Unit 6

A test of 𝐻0:𝑝=0.5 versus 𝐻𝑎:𝑝>0.5 based on a sample of size 200 yields the standardized test statistic 𝑧=2.19. Assume that the conditions for performing inference are met. Determine the value of 𝑝̂ the sample proportion of successes.

0.5774

Which of the following is the critical value for calculating a 94% confidence interval for a population proportion? Enter the critical value to 3 decimal places.

1.88

Many television viewers express doubts about the validity of certain commercials. In an attempt to answer their critics, Timex Group USA wishes to estimate the true proportion 𝑝 of all consumers who believe what is shown in Timex television commercials. What is the smallest number of consumers that Timex can survey to guarantee a margin of error of 0.05 or less at a 99% confidence level?

700

You are thinking about opening a restaurant and are searching for a good location. From research you have done, you know that the mean income of those living near the restaurant must be over $85,000 to support the type of upscale restaurant you wish to open. You decide to take a simple random sample of 50 people living near one potential location. Based on the mean income of this sample, you will perform a test of𝐻0:𝜇𝐻𝑎:𝜇=$85,000>$85,000where 𝜇 is the true mean income in the population of people who live near the restaurant. Which of the following is 𝑛𝑜𝑡 true regarding Type I and Type II errors in this setting?

All are true

At a baseball game, 42 of 65 randomly selected people own an iPod. At a rock concert occurring at the same time across town, 34 of 52 randomly selected people own an iPod. A researcher wants to test the claim that the proportion of iPod owners at the two venues is different. A 90% confidence interval for the difference (Game - Concert) in population proportions is (−0.154, 0.138). Which of the following gives the correct outcome of the researcher's test of the claim?

Because the interval includes 0, the researcher cannot conclude that the proportion of iPod owners at the two venues is different.

A radio talk show host with a large audience is interested in the proportion 𝑝 of adults in his listening area who think the drinking age should be lowered to eighteen. To find this out, he poses the following question to his listeners: "Do you think that the drinking age should be reduced to eighteen in light of the fact that 18-year-olds are eligible for military service?" He asks listeners to go to his website and vote "Yes" if they agree the drinking age should be lowered and "No" if not. Of the 100 people who voted, 70 answered "Yes." Which of the following conditions are violated?

I only

When constructing a confidence interval for a difference between two population proportions, why is it important to check that the number of successes and the number of failures in each sample is at least 10? Select the correct answer.

So we can assume the sampling distribution of 𝑝̂1−𝑝̂2 is approximately Normal.

A New York Times/CBS News Poll asked a random sample of U.S. adults the question "Do you favor an amendment to the Constitution that would permit organized prayer in public schools?" Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). What would happen to the length of the interval if the confidence level were increased to 99%?

The interval would increase.

A telephone poll of an SRS of 1234 adults found that 62% are generally satisfied with their lives. The announced margin of error for the poll was 3%. Does the margin of error account for the fact that some adults do not have telephones?

No; the margin of error only accounts for sampling variability.

Earlier in this section, you read about an experiment comparing surgery and observation as treatments for men with prostate cancer. After 20 years, 𝑝̂𝑆 = 141364 = 0.387 of the men who were assigned to surgery were still alive and 𝑝̂𝑂 = 122367 = 0.332 of the men who were assigned to observation were still alive. Which of the following is the 95% confidence interval for 𝑝̂𝑆−𝑝̂𝑂?

(0.387−0.332)±1.960.387(0.613)364+0.332(0.668)367‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√(0.387−0.332)±1.960.387(0.613)364+0.332(0.668)367‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

Thirty-five people from a random sample of 125 workers from Company A admitted to using sick leave when they weren't really ill. Seventeen employees from a random sample of 68 workers from Company B admitted that they had used sick leave when they weren't ill. Which of the following is a 95% confidence interval for the difference in the proportions of workers at the two companies who would admit to using sick leave when they weren't ill?

0.03±1.96(0.28)(0.72)125+(0.25)(0.75)68‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

The Gallup Poll interviews 1600 people. Of these,18% say that they jog regularly. The news report adds: "The poll had a margin of error of plus or minus three percentage points at a 95% confidence level." You can safely conclude that

95% of all Gallup Poll samples like this one give answers within ±3% of the true population value.

Slow response times by paramedics, firefighters, and policemen can have serious consequences for accident victims. In the case of life-threatening injuries, victims generally need medical attention within 8 minutes of the accident. Several cities have begun to monitor emergency response times. In one such city, emergency personnel took more than 8 minutes to arrive on 22% of all calls involving life-threatening injuries last year. The city manager shares this information and encourages these first responders to "do better." After 6 months, the city manager selects an SRS of 400 calls involving life-threatening injuries and examines the response times. She then performs a test at the a 𝛼=0.05 level of𝐻0:𝑝𝐻𝑎:𝑝=0.22<0.22 where 𝑝 is the true proportion of calls involving life-threatening injuries during this 6-month period for which emergency personnel took more than 8 minutes to arrive. Which of the following is true regarding Type I and Type II errors in this setting?

A Type I error would be finding convincing evidence that the proportion of all calls in which first responders took more than 8 minutes to arrive had decreased when it really hasn't.

Slow response times by paramedics, firefighters, and policemen can have serious consequences for accident victims. In the case of life-threatening injuries, victims generally need medical attention within 8 minutes of the accident. Several cities have begun to monitor emergency response times. In one such city, emergency personnel took more than 8 minutes to arrive on 22% of all calls involving life-threatening injuries last year. The city manager shares this information and encourages these first responders to "do better." After 6 months, the city manager selects an SRS of 400 calls involving life-threatening injuries and examines the response times. She then performs a test at the a 𝛼=0.05 level of𝐻0:𝑝𝐻𝑎:𝑝=0.22<0.22where 𝑝 is the true proportion of calls involving life-threatening injuries during this 6-month period for which emergency personnel took more than 8 minutes to arrive. Which type of error is more serious in this case? Justify your answer.

A Type I error would be more serious because the city would overestimate the ability of the emergency personnel to get to the scene quickly and people may end up dying.

In the population of people in the United States, about 10% are left-handed. After bumping elbows at lunch with several left-handed students, Simon wondered if more than 10% of students at his school are left-handed. To investigate, he selected an SRS of 50 students and found 8 lefties (𝑝̂=850=0.16). We want to test 𝐻0:𝑝𝐻𝑎:𝑝=0.10>0.10 where𝑝= the true proportion of all students at Simon's school that are left-handed. To determine if these data provide convincing evidence that more than 10% of the students at Simon's school are left-handed, 200 trials of a simulation were conducted. Each dot in the graph shows the proportion of students that are left-handed in a random sample of 50 students, assuming that each student has a 10% chance of being left handed. Based upon the simulation results, 24 of the 200 simulated trials yielded a sample proportion of 0.16 or greater, so the 𝑃-value is approximately 24/200 = 0.12.

Based upon the simulation results the 𝑃-value is approximately 24/100 = 0.24. Assuming that the true proportion of all students at Simon's school that are left-handed is 0.10, there is a 0.24 probability of getting a sample proportion of 0.16 or greater just by chance in a random sample of 50 students.

To estimate the difference in the proportion of students at high school A and high school B who drive themselves to school, a district administrator selected a random sample of 100 students from each school. At school A, 23 of the students said they drive themselves; at school B, 29 of the students said they drive themselves. A 90% confidence interval for 𝑝̂𝐴−𝑝̂𝐵 is −0.16 to 0.04. Based on this interval, which conclusion is best?

Because 0 is in the interval, there is not convincing evidence of a difference in the population proportions.

Slow response times by paramedics, firefighters, and policemen can have serious consequences for accident victims. In the case of life-threatening injuries, victims generally need medical attention within 8 minutes of the accident. Several cities have begun to monitor emergency response times. In one such city, emergency personnel took more than 8 minutes to arrive on 22% of all calls involving life-threatening injuries last year. The city manager shares this information and encourages these first responders to "do better." After 6 months, the city manager selects an SRS of 400 calls involving life-threatening injuries and examines the response times. She then performs a test at the a level of𝐻0:𝑝𝐻𝑎:𝑝=0.22<0.22 where 𝑝 is the true proportion of calls involving life-threatening injuries during this 6-month period for which emergency personnel took more than 8 minutes to arrive. Which of the following statements regarding the seriousness of the consequences and the 𝛼 level is true?

Because the consequences for a Type I error are more serious, the α level should be lower.

The Pew Internet and American Life Project asked a random sample of U.S. adults, "Do you ever . . . use Twitter or another service to share updates about yourself or to see updates about others?" According to Pew, the resulting 95% confidence interval is (0.123, 0.177). Based on the confidence interval, is there convincing evidence that the true proportion of U.S. adults who would say they use Twitter or another service to share updates differs from 0.17?

Because the value 0.17 is included in the confidence interval, it is a plausible value for the true proportion of U.S. adults who would say they use Twitter. We do not have convincing evidence that the true proportion of U.S. adults who would say they use Twitter differs from 0.17.

A New York Times/CBS News Poll asked a random sample of U.S. adults the question "Do you favor an amendment to the Constitution that would permit organized prayer in public schools?" Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). Based on this poll, a reporter claims that more than two-thirds of U.S. adults favor such an amendment. Use the confidence interval to evaluate the reporter's claim.

Because the value 23=0.667 (and values less than 23) are in this interval, it is plausible that 23 or less of the population favor such an amendment. Thus, there is not convincing evidence that more than 23 of U.S. adults favor such an amendment.

Many new products introduced into the market are targeted toward children. The choice behavior of children with regard to new products is of particular interest to companies that design marketing strategies for these products. As part of one study, randomly selected children in different age groups were compared on their ability to sort new products into the correct product category (milk or juice). Here are some of the data: Researchers want to know if a greater proportion of 6- to 7-year-olds can sort correctly than 4- to 5-year-olds. So 𝐻0:𝑝1−𝑝2=0,𝐻𝑎:𝑝1−𝑝2<0 where 𝑝1= the true proportion of 4- to 5-year-olds who would sort correctly and 𝑝2=the true proportion of 6- to 7-year-olds who would sort correctly. The test statistic is 𝑧=-3.45 and the 𝑃-value is 0.0003. What conclusion should you make?

Because the 𝑃-value of 0.0003<𝛼=0.05, we reject 𝐻0. We have convincing evidence that the true proportion of 4- to 5-year-olds who would sort correctly is less than the true proportion of 6- to 7-year-olds who would sort correctly.

Jason reads a report that says 80% of U.S. high school students have a computer at home. He believes the proportion is smaller than 0.80 at his large rural high school. Jason chooses an SRS of 60 students and finds that 41 have a computer at home. He would like to carry out a test at the 𝛼=0.05 significance level of 𝐻0:𝑝𝐻𝑎:𝑝=0.80<0.80 where 𝑝= the true proportion of all students at Jason's high school who have a computer at home. What conclusion would you make?

Because the 𝑃-value of 0.0116<𝛼=0.05, we reject 𝐻0. We have convincing evidence that the true proportion of all students at this large rural high school who have a computer at home is less than 0.80.

A surprising number of young adults (ages 19 to 25) still live in their parents' homes. The National Institutes of Health surveyed independent random samples of 2253 men and 2629 women in this age group.7 The survey found that 986 of the men and 923 of the women lived with their parents. A 95% confidence interval for the true difference in the proportions of young men and young women who live in their parents' home is 0.051 to 0.123. Interpret the confidence level.

If we were to select many independent random samples of 2253 men and 2629 women from the population of all young adults (ages 19 to 25) and construct a 99% confidence interval for the difference in the true proportions each time, about 99% of the intervals would capture the difference in the true proportions of men and women aged 19 to 25 who live in their parents' homes.

A New York Times/CBS News Poll asked a random sample of U.S. adults the question "Do you favor an amendment to the Constitution that would permit organized prayer in public schools?" Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). Interpret the confidence level.

If we were to select many random samples of the same size from the population of U.S. adults and construct a 95% confidence interval using each sample, about 95% of the intervals would capture the true proportion of all U.S. adults who would favor an amendment to the Constitution that would permit organized prayer in public schools.

In a random sample of 100 students from a large high school, 37 regularly bring a reusable water bottle from home. Which of the following gives the correct value and interpretation of the standard error of the sample proportion?

In samples of size 100 from this school, the sample proportion of students who bring a reusable water bottle from home typically varies by about 0.048 from the true proportion.

A Gallup poll found that only 28% of American adults expect to inherit money or valuable possessions from a relative. The poll's margin of error was ±3 percentage points at a 95% confidence level. Suppose that Gallup wanted to cut the margin of error in half from 3 percentage points to 1.5 percentage points. How should they adjust their sample size?

Multiply the sample size by 4.

A CBS News poll asked 606 randomly selected women and 442 randomly selected men, "Do you think putting a special tax on junk food would encourage more people to lose weight?" 170 of the women and 102 of the men said "Yes." A 99% confidence interval for the difference (Women - Men) in the true proportion of people in each population who would say "Yes" is -0.020 to 0.120. Does the confidence interval provide convincing evidence that the two population proportions are different? Does the confidence interval provide convincing evidence that the two population proportions are different? No . Because the interval includes 0 as a plausible value for the difference (Women - Men) in the true proportion of people in each population who would say "Yes," we do not have convincing evidence that the two population proportions are different. Does the confidence interval provide convincing evidence that the two population proportions are equal? Explain your answer.

No. Even though 0 (no difference) is a plausible value for the difference there are many other plausible values in the interval that would suggest that the two proportions are not equal.

A recent study of peanut allergies—the LEAP trial—explored the relationship between early exposure to peanuts and the subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. We want test 𝐻0:𝑝1−𝑝2=0,𝐻𝑎:𝑝1−𝑝2≠0 where 𝑝1= the true proportion of children like the ones in this study who are exposed to peanut butter as infants that are allergic to peanuts at age 5 and 𝑝2= the true proportion of children like the ones in this study who are not exposed to peanut butter as infants that are allergic to peanuts at age 5 using 𝛼=0.05. Because the 𝑃-value of this test is approximately 0<𝛼=0.05, we reject 𝐻0. There is convincing evidence that the true proportion of children like the ones in this study who are exposed to peanut butter as infants that are allergic to peanuts at age 5 is different than the true proportion of children like the ones in this study who are not exposed to peanut butter as infants that are allergic to peanuts at age 5. Should you generalize this result to all infants? Why or why not?

No. This experiment likely had parents who volunteered their infants as subjects. When subjects are not randomly selected, we should not generalize the results of an experiment to some larger population of interest.

A confidence interval for a difference in proportions is −0.077 to 0.013. What are the point estimate and the margin of error for this interval?

Point estimate:-0.032 Margin of error: 0.045

Does providing additional information affect responses to a survey question? Two statistics students decided to investigate this issue by asking different versions of a question about texting and driving. Fifty mall shoppers were divided into two groups of 25 at random. The first group was asked version A and the other half were asked version B. The students believed that version A would result in more "Yes" answers. Here are the actual questions: Version A: A lot of people text and drive. Are you one of them? Version B: About 6000 deaths occur per year due to texting and driving. Knowing the potential consequences, do you text and drive? Of the 25 shoppers assigned to version A, 18 admitted to texting and driving. Of the 25 shoppers assigned to version B, only 14 admitted to texting and driving. We want to test 𝐻0:𝑝1−𝑝2=0,𝐻𝑎:𝑝1−𝑝2>0 where 𝑝2= the true proportion of subjects like these who would admit to texting and driving when asked Version A of the question and 𝑝2= the true proportion of subjects like these who would admit to texting and driving when asked Version B. What is the value of the pooled sample proportion? You should you not use the methods of this section to calculate the 𝑃-value because

P̂𝐶 = 0.64 the Large Counts condition is not met. The number of subjects who did not admit to texting and driving when asked Version A, 25(1−0.64)=9, is less than 10.

Many new products introduced into the market are targeted toward children. The choice behavior of children with regard to new products is of particular interest to companies that design marketing strategies for these products. As part of one study, randomly selected children in different age groups were compared on their ability to sort new products into the correct product category (milk or juice). Here are some of the data: Researchers want to know if a greater proportion of 6- to 7-year-olds can sort correctly than 4- to 5-year-olds. So 𝐻0:𝑝1−𝑝2=0,𝐻𝑎:𝑝1−𝑝2<0 where 𝑝1= the true proportion of 4- to 5-year-olds who would sort correctly and 𝑝2=the true proportion of 6- to 7-year-olds who would sort correctly. Age groupnNumber whosorted correctly4- to 5-year-olds50106- to 7-year-olds5328 Calculate the standardized test statistic and 𝑃-value.

Standardized test statistic: 𝑧=(0.20−0.528)−0(0.369)(0.631)50+(0.369)(0.631)53‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√ 𝑃-value: 0.0003

A marketing assistant for a technology firm plans to randomly select 1000 customers to estimate the proportion who are satisfied with the firm's performance. Based on the results of the survey, the assistant will construct a 95% confidence interval for the proportion of all customers who are satisfied. The marketing manager, however, says that the firm can afford to survey only 250 customers. How will this decrease in sample size affect the margin of error and confidence interval?

True: False:

A New York Times/CBS News Poll asked a random sample of U.S. adults the question "Do you favor an amendment to the Constitution that would permit organized prayer in public schools?" Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). Interpret the confidence interval.

We are 95% confident that the interval from 0.63 to 0.69 captures the true proportion of U.S. adults who favor an amendment to the Constitution that would permit organized prayer in public schools.

According to a recent Pew Research Center report, many American adults have made money by selling something online. In a random sample of 4579 American adults, 914 reported that they earned money by selling something online in the previous year. The 98% confidence interval resulting from this survey is 0.1996±0.0137. Interpret the confidence interval.

We are 98% confident that the interval from 0.1859 to 0.2133 captures 𝑝 = the true proportion of American adults who have earned money by selling something online in the previous year.

Members of the city council want to know if a majority of city residents supports a 1% increase in the sales tax to fund road repairs. To investigate, they survey a random sample of 300 city residents and use the results to test the following hypotheses: 𝐻0:𝑝𝐻𝑎:𝑝=0.50>0.50 where 𝑝 is the proportion of all city residents who support a 1% increase in the sales tax to fund road repairs. A Type I error in the context of this study occurs if the city council

finds convincing evidence that a majority of residents supports the tax increase, when in reality at most 50% of city residents support the increase.

A test of 𝐻0:𝑝=0.5 versus 𝐻𝑎:𝑝>0.5 based on a sample of size 200 yields the standardized test statistic 𝑧=2.19. Assume that the conditions for performing inference are met. This test is

significant at 𝛼=0.05 but not at 𝛼=0.01 .

Many new products introduced into the market are targeted toward children. The choice behavior of children with regard to new products is of particular interest to companies that design marketing strategies for these products. As part of one study, randomly selected children in different age groups were compared on their ability to sort new products into the correct product category (milk or juice). Given in the table are some of the data. Researchers want to know if a greater proportion of 6- to 7-year-olds can sort correctly than 4- to 5-year-olds. Age groupnNumber whosorted correctly4- to 5-year-olds50106- to 7-year-olds5328 State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest.

𝐻0:𝑝1−𝑝2=0,𝐻𝑎:𝑝1−𝑝2<0 where 𝑝1= the true proportion of 4- to 5-year-olds who would sort correctly and 𝑝2= the true proportion of 6- to 7-year-olds who would sort correctly.

In the population of people in the United States, about 10% are left-handed. After bumping elbows at lunch with several left-handed students, Simon wondered if more than 10% of students at his school are left-handed. To investigate, he selected an SRS of 50 students and found 8 lefties (𝑝̂=850=0.16). What are the appropriate hypotheses for performing this significance test?

𝐻0:𝑝=0.10,𝐻𝑎:𝑝>0.10, where 𝑝 = the true proportion of all students at Simon's school that are left-handed

A change is made that should improve student satisfaction with the parking situation at a local high school. Before the change, 37% of students approve of the parking that's provided. The null hypothesis 𝐻0:𝑝>0.37 is tested against the alternative 𝐻0:𝑝=0.37. The given hypotheses are incorrect. Give the correct hypotheses.

𝐻0:𝑝=0.37;𝐻𝑎:𝑝>0.37, where 𝑝 = the true proportion of all students at this high school that approve of the parking that is provided.

A researcher suspects that the mean birth weights of babies whose mothers did not see a doctor before delivery is less than 3000 grams. The researcher states the hypotheses as 𝐻0:𝑥¯=3000 grams 𝐻𝑎:𝑥¯<3000 grams. These hypotheses are incorrect. What are the appropriate hypotheses for performing this significance test?

𝐻0:𝜇=3000 grams; 𝐻𝑎:𝜇<3000 grams, where 𝜇 = the true mean birth weights of all babies whose mothers did not see a doctor before delivery.

A New York Times/CBS News Poll asked a random sample of U.S. adults the question "Do you favor an amendment to the Constitution that would permit organized prayer in public schools?" Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). What is the point estimate that was used to construct this interval? What is the margin of error?

𝑝̂=0.66 M.E. = 0.03

Jason reads a report that says 80% of U.S. high school students have a computer at home. He believes the proportion is smaller than 0.80 at his large rural high school. Jason chooses an SRS of 60 students and finds that 41 have a computer at home. He would like to carry out a test at the 𝛼=0.05 significance level of 𝐻0:𝑝𝐻𝑎:𝑝=0.80<0.80 where 𝑝= the true proportion of all students at Jason's high school who have a computer at home. Calculate the standardized test statistic and the 𝑃-value.

𝑧=−2.27;𝑃-value = 0.0116

According to a recent Pew Research Center report, many American adults have made money by selling something online. In a random sample of 4579 American adults, 914 reported that they earned money by selling something online in the previous year. Assume the conditions for inference are met. Determine the critical value 𝑧∗ for a 98% confidence interval for a proportion.

𝑧∗=2.326

According to a recent Pew Research Center report, many American adults have made money by selling something online. In a random sample of 4579 American adults, 914 reported that they earned money by selling something online in the previous year. Assume the conditions for inference are met. Construct a 98% confidence interval for the proportion of all American adults who would report having earned money by selling something online in the previous year.

(0.1859, 0.2133)

A 2016 survey of 1480 randomly selected U.S. adults found that 55% of respondents agreed with the following statement: "Organic produce is better for health than conventionally grown produce." Construct and interpret a 99% confidence interval for the proportion of all U.S. adults who think that organic produce is better for health than conventionally grown produce.

We are 99% confident that the interval from 0.517 to 0.583 captures 𝑝 = the true proportion of U.S. adults who would agree with the statement: "Organic produce is better for health than conventionally grown produce."

What conclusion would you make?

We would fail to reject 𝐻0 at the 𝛼 = 0.05 level. We do not have convincing evidence that the true proportion of all students at Simon's school that are left-handed is greater than 0.10.