 100 grams (about 3 ounces) of wheat flour is 75. If the FDA finds more than this number, they conduct further tests to determine if the flour is too contaminated by insect parts to be fit for human consumption.
The null hypothesis is that the mean number of insect parts per 100 grams is 75. The alternative hypothesis is that the mean number of insect parts per 100 grams is greater than 75.
Is the following a Type I error or a Type II error or neither?
The test fails to show that the mean number of insect parts is greater than 75 per 100 grams when it is.

A. Type I error
B. Type II error
C. Neither
B. Type II error
Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken by CHDS is the age of first time expectant mothers. Suppose that CHDS finds the average age for a first time mother is 26 years old. Suppose also that, in 2015, a random sample of 50 expectant mothers have mean age of 26.5 years old, with a standard deviation of 1.9 years. At the 5% significance level, we conduct a one-sided T-test to see if the mean age in 2015 is significantly greater than 26 years old. Statistical software tells us that the p-value = 0.034.
Which of the following is the most appropriate conclusion?

A. There is a 3.4% chance that a random sample of 50 expectant mothers will have a mean age of 26.5 years old or greater if the mean age for a first time mother is 26 years old.
B. There is a 3.4% chance that mean age for all expectant mothers is 26 years old in 2015.
C. There is a 3.4% chance that mean age for all expectant mothers is 26.5 years old in 2015.
D. There is 3.4% chance that the population of expectant mothers will have a mean age of 26.5 years old or greater in 2015 if the mean age for all expectant mothers was 26 years old in 1959.
A. There is a 3.4% chance that a random sample of 50 expectant mothers will have a mean age of 26.5 years old or greater if the mean age for a first time mother is 26 years old.
A researcher conducts an experiment on human memory and recruits 15 people to participate in her study. She performs the experiment and analyzes the results. She uses a t-test for a mean and obtains a p-value of 0.17.
Which of the following is a reasonable interpretation of her results?

A. This suggests that her experimental treatment has no effect on memory.
B. If there is a treatment effect, the sample size was too small to detect it.
C. She should reject the null hypothesis.
D. There is evidence of a small effect on memory by her experimental treatment.
B. If there is a treatment effect, the sample size was too small to detect it.
A group of 53 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic drinks they have in a typical week. The purpose of this study was to compare the drinking habits of the students at the college to the drinking habits of college students in general. In particular, the dean of students, who initiated this study, would like to check whether the mean number of alcoholic drinks that students at his college in a typical week differs from the mean of U.S. college students in general, which is estimated to be 4.73.
The group of 53 students in the study reported an average of 4.90 drinks per with a standard deviation of 3.74 drinks.
Find the p-value for the
hypothesis test. =0.7420

TWO-TAILED
N= 53
X=4.90
S=3.74
M=4.73
T*= X-M/S/SQRT(N)
P-VALUE =T.DIST.2T(T*,N-1)
Commute times in the U.S. are heavily skewed to the right. We select a random sample of 490 people from the 2000 U.S. Census who reported a non-zero commute time.
In this sample, the mean commute time is 28.3 minutes with a standard deviation of 18.7 minutes. Can we conclude from this data that the mean commute time in the U.S. is less than half an hour? Conduct a hypothesis test at the 5% level of significance.
What is the p-value for this hypothesis test?
What is the p-value for this hypothesis test?
=0.0224

LEFT TAILED
N= 490
X=28.5
S=18.7
M=30
T*= X-M/S/SQRT(N
P-VALUE =T.DIST(T*,60-1, 1)
Dean Halverson recently read that full-time college students study 20 hours each week. She decides to do a study at her university to see if there is evidence that students study an average of more than 20 hours each week. A random sample of 32 students were asked to keep a diary of their activities over a period of several weeks. It was found that the average number of hours that the 32 students studied each week was 20.9 hours. The sample standard deviation of 4.2 hours.
Find the p-value.
Find the p-value.
0.1173

RIGHT TAILED
N= 32
X=20.9
S=4.2
M=20
T*= X-M/S/SQRT(N
P-VALUE =T.DIST.RT(T*,N-1)
A medical researcher is studying the effects of a drug on blood pressure. Subjects in the study have their blood pressure taken at the beginning of the study. After being on the medication for 4 weeks, their blood pressure is taken again. The change in blood pressure is recorded and used in doing the hypothesis test.
Change: Final Blood Pressure - Initial Blood Pressure
The researcher wants to know if there is evidence that the drug reduces blood pressure. At the end of 4 weeks, 31 subjects in the study had an average change in blood pressure of -2.7 with a standard deviation of 5.3.
Find the p-value for the hypothesis test.
Find the p-value for the hypothesis test.
=0.004

RIGHT TAILED
N= 31
X=-2.7
S=5.3
M=0
T*= X-M/S/SQRT(N
P-VALUE =T.DIST.RT(T*,N-1)
Find the p-value for the hypothesis test. A random sample of size 51 is taken. The sample has a mean of 423 and a standard deviation of 84.
H0: µ = 400
Ha: µ > 400
The p-value for the hypothesis test is
The p-value for the hypothesis test is
=0.0281
RIGHT TAILED
N= 51
X=423
S=84
M=400
T*= X-M/S/SQRT(N
P-VALUE =T.DIST.RT(T*,N-1)
Find the p-value for the hypothesis test. A random sample of size 51 is taken. The sample has a mean of 423 and a standard deviation of 84.
H0: µ = 400
Ha: µ ≠ 400
The p-value for the hypothesis test is
The p-value for the hypothesis test is
=0.0561
TWO-TAILED
N= 51
X=423
S=84
M=400
T*= X-M/S/SQRT(N)
P-VALUE =T.DIST.2T(T*,N-1)
Find the p-value for the hypothesis test. A random sample of size 51 is taken. The sample has a mean of 392 and a standard deviation of 83.
H0: µ = 400
Ha: µ< 400
The p-value for the hypothesis test is
The p-value for the hypothesis test is
=0.2472
LEFT TAILED
N= 51
X=392
S=83
M=400
T*= X-M/S/SQRT(N
P-VALUE =T.DIST(T*,N-1, 1)
Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken by CHDS is the weight increase (in pounds) for expectant mothers in the second trimester.
In a fictitious study, suppose that CHDS finds the average weight increase in the second trimester is 14 pounds. Suppose also that, in 2015, a random sample of 39 expectant mothers have mean weight increase of 15.9 pounds in the second trimester, with a standard deviation of 6.0 pounds.
A hypothesis test is done to see if there is evidence that weight increase in the second trimester is greater than 14 pounds.
Find the p-value for the hypothesis test
Find the p-value for the hypothesis test
=0.0276

RIGHT TAILED
N= 39
X=15.9
S=6.0
M=15
T*= X-M/S/SQRT(N)
P-VALUE =T.DIST.RT(T*,N-1)
In a study at West Virginia University Hospital, researchers investigated smoking behavior of cancer patients to create a program to help patients stop smoking. They published the results in Smoking Behaviors Among Cancer Survivors (January 2009 issue of the Journal of Oncology Practice.) In this study, the researchers sent a 22-item survey to 1,000 cancer patients. They collected demographic information (age, sex, ethnicity, zip code, level of education), clinical and smoking history, and information about quitting smoking.
The questionnaire filled out by cancer patients at West Virginia University Hospital also asked patients if they were current smokers. The current smoker rate for female cancer patients was 11.6%. 95 female respondents were included in the analysis. For male cancer patients, the current smoker rate was 10.4%, and 67 male respondents were included in the analysis.
Suppose that these current smoker rates are the true parameters for all cancer patients.
Can we use a normal model for the sampling distribution of differences in proportions?

A. Yes, a normal model is a good fit for this sampling distribution.
B. No, a normal model is not a good fit for this sampling distribution.
...
The histogram below is from a simulation of the difference between two population proportions. The mean and the standard error are calculated using the population proportion values and the sample size values.
Which of the following is the best estimate of the population proportions?

A. The population proportions are 0.65 and 0.60.
B. The population proportions are 0.40 and 0.60.
C. The population proportions are 0.10 and 0.20.
D. The population proportions are 0.70 and 0.55.
D. The population proportions are 0.70 and 0.55.
The histogram below is from a simulation of the difference between two population proportions. The mean and the standard error are calculated using the population proportion values and the sample size values.
Which of the following is the best estimate of the standard error?

A. The standard error is -0.15.
B. The standard error is 0.15.
C. The standard error is -0.05.
D. The standard error is 0.05.
D. The standard error is 0.05.
Skittles: The popular Skittles candy comes in 5 colors. According to the Skittles website, the colors are evenly distributed in the population of Skittle candies. So each color makes up 20% of the population.
Suppose that we buy 2 small bags of Skittles. We determine the percentage of green Skittles in one bag and the percentage of orange Skittles in the other bag. Suppose that each bag contains 40 candies. We define the difference in sample proportions as "green" minus "orange."
Which of the following will give a sample difference of 0.10?

A. 10 green and 6 orange
B. 4 green and 8 orange
C. 10 green and 10 orange
A. 10 green and 6 orange
In a study at West Virginia University Hospital, researchers investigated smoking behavior of cancer patients to create a program to help patients stop smoking. They published the results in Smoking Behaviors Among Cancer Survivors (January 2018, Journal of Oncology Practice). In this study, the researchers sent a 22-item survey to 1323 cancer patients. They collected demographic information (age, sex, ethnicity, zip code, level of education), clinical and smoking history, and information about quitting smoking.
Of the 1323 patients who were mailed surveys, 265 patients responded. For various reasons, researchers used only 238 of the completed surveys. 35 out of 128 female cancer patients reported being past smokers, and 62 out of 110 male cancer patients reported being past smokers. Calculate the difference between the corresponding sample proportions ^p1−^p2 (female minus male). Round the answer to 4 decimal places.
=-0.2902

P1= 35/128
P2=62/110
=P1-P2
In 2014, students in an advanced Statistics course at UC Berkeley conducted an anonymous survey about use of cognition-enhancing drugs among college males. One survey group of males included members from a fraternity, and the other survey group of males included no fraternity members.
The standard error formula for the difference between sample proportions is

Calculate the standard error for a survey comparing proportions of cognition-enhancing drug use of fraternity members to non-fraternity members, where p1 = 0.3, n1 = 132, p2 = 0.22, n2 = 89. Round all calculations to the thousandth decimal place.
=0.059
SQAURE ROOT LAST
EQUATION AND VALUES ARE GIVEN
In 2014, students in an advanced Statistics course at UC Berkeley conducted an anonymous survey about use of cognition-enhancing drugs among college males. One survey group of males included members from a fraternity, and the other survey group of males included no fraternity members.
The graph below represents the sampling distribution of differences between the proportion of fraternity members who have used cognition-enhancing drugs and the proportion of non-fraternity members who have used cognition-enhancing drugs. The difference in proportions is p1−p2=0.05 (fraternity members minus non-fraternity members).
Consider simulating this data with 5000 samples from each of the fraternity member group and the non-fraternity member group. Based on the sampling distribution of difference in proportions, which of the following results for p1−p2 would be most unusual?

A. 0.0
B. -0.06
C. 0.29
D. -0.1
C. 0.29
In the article "Coffee, Caffeine, and Risk of Depression Among Women" in the September 2011 edition of the Archives of Internal Medicine, researchers investigated the relationship between caffeine consumption and depression among women.
The participants in this study were older, with substantially lower rates of depression when compared to female teens. Researchers compared two groups of women (among others) in this study: those who do not drink coffee and those who routinely drink 4 or more cups of coffee each day.
For the following question, a coffee drinker is a woman who drinks four or more cups each day.
Suppose that the difference between the proportions of coffee drinkers and non-coffee drinkers who are depressed is equal to −0.02. (So the statistic is p-hat for coffee drinkers minus p-hat for non-coffee drinkers.)
Which group has a higher depression rate?

A. Coffee Drinkers
B. Non-Coffee Drinkers
B. Non-Coffee Drinkers
In 2014, students in an advanced Statistics course at UC Berkeley conducted an anonymous survey about use of cognition-enhancing drugs among college males. One survey group of males included members from a fraternity, and the other survey of males group included no fraternity members.
We can use a simulation model to estimate the true difference between the proportion of fraternity members who have used cognition-enhancing drugs and the proportion of non-fraternity members who have used cognition-enhancing drugs. The graph below represents the sampling distribution of differences.

The difference in proportions is p1 - p2 = 0.05 (fraternity members minus non-fraternity members), and the standard error is 0.08. The z-score for this difference in sample proportions is 0.885.
Which of the following is the most appropriate conclusion?

A.There is a statistically significant difference between the proportion of fraternity members who have used cognition-enhancing drugs and the proportion of non-fraternity members who have used cognition-enhancing drugs.
B. There is not a statistically significant difference between the proportion of fraternity members who have used cognition-enhancing drugs and the proportion of non-fraternity members who have used cognition-enhancing drugs.
C. We do not have enough information to make a conclusion about the difference between the proportion of fraternity members who have used cognition-enhancing drugs and the proportion of non-fraternity members who have used cognition-enhancing drugs.
B. There is not a statistically significant difference between the proportion of fraternity members who have used cognition-enhancing drugs and the proportion of non-fraternity members who have used cognition-enhancing drugs.
So the statistic is the sample proportion for coffee drinkers minus the sample proportion for non-coffee drinkers.

A. Graph A
B. Graph B
C. Graph C
A. Graph A

BLUE GRAPHS
Cheating: According to a report on academic cheating by ETS, the Educational Testing Service, college students majoring in Business and Engineering are more likely to cheat than students in other majors. For a statistics project, a community college student at Diablo Valley College (DVC) decides to investigate cheating in two popular majors at DVC: business and nursing. She convinces professors who teach business and nursing courses to distribute a short anonymous survey in their classes. The question about cheating is one of many other questions about college life.
From this data, the student plans to calculate a confidence interval for the difference in two population proportions. Which is the most reasonable description of the two populations for her conclusion?

A. business and nursing students in the classes that were surveyed
B. business and nursing students at DVC
C. business and nursing students at community colleges
D. business and nursing students at community colleges the year of the study
B. business and nursing students at DVC
Quit Smoking:The New England Journal of Medicine published the results of a double-blind, placebo-controlled experiment to study the effect of nicotine patches versus a placebo on quitting smoking.
With the data from the experiment we calculate the sample difference in the "quit smoking" rates for the nicotine treatment group and the placebo group ("treatment" minus "placebo"). We get 0.8%=0.008. The 99% confidence interval based on this sample difference is -0.1036 to 0.0882.
We also calculate a 90% confidence interval. How does the 90% confidence interval compare to the 99% confidence interval? For a 90% confidence interval, the margin of error (MOE) will:

A. Increase
B. Decrease
C. Stay the same
B. Decrease
In March 2015, the Public Policy Institute of California (PPIC) surveyed 7525 likely voters living in California. This is the 148th PPIC research poll, and is part of a survey series that started in 1998. PPIC researchers find that 3266 survey participants are registered Democrats and 2137 survey participants are registered Republicans. PPIC is interested in the difference in approval rate for the governor between registered Democrats and registered Republicans in California.
Participants were asked: "Overall, do you approve or disapprove of the way the governor of California is handling his job?" 2450 of the registered Democrats answered "Yes" and 684 of the registered Republicans answered "Yes."
True or false? A normal model is an appropriate fit for the sampling distribution of sample differences.

A. True
B. False
A. True
In March 2015, the Public Policy Institute of California (PPIC) surveyed 7,525 likely voters living in California. In the survey, respondents were asked about global warming. PPIC researchers are interested in the difference in viewpoints across racial/ethnic groups. PPIC results show that 75% of Latinos view global warming as a serious problem, and 46% of whites view global warming as a serious problem.
Using the data from the survey, we calculate the sample difference in global warming viewpoints for Latino respondents and white respondents to be 29% = 0.29.
Researchers can use the 29% sample difference to draw a conclusion about which populations?

A. Latinos and whites living in the United States who are likely voters
B. Latinos and whites who live in California
C. Latinos and whites living in California who are likely to vote
D. Latinos and whites who live in California and are concerned about global warming
C. Latinos and whites living in California who are likely to vote
Quit Smoking: The New England Journal of Medicine published the results of a double-blind, placebo-controlled experiment to study the effect of nicotine patches and the antidepressant bupropion on quitting smoking.
With the data from the experiment we calculate the sample difference in the "quit smoking" rates for the nicotine treatment group and the placebo group ("treatment" minus "placebo"). We get 0.8% = 0.008. The 99% confidence interval based on this sample difference is -0.1036 to 0.0882.
Which of the following is a valid conclusion?

A. We are 99% confident that the proportion of smokers who will quit smoking with a nicotine patch treatment is about 8.8 to 10.4% higher than the proportion who quit smoking with a placebo treatment.
B. The nicotine patch treatment is not effective in helping smokers quit. We are 99% confident that smokers using a nicotine patch treatment have a "quit smoking" rate that is anywhere from 10.4% lower to 8.8% higher than the success rate of smokers with a placebo treatment.
B. The nicotine patch treatment is not effective in helping smokers quit. We are 99% confident that smokers using a nicotine patch treatment have a "quit smoking" rate that is anywhere from 10.4% lower to 8.8% higher than the success rate of smokers with a placebo treatment.
Quit Smoking: The New England Journal of Medicine published the results of a double-blind, placebo-controlled experiment to study the effect of nicotine patches and the antidepressant bupropion on quitting smoking.
With the data from the experiment we calculate the sample difference in the "quit smoking" rates for the nicotine treatment group and the placebo group ("treatment" minus "placebo"). We get 0.8% = 0.008. Which of the following is an appropriate conclusion based on this finding?

A. In this experiment the nicotine treatment had a higher success rate than the placebo group, but the improvement was less than 1%.
B. In this experiment, the placebo group had a higher success rate than the nicotine group by 0.8%.
C. Nicotine patches will produce a slightly higher success rate when compared to a placebo, but the difference is not statistically significant.
D. Nicotine patches and the antidepressant bupropion work equally well on "quit smoking" rates.
A. In this experiment the nicotine treatment had a higher success rate than the placebo group, but the improvement was less than 1%.
The difference between teenage female and male depression rates estimated from two samples is 0.06. The estimated standard error of the sampling distribution is 0.03. What is the 95% confidence interval? Use the critical value z = 1.96.
(0, 0.12)
P1= P1-Z*SE
P2= P1+Z*SE
Quit Smoking: Previous studies suggest that use of nicotine-replacement therapies and antidepressants can help people stop smoking. The New England Journal of Medicine published the results of a double-blind, placebo-controlled experiment to study the effect of nicotine patches and the antidepressant bupropion on quitting smoking. The target for quitting smoking was the 8th day of the experiment.
In this experiment researchers randomly assigned smokers to treatments. Of the 174 smokers taking a placebo, 22 stopped smoking by the 8th day. Of the 292 smokers taking only the antidepressant buproprion, 81 stopped smoking by the 8th day. Calculate the estimated standard error for the sampling distribution of differences in sample proportions.
The estimated standard error =
=0.036

N1=
N2=
P1=
P2=
SE=SQRT (P1(1-P1)/N1 +P2(1-P2)/N2
Quit Smoking: Previous studies suggest that use of nicotine-replacement therapies and antidepressants can help people stop smoking. The New England Journal of Medicine published the results of a double-blind, placebo-controlled experiment to study the effect of nicotine patches and the antidepressant bupropion on quitting smoking. The target for quitting smoking was the 8th day of the experiment.
In this experiment researchers randomly assigned smokers to treatments. Of the 162 smokers taking a placebo, 28 stopped smoking by the 8th day. Of the 272 smokers taking only the antidepressant buproprion, 82 stopped smoking by the 8th day.
Calculate the 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment). (The standard error is about 0.0407. Use critical value z = 2.576.)

(, )
Round your answer to three decimal places. Put lower bound in the first box and upper bound in the second box.
(-0.233, -0.024)
P1= P1-Z*SE
P2= P1+Z*SE
Gardeners on the west coast of the United States are investigating the difference in survival rates of two flowering plants in drought climates. Plant A has a survival rate of 0.7 and plant B has a survival rate of 0.48. The standard error of the difference in proportions is 0.091. What is the margin of error for a 99% confidence interval? Use critical value z = 2.576.
MOE =
=0.234
MOE=Z *SE
The 99% confidence interval for the difference in proportions of women and men who support stricter gun control is (0.02, 0.08). (Difference here is "women" minus "men.") Which of the following conclusions about the confidence interval is most appropriate?

A. 99% of the time the gender difference seen in a poll on this issue is between 0.02 and 0.08.
B. We are 99% confident that the difference between the proportions of women and men who support stricter gun control is between 0.02 and 0.08.
C. There is a 99% chance that the difference between the proportions of women and men who support stricter gun control is between 0.02 and 0.08.
B. We are 99% confident that the difference between the proportions of women and men who support stricter gun control is between 0.02 and 0.08.
In 2011, the Institute of Medicine (IOM), a non-profit group affiliated with the US National Academy of Sciences, reviewed a study measuring bone quality and levels of vitamin-D in a random sample from bodies of 675 people who died in good health. 8.5% of the 82 bodies with low vitamin-D levels (below 50 nmol/L) had weak bones. Comparatively, 1% of the 593 bodies with regular vitamin-D levels had weak bones.
Is a normal model a good fit for the sampling distribution?

A. Yes, there are close to equal numbers in each group.
B. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group.
C. No, the groups are not the same size.
D. No, there are not at least 10 people with weak bones and 10 people with strong bones in each group.
D. No, there are not at least 10 people with weak bones and 10 people with strong bones in each group.
Researchers conducted a randomized, double-blind clinical trial to compare the herb St. John's Wort, the antidepressant drug sertraline, and a placebo for the treatment of depression. Of the 113 patients in the herb treatment group, 23.9% showed improvement, compared to 24.8% of the 113 in the sertraline group and 31.9% of the 113 patients in the placebo group. Medscape Medical News published this research in 2002.
Is a normal model a good fit for the sampling distribution?

A. Yes, the sample sizes are large and there are equal numbers in each treatment group.
B. Yes, there are at least 10 people who improved and 10 who did not improve in each randomized treatment group.
C. No, there are not at least 10 successes and failures in each group.
D. No, we do not know if the researchers used random assignment to create treatment groups.
B. Yes, there are at least 10 people who improved and 10 who did not improve in each randomized treatment group.
Students in a discussion of gun control in a sociology class at Foothill Community College argue that Republicans are more likely to oppose gun control than Independents. They use data from an article titled "Gun Control Splits America," published March 23, 2010 in pewresarch.org by the Pew Research Center for the People and the Press. In this study 62% of Republicans and 57% of Independents say that states should not be able to pass laws banning handguns.
The article states that researchers collected data from 1,500 adults nationwide reached on landlines and cell phones. Which issue will invalidate the conclusion of the hypothesis test?

A. The data is not a random sample.
B. Researchers interviewed more Republicans than Independents.
C. The data comes from only 1,500 adults nationwide when there are nearly 309 million adults in the U.S.
A. The data is not a random sample.
A politician claims that a larger proportion of members of the news media are Democrats when compared to the general public. Let p1 represent the proportion of the news media that is Democrat and p2 represent the proportion of the public that is Democrat. What are the appropriate null and alternative hypotheses that correspond to this claim?

A. H0: p1 - p2 = 0; Ha: p1 - p2 < 0
B. H0: p1 - p2 = 0; Ha: p1 - p2 > 0
C. H0: p1 - p2 = 0; Ha: p1 - p2 ≠ 0
B. H0: p1 - p2 = 0; Ha: p1 - p2 > 0
A scientist claims that a smaller proportion of members of the National Academy of Sciences are women when compared to the proportion of women nationwide. Let p1 represent the proportion of women in the National Academy of Sciences and p2 represent the proportion of women nationwide. Which is the correct alternative hypotheses that corresponds to this claim?

A. Ha: p1 - p2 < 0
B. Ha: p1 - p2 > 0
C. Ha: p1 - p2 ≠ 0
A. Ha: p1 - p2 < 0
In the article, Attitudes About Marijuana and Political Views (Psychological Reports, 1973), researchers reported on the use of cannabis by liberals and conservatives during the 1970's.
To test the claim (at 1% significance) that the proportion of voters who smoked cannabis frequently was lower among conservatives, the hypotheses were
In the hypothesis test about cannabis use by conservatives and liberals, the P-value is extremely small.
Which of the following errors is possible in this situation?

A. Type I only
B. Type II only
C. Type I and Type II
D. Neither Type I nor Type II
A. Type I only
In the article Foods, Fortificants, and Supplements: Where Do Americans Get Their Nutrients? researchers analyze the nutrient and vitamin intake from a random sample of 16,110 U.S. residents. Researchers compare the level of daily vitamin intake for vitamin A, vitamin B-6, vitamin B-12, vitamin C, vitamin D, vitamin E and calcium. Unless otherwise stated, all hypothesis tests in the study are conducted at the 5% significance level.
To test the claim (at 5% significance) that the proportion of U.S. residents who consume recommended levels of vitamin A is higher among women than men, researchers set up the following hypotheses:
In this hypothesis test which of the following errors is a Type I error?

A. Researchers conclude that a larger proportion of women consume the recommended daily intake of vitamin A when there is actually no difference between vitamin A consumption for women and men.
B. Researchers conclude that there is no difference between vitamin A consumption for women and men when actually a larger proportion of women consume the recommended daily intake of vitamin A.
A. Researchers conclude that a larger proportion of women consume the recommended daily intake of vitamin A when there is actually no difference between vitamin A consumption for women and men.
In 2011, the Institute of Medicine (IOM), a non-profit group affiliated with the US National Academy of Sciences, reviewed a study measuring bone quality and levels of vitamin-D in a random sample from bodies of 675 people who died in good health. 8.5% of the 82 bodies with low vitamin-D levels (below 50 nmol/L) had weak bones. Comparatively, 1% of the 593 bodies with regular vitamin-D levels had weak bones.
Medical researchers are conducting a double-blind experiment treating high blood pressure patients with a new vitamin supplement. The researchers are conducting a hypothesis test in which a Type I error is very serious, and the Type II error is not very serious. Which level of significance is the best choice?

A. α = 0.01
B. α = 0.05
C. α = 0.10
A. α = 0.01
n the article, Attitudes About Marijuana and Political Views (Psychological Reports, 1973), researchers reported on the use of cannabis by liberals and conservatives during the 1970's.
To test the claim (at 1% significance) that the proportion of voters who smoked cannabis frequently was lower among conservatives, the hypotheses were


In the hypothesis test about cannabis use by conservatives and liberals, the test statistic was z = -4.27, with a corresponding p-value of about 0.00001.

A. The data do not support the claim that a lower proportion of conservatives smoke cannabis when compared to liberals.
B. The data support the claim that the proportion of conservatives who smoke cannabis is no different that the proportion for liberals.
C. The data support the claim that a lower proportion of conservatives smoke cannabis when compared to liberals.
C. The data support the claim that a lower proportion of conservatives smoke cannabis when compared to liberals.
In the article Foods, Fortificants, and Supplements: Where Do Americans Get Their Nutrients? researchers analyze the nutrient and vitamin intake from a random sample of 16,110 U.S. residents. Researchers compare the level of daily vitamin intake for vitamin A, vitamin B-6, vitamin B-12, vitamin C, vitamin D, vitamin E and calcium. Unless otherwise stated, all hypothesis tests in the study are conducted at the 5% significance level.
For the claim that the proportion of U.S. residents who consumed recommended levels of vitamin A is higher among women than men, the null and alternative hypotheses are:


The p-value is 0.08, and researchers conduct this test at a 5% level of significance. Which of the following is the correct conclusion?

A. Reject H0 , and support Ha .
B. Support H0 , and reject Ha .
C. Fail to Reject H0 , do not support Ha .
C. Fail to Reject H0 , do not support Ha .
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