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The pesticide diazinon is commonly used to treat infestations of the German cockroach, Blattella germanica. A study investigated the persistence of this pesticide on various types of surfaces. Researchers applied a 0.5% emulsion of diazinon to glass and plasterboard. After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On glass, 18 cockroaches died, while on plasterboard, 25 died. If p1 and p2 are the true proportions of cockroaches like these that would die within 48 hours on glass treated with diazinon and on plasterboard treated with diazinon, respectively, check if the conditions for calculating a confidence interval for p1 – p2 are met.
Random: Random assignment. Large Counts: 25, 11, 18, 18 ≥ 10.
A company that records and sells rewritable DVDs wants to compare the reliability of DVD fabricating machines produced by two different manufacturers. They randomly select 500 DVDs produced by each fabricator and find that 484 of the disks produced by the first machine are acceptable and 480 of the disks produced by the second machine are acceptable. If p1 and p2 are the proportions of acceptable DVDs produced by the first and second machines, respectively, check if the conditions for calculating a confidence interval for p1 – p2 are met.
Random: Independent random samples. 10%: 500 < 10% of all DVDs produced by each machine. Large Counts: 480, 20, 484, 16 ≥ 10.
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. The survey found that 986 of the men and 923 of the women lived with their parents. Construct and interpret a 99% confidence interval for the difference in the true proportions of men and women aged 19 to 25 who live in their parents’ homes.
S: p1 = true proportion of young men who live in their parents’ home and p2 = that of young women . . . P: Two-sample z interval for p1 - p2. Random: Independent random samples. 10%: 2253 < 10% of all young men and 2629 < 10% of all young women. Large Counts: 986, 1267, 923, 1706 ≥ 10. D: (0.051, 0.123) C: We are 99% confident that the interval from 0.051 to 0.123 captures p1 - p2 = the true difference in the proportions of young men and young women who live in their parents’ home.
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. The survey found that 986 of the men and 923 of the women lived with their parents. Does your interval from part (a) give convincing evidence of a difference between the population proportions? Justify your answer.
Because the interval does not contain 0, there is convincing evidence that the true proportion of young men who live at their parents’ home is different from the true proportion of young women who do.
In a Pew Research poll, 287 out of 522 randomly selected U.S. men were able to identify Egypt when it was highlighted on a map of the Middle East. When 520 randomly selected U.S. women were asked, 233 were able to do so. Construct and interpret a 95% confidence interval for the difference in the true proportions of U.S. men and U.S. women who can identify Egypt on a map.
S: pM= true proportion of men who can identify Egypt on a map and pW = that of women . . . P: Two-sample z-interval for pM - pW. Random: Independent random samples. 10%: 522 ≥ 10% of all men and 520 < 10% of all women. Large Counts: 287, 235, 233, 287 ≥ 10. D: (0.041, 0.162) C: We are 95% confident that the interval from 0.041 to 0.162 captures p1 - p2 5 the true difference in the proportions of men and women who can identify Egypt on a map. (b) Because the interval does not contain 0, there is convincing evidence that the true proportion of men who can identify Egypt on a map is different from the true proportion of women who can identify Egypt on a map.
In a Pew Research poll, 287 out of 522 randomly selected U.S. men were able to identify Egypt when it was highlighted on a map of the Middle East. When 520 randomly selected U.S. women were asked, 233 were able to do so. Based on your interval, is there convincing evidence of a difference in the true proportions of U.S. men and women who can identify Egypt on a map? Justify your answer.
Because the interval does not contain 0, there is convincing evidence that the true proportion of men who can identify Egypt on a map is different from the true proportion of women who can identify Egypt on a map.
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. The survey found that 986 of the men and 923 of the women lived with their parents. The 99% confidence interval was found to be (0.051, 0.123). Interpret the confidence level for the interval.
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.
In a Pew Research poll, 287 out of 522 randomly selected U.S. men were able to identify Egypt when it was highlighted on a map of the Middle East. When 520 randomly selected U.S. women were asked, 233 were able to do so. The 95% confidence interval was found to be (0.041, 0.162). Interpret the confidence level for the interval.
If we were to select many independent random samples of 522 men and 520 women from the population of all men and women in the U.S. and construct a 95% confidence interval for the difference in the true proportions each time, about 95% of the intervals would capture the difference in the true proportions of U.S. men and U.S. women who can identify Egypt on a map.
Does the appearance of the interviewer influence how people respond to a survey question? Ken (white, with blond hair) and Hassan (darker, with Middle Eastern features) conducted an experiment to address this question. They took turns (in a random order) walking up to people on the main street of a small town, identifying themselves as students from a local high school, and asking them, “Do you support President Obama’s decision to launch airstrikes in Iraq?” Of the 50 people Hassan spoke to, 11 said “Yes,” while 21 of the 44 people Ken spoke to said “Yes.” Construct and interpret a 90% confidence interval for the difference in the proportions of people like these who would say they support President Obama’s decision when asked by Hassan versus when asked by Ken.
S: p1 = true proportion of people like the ones in this study who would say they support President Obama’s decision when asked by Hassan and p2 = true proportion . . . when asked by Ken. P: Two-sample z-interval for p1 - p2. Random: Random assignment. Large Counts: 11, 39, 21, 23 are ≥ 10. D: (20.414, 20.100) C: We are 90% confident that the interval from 20.414 to 20.100 captures p1 - p2 = the true difference in the proportions of people like the ones in this study who would say they support President Obama’s decision when asked by Hassan or by Ken.
Nicotine patches are often used to help smokers quit. Does giving medicine to fight depression help? A randomized double-blind experiment assigned 244 smokers to receive nicotine patches and another 245 to receive both a patch and the antidepressant drug bupropion. After a year, 40 subjects in the nicotine patch group had abstained from smoking, as had 87 in the patch-plus-drug group. Construct and interpret a 99% confidence interval for the difference in the true proportion of smokers like these who would abstain when using bupropion and a nicotine patch and the proportion who would abstain when using only a patch.
S: p1 = true proportion of smokers like these who would abstain when using only a patch and p2 = true proportion . . . when using both. P: Two-sample z-interval for p1 - p2. Random: Random assignment. Large Counts: 40, 204, 87, 158 are ≥ 10. D: (20.291, 20.092) C: We are 99% confident that the interval from 120.291, 20.0922 captures p1 - p2 = the true difference in the proportion of smokers like these who would abstain when using only a patch and the proportion of smokers like these who would abstain when using bupropion and a nicotine patch.
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.”26 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? Explain your answer.
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.
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.”26 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 equal? Explain your answer.
No; because 0 is captured in the interval, it is plausible that the population proportions are equal, but this does not provide convincing evidence that the two proportions are equal.
An association of Christmas tree growers in Indiana wants to know if there is a difference in preference for natural trees between urban and rural households. So the association sponsored a survey of Indiana households that had a Christmas tree last year to find out. In a random sample of 160 rural households, 64 had a natural tree. In a separate random sample of 261 urban households, 89 had a natural tree. A 95% confidence interval for the difference (Rural – Urban) in the true proportion of households in each population that had a natural tree is -0.036 to 0.154. Does the confidence interval provide convincing evidence that the two population proportions are different? Explain your answer.
Because the interval includes 0 as a plausible value for the difference (Rural – Urban) in the true proportion of households that have a natural tree, we do not have convincing evidence that the two population proportions are different.
An association of Christmas tree growers in Indiana wants to know if there is a difference in preference for natural trees between urban and rural households. So the association sponsored a survey of Indiana households that had a Christmas tree last year to find out. In a random sample of 160 rural households, 64 had a natural tree. In a separate random sample of 261 urban households, 89 had a natural tree. A 95% confidence interval for the difference (Rural – Urban) in the true proportion of households in each population that had a natural tree is -0.036 to 0.154. Does the confidence interval provide convincing evidence that the two population proportions are equal? Explain your answer.
No; because 0 is captured in the interval, it is plausible that the population proportions are equal, but this does not provide convincing evidence that the two proportions are equal.
Earlier in this section, you read about an experiment comparing surgery and observation as treatments for men with prostate cancer. After 20 years, p^s = 141/364 = 0.387 of the men who were assigned to surgery were still alive and p^o = 122/367 = 0.332 of the men who were assigned to observation were still alive. Which of the following is the 95% confidence interval for ps-po? (a) (141 – 122) ± 1.96 sqrt(((141*223)/364) + ((122*245)/367)) (b) (141 – 122) ± 1.96 (sqrt(((141*223)/364)) + sqrt(((122*245)/367))) (c) (0.387 – 0.332) ± 1.96 sqrt(((0.387*0.613)/364) + ((0.332*0.668)/367)) (d) (0.387 – 0.332) ± 1.96 (sqrt(((0.387*0.613)/364)) + sqrt(((0.332*0.668)/367))) (e) (0.387 – 0.332) ± 1.96 sqrt(((0.387*0.613)/364) – ((0.332*0.668)/367))
c
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? (a) So we can generalize the results to the populations from which the samples were selected. (b) So we can assume that the two samples are independent. (c) So we can assume that the observations within each sample are independent. (d) So we can assume the sampling distribution of p^1 – p^2 is approximately Normal. (e) So we can assume that population 1 and population 2 are approximately Normal.
d
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 pA – pB is -0.16 to 0.04. Based on this interval, which conclusion is best? (a) Because -0.06 is in the interval, there is convincing evidence of a difference in the population proportions. (b) Because 0 is in the interval, there is convincing evidence of a difference in the population proportions. (c) Because -0.06 is in the interval, there is not convincing evidence of a difference in the population proportions. (d) Because 0 is in the interval, there is not convincing evidence of a difference in the population proportions. (e) Because most of the interval is negative, there is convincing evidence that a greater proportion of students at high school B drive themselves to school.
d
A Gallup poll conducted telephone interviews with a random sample of adults aged 18 and older. Data were obtained for 1000 people. Of these, 370 said that football is their favorite sport to watch on television. Define the parameter p in this setting.
p= the proportion of all adults aged 18 and older who would say that football is their favorite sport to watch on television.
A Gallup poll conducted telephone interviews with a random sample of adults aged 18 and older. Data were obtained for 1000 people. Of these, 370 said that football is their favorite sport to watch on television. What point estimator will you use to estimate p? What is the value of the point estimate?
Point estimator = p^; p^ = 0.37
A Gallup poll conducted telephone interviews with a random sample of adults aged 18 and older. Data were obtained for 1000 people. Of these, 370 said that football is their favorite sport to watch on television. Do you believe that the value of the point estimate is equal to the value of p? Explain your answer.
Because of sampling variability, I doubt that the value of the point estimate is exactly equal to the value of p.
Are you a sports fan? That’s the question the Gallup polling organization asked a random sample of 1527 U.S. adults. Gallup reported that a 95% confidence interval for the proportion of all U.S. adults who are sports fans is 0.565 to 0.615. Calculate the point estimate and the margin of error.
The point estimate is (0.565 + 0.615)/2 = 0.59. The margin of error is 0.615 - 0.59 = 0.025.
Are you a sports fan? That’s the question the Gallup polling organization asked a random sample of 1527 U.S. adults. Gallup reported that a 95% confidence interval for the proportion of all U.S. adults who are sports fans is 0.565 to 0.615. Interpret the confidence level.
If we were to select many random samples of size 1527 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 are sports fans.
Are you a sports fan? That’s the question the Gallup polling organization asked a random sample of 1527 U.S. adults. Gallup reported that a 95% confidence interval for the proportion of all U.S. adults who are sports fans is 0.565 to 0.615. Based on the interval, is there convincing evidence that a majority of U.S. adults are sports fans? Explain your answer.
Yes; because the confidence interval lies entirely above 0.5, there is convincing evidence that a majority of U.S. adults are sports fans.
Explain how increasing the confidence level would affect the margin of error of a confidence interval, if all other things remained the same.
The margin of error must get larger to increase the capture rate of the intervals.
Explain how quadrupling the sample size would affect the margin of error of a confidence interval, if all other things remained the same.
The margin of error will decrease by a factor of 2.
A random digit dialing telephone survey of 880 drivers asked, “Recalling the last 10 traffic lights you drove through, how many of them were red when you entered the intersections?” Of the 880 respondents, 171 admitted that at least one light had been red. Construct and interpret a 95% confidence interval for the population proportion.
S: p5 the true proportion of all drivers who have run at least one red light in the last 10 intersections they have entered. P: One-sample z interval. Random: Random sample. 10%: 880 , 10% of all drivers. Large Counts: 171 $ 10 and 709 $ 10. D: (0.168, 0.220). C: We are 95% confident that the interval from 0.168 to 0.220 captures the true proportion of all drivers who have run at least one red light in the last 10 intersections they have entered.
A random digit dialing telephone survey of 880 drivers asked, “Recalling the last 10 traffic lights you drove through, how many of them were red when you entered the intersections?” Of the 880 respondents, 171 admitted that at least one light had been red. Nonresponse is a practical problem for this survey—only 21.6% of calls that reached a live person were completed. Another practical problem is that people may not give truthful answers. What is the likely direction of the bias: Do you think more or fewer than 171 of the 880 respondents really ran a red light? Why? Are these sources of bias included in the margin of error?
It is likely that more than 171 respondents have run red lights. We would not expect very many people to claim they have run red lights when they have not, but some people will deny running red lights when they have. The margin of error does not account for these sources of bias, only sampling variability.
The Gallup Poll plans to ask a random sample of adults whether they attended a religious service in the past 7 days. How large a sample would be required to obtain a margin of error of at most 0.01 in a 99% confidence interval for the population proportion who would say that they attended a religious service?
Solving 2.576 sqrt((0.5(0.5))/ n) ≤ 0.01gives n = 16,590 adults.
As part of the Pew Internet and American Life Project, researchers conducted two surveys. The first survey asked a random sample of 1060 U.S. teens about their use of social media. A second survey posed similar questions to a random sample of 2003 U.S. adults. In these two studies, 71.0% of teens and 58.0% of adults used Facebook. 30 Let pT = the true proportion of all U.S. teens who use Facebook and pA = the true proportion of all U.S. adults who use Facebook. Calculate and interpret a 99% confidence interval for the difference in the true proportions of U.S. teens and adults who use Facebook.
S: 99% CI for pT - pA, where pT5 the true proportion of U.S. teens who use Facebook and pA 5 the true proportion of U.S. adults who use Facebook. P: Two-sample z-interval for pT = pA. Random: The data come from independent random samples of 1060 U.S. teens and 2003 U.S. adults. Large Counts: 1060(0.71) < 753, 1060(1 - 0.71) < 297, 2003(0.58) < 1162, and 2003(1 - 0.58) < 841 are all $ 10. D: p^T= 0.71 and p^A = 0.58(0.084, 0.176) C: We are 99% confident that the interval from 0.084 to 0.176 captures pT - pA = the difference in the true proportions of teens and adults who use Facebook.
As part of the Pew Internet and American Life Project, researchers conducted two surveys. The first survey asked a random sample of 1060 U.S. teens about their use of social media. A second survey posed similar questions to a random sample of 2003 U.S. adults. In these two studies, 71.0% of teens and 58.0% of adults used Facebook. 30 Let pT = the true proportion of all U.S. teens who use Facebook and pA = the true proportion of all U.S. adults who use Facebook. A 99% confidence interval was found to be (0.084, 0.176). Based on this confidence interval, is there convincing evidence of a difference in the population proportions? Explain your answer.
Because the interval does not include 0 as a plausible value for the difference in the true proportion of U.S. teens and U.S. adults who use Facebook, we do have convincing evidence that the two population proportions are different.
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 3 percentage points at a 95% confidence level.” You can safely conclude that (a) 95% of all Gallup Poll samples like this one give answers within ±3% of the true population value. (b) the percent of the population who jog is certain to be between 15% and 21%. (c) 95% of the population jog between 15% and 21% of the time. (d) we can be 95% confident that the sample proportion is captured by the confidence interval. (e) if Gallup took many samples, 95% of them would find that 18% of the people in the sample jog.
a
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? (a) −0.032, 0.045 (b) −0.032, 0.090 (c) −0.032, 0.180 (d) −0.045, 0.032 (e) −0.045, 0.090
a
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? (a) In samples of size 100 from this school, the sample proportion of students who bring a reusable water bottle from home will be at most 0.095 from the true proportion. (b) In samples of size 100 from this school, the sample proportion of students who bring a reusable water bottle from home will be at most 0.048 from the true proportion. (c) 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 0.095 from the true proportion. (d) 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. (e) There is not enough information to calculate the standard error.
d
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. Which of the following 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? (a) 550 (b) 600 (c) 650 (d) 700 (e) 750
d
Which of the following is the critical value for calculating a 94% confidence interval for a population proportion? (a) 1.555 (b) 1.645 (c) 1.881 (d) 1.960 (e) 2.576
c
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 18. To find this out, he poses the following question to his listeners: “Do you think that the drinking age should be reduced to 18 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) Random II) 10% III) Large Counts (a) I only (b) II only (c) III only (d) I and II only (e) I, II, and III
a
A marketing assistant for a technology firm plans to randomly select 1000 customers to estimate the pro-portion 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? (a) The margin of error will be about 4 times larger. (b) The margin of error will be about 2 times larger. (c) The margin of error will be about the same size. (d) The margin of error will be about half as large. (e) The margin of error will be about one-fourth as large.
b
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? (a) 0.03 ± sqrt(((0.28*0.72)/125) + ((0.25*0.75)/68)) (b) 0.03 ± 1.96 sqrt(((0.28*0.72)/125) + ((0.25*0.75)/68)) (c) 0.03 ± 1.96 sqrt(((0.28*0.72)/125) - ((0.25*0.75)/68)) (d) 57 ± 1.96 sqrt(((0.28*0.72)/125) + ((0.25*0.75)/68)) (e) 57 ± 1.96 sqrt(((0.28*0.72)/125) - ((0.25*0.75)/68))
b
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? (a) Yes; the margin of error accounts for all sources of error in the poll. (b) Yes; taking an SRS eliminates any possible bias in estimating the population proportion. (c) Yes; the margin of error accounts for undercoverage but not nonresponse. (d) No; the margin of error accounts for nonresponse but not undercoverage. (e) No; the margin of error only accounts for sampling variability.
e
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? (a) Because the interval includes 0, the researcher can conclude that the proportion of iPod owners at the two venues is the same. (b) Because the center of the interval is −0.008, the researcher can conclude that a higher proportion of people at the rock concert own iPods than at the baseball game. (c) Because the interval includes 0, the researcher cannot conclude that the proportion of iPod owners at the two venues is different. (d) Because the interval includes −0.008, the researcher cannot conclude that the proportion of iPod owners at the two venues is different. (e) Because the interval includes more negative than positive values, the researcher can conclude that a higher proportion of people at the rock concert own iPods than at the baseball game.
c
The U.S. Forest Service is considering additional restrictions on the number of vehicles allowed to enter Yellowstone National Park. To assess public reaction, the service asks a random sample of 150 visitors if they favor the proposal. Of these, 89 say “Yes.” Construct and interpret a 99% confidence interval for the proportion of all visitors to Yellowstone who favor the restrictions.
STATE: p= the true proportion of all visitors to Yellowstone who would say they favor the restrictions. PLAN: One-sample z interval. Random: The visitors were selected randomly. 10%: n= 150 is less than 10% of all visitors to Yellowstone National Park. Large Counts: np^ = 89 ≥ 10 and n(1 - p^) = 61 ≥ 10. DO: p^ = 89/150 = 0.593; (0.490, 0.696). CONCLUDE: We are 99% confident that the interval from 0.490 to 0.696 captures p5 the true proportion of all visitors who would say that they favor the restrictions.
The U.S. Forest Service is considering additional restrictions on the number of vehicles allowed to enter Yellowstone National Park. To assess public reaction, the service asks a random sample of 150 visitors if they favor the proposal. Of these, 89 say “Yes.” A 99% confidence was found to be (0.490, 0.696) based on this, is there convincing evidence that more than half of all visitors to Yellowstone National Park favor the proposal? Justify your answer.
Because there are values less than 0.50 in the confidence interval, the U.S. Forest Service cannot conclude that more than half of visitors to Yellowstone National Park favor the proposal. It is plausible that only 49% favor the proposal.
For some people, mistletoe is a symbol of romance. Mesquite trees, however, have no love for the parasitic plant that attaches itself and steals nutrients from the tree. To estimate the proportion of mesquite trees in a desert park that are infested with mistletoe, a random sample of mesquite trees was randomly selected. After inspecting the trees in the sample, the park supervisor calculated a 95% confidence interval of 0.2247 to 0.5753. Interpret the confidence level.
If we were to select many random samples of the size used in this study from the same population of mesquite trees and construct a 95% confidence interval using each sample, about 95% of the intervals would capture the true proportion of all mesquite trees in this park that are infested with mistletoe.
For some people, mistletoe is a symbol of romance. Mesquite trees, however, have no love for the parasitic plant that attaches itself and steals nutrients from the tree. To estimate the proportion of mesquite trees in a desert park that are infested with mistletoe, a random sample of mesquite trees was randomly selected. After inspecting the trees in the sample, the park supervisor calculated a 95% confidence interval of 0.2247 to 0.5753. Calculate the sample size used to create this interval.
Point estimate = 0.4 and margin of error = 0.1753; 0.1753 = 1.96 sqrt((0.4(0.62))/n); n = 30
Do “props” make a difference when researchers interact with their subjects? Emily and Madi asked 100 people if they thought buying coffee at Star-bucks was a waste of money. Half of the subjects were asked while Emily and Madi were holding cups from Starbucks, and the other half of the subjects were asked when the girls were empty handed. The choice of holding or not holding the cups was deter-mined at random for each subject. When holding the cups, 19 of 50 subjects agreed that buying cof-fee at Starbucks was a waste of money. When they weren’t holding the cups, 23 of 50 subjects said it was a waste of money. Calculate and interpret a 90% confidence interval for the difference in the proportion of people like the ones in this experiment who would say that buying coffee from Starbucks is a waste of money when asked by interviewers holding or not holding a cup from Starbucks.
STATE: 90% CI for p1 - p2, where p1 = the true proportion of people like these who would say that buying coffee at Starbucks is a waste of money when the girls hold cups from Starbucks and p2 = the true proportion of people like these who would say that buying coffee at Starbucks is a waste of money when the girls were empty handed. PLAN: Two-sample z-interval for p1 - p2. Random: Subjects were randomly assigned to be asked while the girls held Starbucks cups or were empty handed. Large Counts: 19, 50 - 19 = 31, 23, 50 - 23 = 27 are all ≥ 10. DO: p1 = 19/50 = 0.38 and p2 =23/50 =0.46; (20.242, 0.082). CONCLUDE: We are 90% confident that the interval from 20.242 to 0.082 captures p1 - p2 = the difference (Holding – Empty handed) in the true proportions of people like these who would say that buying coffee at Starbucks is a waste of money.
For the following scenario, state appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest: Mr. Tabor believes that less than 75% of the students at his school completed their math homework last night. The math teachers inspect the homework assignments from a random sample of 50 students at the school.
H0: p=0.75; Ha: p< 0.75, where p= the true proportion of the students at Mr. Tabor’s school who completed their math homework last night.
For the following scenario, state appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest: A Gallup poll report revealed that 72% of teens said they seldom or never argue with their friends. 3 Yvonne wonders whether this result holds true in her large high school, so she surveys a random sample of 150 students at her school.
H0: p= 0.72; Ha: p≠ 0.72, where p= the true proportion of teens in Yvonne’s school who rarely or never argue with their friends.
For the following scenario, state appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest: One company’s bottles of grape-fruit juice are filled by a machine that is set to dispense an average of 180 milliliters (ml) of liquid. A quality-control inspector must check that the machine is working properly. The inspector takes a random sample of 40 bottles and measures the volume of liquid in each bottle.
H0: µ = 180; Ha: µ ≠ 180, where m = the true mean volume of liquid dispensed by the machine.
For the following scenario, state appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest: The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ attitudes toward school and study habits. Scores range from 0 to 200. Higher scores indicate better attitudes and study habits. The mean score for U.S. college students is about 115. A teacher suspects that older students have better attitudes toward school, on average. She gives the SSHA to an SRS of 45 of the over 1000 students at her college who are at least 30 years of age.
H0: µ = 115; Ha: µ > 115, where m = the true mean score on the SSHA for all students at least 30 years of age at the teacher’s college.
For the following scenario, state appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest: During the winter months, the temperatures at the Starneses’ Colorado cabin can stay well below freezing (32°F or 0°C) for weeks at a time. To prevent the pipes from freezing, Mrs. Starnes sets the thermostat at 50°F. The manufacturer claims that the thermostat allows variation in home temperature of σ =°3F. Mrs. Starnes suspects that the manufacturer is overstating the consistency of the thermostat.
H0: σ = 3; Ha: σ > 3, where s = the true standard deviation of the temperature in the cabin.
For the following scenario, state appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest: When ski jumpers take off, the distance they fly varies considerably depending on their speed, skill, and wind conditions. Event organizers must position the landing area to allow for differences in the distances that the athletes fly. For a particular competition, the organizers estimate that the variation in distance flown by the athletes will be σ=10 meters. An experienced jumper thinks that the organizers are underestimating the variation.
H0: σ = 10; Ha: σ > 10, where s = the true standard deviation of the distance jumped by the ski jumpers.
Explain what’s wrong with the stated hypotheses. Then give correct hypotheses. 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 Ho: p>0.37 is tested against the alternative: Ha: p=0.37.
The null hypothesis is always that there is “no difference” or “no change”; the alternative hypothesis is what we suspect is true. These ideas are reversed in the stated hypotheses. Correct: H0: p= 0.37; Ha: p> 0.37.
Explain what’s wrong with the stated hypotheses. Then give correct hypotheses. 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 Ho: x_bar = 3000 grams, Ha: x_bar < 3000 grams
Hypotheses are always about population parameters. However, the stated hypotheses are in terms of the sample statistic. Correct: H0: µ = 3000 grams; Ha: µ < 3000 grams.
Explain what’s wrong with the stated hypotheses. Then give correct hypotheses. A change is made that should improve student satisfaction with the parking situation at your school. Before the change, 37% of students approve of the parking that’s provided. The null hypothesis Ho: p^= 0.37 is tested against the alternative Ha: p^> 0.37.
Hypotheses are always about population parameters, but the stated hypotheses are about the sample statistic. Correct: H0: p= 0.37; Ha: p> 0.37.
Explain what’s wrong with the stated hypotheses. Then give correct hypotheses. 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 Ho: μ =3000 grams, Ha: μ ≤ 2999 grams
Values in both hypotheses must be the same; the null hypothesis should have 5 and the alternative should have <. Correct: H0: µ=3000 grams; Ha: µ<3000 grams.
Mr. Tabor believes that less than 75% of the students at his school completed their math homework last night. The math teachers inspect the homework assignments from a random sample of 50 students at the school. The math teachers inspect the homework assignments from a random sample of 50 students at the school. Only 68% of the students completed their math homework. A significance test yields a P-value of 0.1265. Explain what it would mean for the null hypothesis to be true in this setting.
If H0: p= 0.75 is true, then the proportion of all students at Mr. Tabor’s school who completed their homework last night is 0.75.
Mr. Tabor believes that less than 75% of the students at his school completed their math homework last night. The math teachers inspect the homework assignments from a random sample of 50 students at the school. The math teachers inspect the homework assignments from a random sample of 50 students at the school. Only 68% of the students completed their math homework. A significance test yields a P-value of 0.1265. Interpret the P -value.
Assuming the proportion is 0.75, there is a 0.1265 probability of getting a sample proportion of 0.68 or less by chance in a random sample of 50 students at the school.
The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ attitudes toward school and study habits. Scores range from 0 to 200. Higher scores indicate better attitudes and study habits. The mean score for U.S. college students is about 115. A teacher suspects that older students have better attitudes toward school, on average. She gives the SSHA to an SRS of 45 of the over 1000 students at her college who are at least 30 years of age. Explain what it would mean for the null hypothesis to be true in this setting.
If H0: µ = 115 is true, then the true mean score is 115.
The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ attitudes toward school and study habits. Scores range from 0 to 200. Higher scores indicate better attitudes and study habits. The mean score for U.S. college students is about 115. A teacher suspects that older students have better attitudes toward school, on average. She gives the SSHA to an SRS of 45 of the over 1000 students at her college who are at least 30 years of age. Interpret the P -value.
Assuming the true mean score is 115, there is a 0.0101 probability of getting a sample mean of 125.7 or greater just by chance in an SRS of 45 older students.
One company’s bottles of grape-fruit juice are filled by a machine that is set to dispense an average of 180 milliliters (ml) of liquid. A quality-control inspector must check that the machine is working properly. The inspector takes a random sample of 40 bottles and measures the volume of liquid in each bottle. The mean amount of liquid in the bottles is 179.6 ml and the standard deviation is 1.3 ml. A significance test yields a P -value of 0.0589. Interpret the P -value.
Assuming the true mean volume of liquid dispensed by the machine is 180 ml, there is a 0.0589 probability of getting a sample mean at least as far from 180 as 179.6 (in either direction) by chance in a random sample of 40 bottles filled by the machine.
A Gallup poll report revealed that 72% of teens said they seldom or never argue with their friends. 3 Yvonne wonders whether this result holds true in her large high school, so she surveys a random sample of 150 students at her school. Yvonne finds that 96 of the 150 students (64%) say they rarely or never argue with friends. A significance test yields a P-value of 0.0291. Interpret the P-value.
Assuming the true proportion of teens in Yvonne’s school who rarely or never argue with their friends is 0.72, there is a 0.0291 probability of getting a sample proportion at least as far from 0.72 as 0.64 (in either direction) by chance in a random sample of 150 students from her school.
A student performs a test of Ho: μ = 100 versus Ha: μ > 100 and gets a P-value of 0.044. The student says, “There is a 0.044 probability of getting the sample result I did by chance alone.” Explain why the student’s explanation is wrong.
The student forgot to include the conditions and the direction in the interpretation. Assuming the null hypothesis is true, there is a 0.044 probability of getting the sample result I did or one even larger by chance alone.
A student performs a test of Ho: p=0.3 versus Ha: p<0.3 and gets a P-value of 0.22. The student says, “This means there is about a 22% chance that the null hypothesis is true.” Explain why the student’s explanation is wrong.
Either H0 is true or H0 is false. Assuming the null hypothesis is true, there is a 0.22 probability of getting a sample proportion as small as or smaller than the one observed just by chance.
Mr. Tabor believes that less than 75% of the students at his school completed their math homework last night. The math teachers inspect the homework assignments from a random sample of 50 students at the school. The math teachers inspect the homework assignments from a random sample of 50 students at the school. Only 68% of the students completed their math homework. A significance test yields a P-value of 0.1265. What conclusion would you make at the α=0.05 level?
Because the P-value of 0.1265 is greater than α= 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of students at Mr. Tabor’s school who completed their math homework last night is less than 0.75.
The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ attitudes toward school and study habits. Scores range from 0 to 200. Higher scores indicate better attitudes and study habits. The mean score for U.S. college students is about 115. A teacher suspects that older students have better attitudes toward school, on average. She gives the SSHA to an SRS of 45 of the over 1000 students at her college who are at least 30 years of age. In the study of older students’ attitudes, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. A significance test yields a P -value of 0.0101. What conclusion would you make at the α=0.05 level?
Because the P-value of 0.0101 is less than α= 0.05, we reject H0. We have convincing evidence that the true mean score on the SSHA for all students at least 30 years of age at the teacher’s college is greater than 115.
One company’s bottles of grape-fruit juice are filled by a machine that is set to dispense an average of 180 milliliters (ml) of liquid. A quality-control inspector must check that the machine is working properly. The inspector takes a random sample of 40 bottles and measures the volume of liquid in each bottle. The mean amount of liquid in the bottles is 179.6 ml and the standard deviation is 1.3 ml. A significance test yields a P -value of 0.0589. Interpret the P -value. What conclusion would you make at the α=0.10 level?
Because the P-value of 0.0589 is less than α = 0.10, we reject H0. We have convincing evidence that the true mean volume of liquid dispensed differs from 180 ml.
One company’s bottles of grape-fruit juice are filled by a machine that is set to dispense an average of 180 milliliters (ml) of liquid. A quality-control inspector must check that the machine is working properly. The inspector takes a random sample of 40 bottles and measures the volume of liquid in each bottle. The mean amount of liquid in the bottles is 179.6 ml and the standard deviation is 1.3 ml. A significance test yields a P -value of 0.0589. Interpret the P -value. Would your conclusion change if a 5% significance level was used instead of a 10% significance level? Explain your reasoning.
Yes; because the P-value of 0.0589 > α = 0.05, we would fail to reject H0. We lack convincing evidence that the true mean volume differs from 180 ml.
A Gallup poll report revealed that 72% of teens said they seldom or never argue with their friends. 3 Yvonne wonders whether this result holds true in her large high school, so she surveys a random sample of 150 students at her school. Yvonne finds that 96 of the 150 students (64%) say they rarely or never argue with friends. A significance test yields a P-value of 0.0291. Interpret the P-value. What conclusion would you make at the α=0.01 level?
Because the P-value of 0.0291 > α=0.01, we fail to reject H0. We lack convincing evidence that the true proportion of teens in Yvonne’s school who rarely or never argue differs from 0.72.
A Gallup poll report revealed that 72% of teens said they seldom or never argue with their friends. 3 Yvonne wonders whether this result holds true in her large high school, so she surveys a random sample of 150 students at her school. Yvonne finds that 96 of the 150 students (64%) say they rarely or never argue with friends. A significance test yields a P-value of 0.0291. Interpret the P-value. Would your conclusion change if a 5% significance level was used instead of a 1% significance level? Explain your reasoning.
Yes; because the P-value of 0.0291< α = 0.05, we would reject H0. We have convincing evidence that the true proportion differs from 0.72.
A student performs a test of H0: p= 0.75 versus Ha: p< 0.75 at the α=0.05 significance level and gets a P-value of 0.22. The student writes: “Because the P-value is large, we accept H0. The data provide convincing evidence that the null hypothesis is true.” Explain what is wrong with this conclusion.
It is never correct to “accept the null hypothesis.” If the P-value is large, the data do not provide convincing evidence that the alternative hypothesis is true. Lacking evidence for the alternative hypothesis does not provide convincing evidence that the null hypothesis is true.
A student performs a test of H0: μ =12 versus Ha μ ≠ 12 at the α=0.05 significance level and gets a P-value of 0.01. The student writes: “Because the P -value is small, we reject H0. The data prove that Ha is true.” Explain what is wrong with this conclusion.
The data never “prove” a hypothesis true, no matter how large or small the P-value. There can always be error.
The mean weight of loaves of bread produced at the bakery where you work is supposed to be 1 pound. You are the supervisor of quality control at the bakery, and you are concerned that new employees are producing loaves that are too light. Suppose you weigh an SRS of bread loaves and find that the mean weight is 0.975 pound. State appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest.
H0: µ = 1, Ha: µ < 1, where µ = the true mean weight (in pounds) of bread loaves produced.
The mean weight of loaves of bread produced at the bakery where you work is supposed to be 1 pound. You are the supervisor of quality control at the bakery, and you are concerned that new employees are producing loaves that are too light. Suppose you weigh an SRS of bread loaves and find that the mean weight is 0.975 pound. Explain why there is some evidence for the alternative hypothesis.
Some evidence for the alternative hypothesis exists because the mean weight of an SRS of bread loaves is only 0.975 pound—less than the suspected mean (1 pound).
The mean weight of loaves of bread produced at the bakery where you work is supposed to be 1 pound. You are the supervisor of quality control at the bakery, and you are concerned that new employees are producing loaves that are too light. Suppose you weigh an SRS of bread loaves and find that the mean weight is 0.975 pound. The P -value for the test is 0.0806. Interpret the P -value.
Assuming the true mean weight of bread loaves is 1 pound, a 0.0806 probability exists of getting a sample mean of 0.975 pound or less by chance in a random sample of loaves.
The mean weight of loaves of bread produced at the bakery where you work is supposed to be 1 pound. You are the supervisor of quality control at the bakery, and you are concerned that new employees are producing loaves that are too light. Suppose you weigh an SRS of bread loaves and find that the mean weight is 0.975 pound. What conclusion would you make at the α 0.01= significance level with a P-value of 0.0806?
Because the P-value of 0.0806 > α = 0.01, we fail to reject H0. We do not have convincing evidence that the true mean weight for all loaves of bread produced is less than 1 pound.
Nationally, the proportion of red cars on the road is 0.12. A statistically minded fan of the Philadelphia Phillies (whose team color is red) wonders if Phillies fans are more likely to drive red cars. One day during a home game, he takes a random sample of 210 cars parked at Citizens Bank Park (the Phillies’ home field), and counts 35 red cars. State appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest.
H0: p = 0.12, Ha: p > 0.12, where p= the true proportion of all cars parked at Phillies home field that are red.
Nationally, the proportion of red cars on the road is 0.12. A statistically minded fan of the Philadelphia Phillies (whose team color is red) wonders if Phillies fans are more likely to drive red cars. One day during a home game, he takes a random sample of 210 cars parked at Citizens Bank Park (the Phillies’ home field), and counts 35 red cars. Explain why there is some evidence for the alternative hypothesis.
Some evidence for the alternative hypothesis exists because the sample proportion of red parked cars is 35/210 = 0.167, which is greater than the national proportion of red cars (0.12).
Nationally, the proportion of red cars on the road is 0.12. A statistically minded fan of the Philadelphia Phillies (whose team color is red) wonders if Phillies fans are more likely to drive red cars. One day during a home game, he takes a random sample of 210 cars parked at Citizens Bank Park (the Phillies’ home field), and counts 35 red cars. The P -value for the test is 0.0187. Interpret the P -value.
Assuming the true proportion of all red cars parked at Phillies home field is 0.12, a 0.0187 probability exists of getting a sample proportion of 35/210 = 0.167 or greater by chance in a random sample of 210 cars.
Nationally, the proportion of red cars on the road is 0.12. A statistically minded fan of the Philadelphia Phillies (whose team color is red) wonders if Phillies fans are more likely to drive red cars. One day during a home game, he takes a random sample of 210 cars parked at Citizens Bank Park (the Phillies’ home field), and counts 35 red cars. What conclusion would you make at the α 0.05= significance level for test with a p-value of 0.0187?
Because the P-value of 0.0187 < α = 0.05, we reject H0. We have convincing evidence that the true proportion of all cars parked at Phillies home field that are red is greater than 0.12.
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 of H0: μ = $85, 000, Ha: μ > $85, 000 where μ is the true mean income in the population of people who live near the restaurant. Describe a Type I error and a Type II error in this setting, and give a possible consequence of each.
Type I: You find convincing evidence that the mean income of all residents near the restaurant exceeds $85,000 when in reality it does not. Consequence: You open in an undesirable location, so your restaurant may go out of business. Type II: You do not find convincing evidence that the mean income of all residents near the restaurant exceeds $85,000 when in reality it does. Consequence: You do not open in a desirable location and lose potential income.
Television networks rely heavily on ratings of TV shows when deciding whether to renew a show for another season. Suppose a network has decided that “Miniature Golf with the Stars” will only be renewed if it can be established that more than 12% of U.S. adults watch the show. A polling company asks a random sample of 2000 U.S. adults if they watch “Miniature Golf with the Stars.” The network uses the data to perform a test of H0: p=0.12, Ha p>0.12 where p is the true proportion of all U.S. adults who watch the show. Describe a Type I error and a Type II error in this setting, and give a possible consequence of each.
Type I: The network finds convincing evidence that the true proportion of U.S. adults who watch “Miniature Golf with the Stars” is greater than 0.12, when it really equals 0.12. Consequence: The show will be renewed when interest does not support it. Type II: The network doesn’t find convincing evidence that the true proportion who watch “Miniature Golf with the Stars” is greater than 0.12, when it really is greater than 0.12. Consequence: The show will not renew when there was a market for it.
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 α=0.05 level of H0 p= 0.22 versus Ha: p< 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. Describe a Type I error and a Type II error 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 hadn’t. A Type II error would be not finding convincing evidence that the proportion of all calls in which first responders took more than 8 minutes to arrive decreased when it really had.
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 α=0.05 level of H0 p= 0.22 versus Ha: p< 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 type of error is more serious in this case? Justify your answer.
A Type I error would be worse because the city would overestimate the ability of the emergency personnel to get to the scene quickly and people may end up dying.
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 α=0.05 level of H0 p= 0.22 versus Ha: p< 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. Do you agree with the manager’s choice of α=0.05? Why or why not?
The probability of a Type I error is α = 0.05; because the consequence may be the difference between life and death, the manager should use a value for a that is lower, such as α = 0.01.
The Environmental Protection Agency (EPA) has determined that safe drinking water should contain at most 1.3 mg/liter of copper, on average. A water supply company is testing water from a new source and collects water in small bottles at each of 30 randomly selected locations. The company performs a test at the α=0.05 significance level of H0: μ =1.3 versus Ha: μ >1.3, where μ is the true mean copper content of the water from the new source. Describe a Type I error and a Type II error in this setting.
A Type I error would be finding convincing evidence that the true mean copper content of the water from the new source is greater than 1.3 mg/liter when it really isn’t. A Type II error would be not finding convincing evidence that the true mean copper content of the water from the new source is greater than 1.3 mg/l when it really is.
The Environmental Protection Agency (EPA) has determined that safe drinking water should contain at most 1.3 mg/liter of copper, on average. A water supply company is testing water from a new source and collects water in small bottles at each of 30 randomly selected locations. The company performs a test at the α=0.05 significance level of H0: μ =1.3 versus Ha: μ >1.3, where μ is the true mean copper content of the water from the new source. Which type of error is more serious in this case? Justify your answer.
A Type II error would be worse because the water would not be safe for drinking, yet would seem safe.
The Environmental Protection Agency (EPA) has determined that safe drinking water should contain at most 1.3 mg/liter of copper, on average. A water supply company is testing water from a new source and collects water in small bottles at each of 30 randomly selected locations. The company performs a test at the α=0.05 significance level of H0: μ =1.3 versus Ha: μ >1.3, where μ is the true mean copper content of the water from the new source. Do you agree with the company’s choice of α=0.05? Why or why not?
Here, we seek to minimize the probability of making a Type II error. As the probability of a Type II error increases, the probability of Type I error decreases; so I would recommend that the company use a value of a greater than 0.05.
Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. The mean time is 18 seconds for one particular maze. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures how long each of 10 mice takes with a loud noise as stimulus. The appropriate hypotheses for the significance test are (a) H0: μ = 18; Ha: µ ≠ 18 (b) H0: μ = 18; Ha: µ > 18 (c) H0: μ > 18; Ha: µ = 18 (d) H0: μ = 18; Ha: µ < 18 (e) H0: x_bar = 18; Ha: x_bar < 18
d
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: p =0.50; Ha: p > 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 (a) finds convincing evidence that a majority of residents supports the tax increase, when in reality there isn’t convincing evidence that a majority supports the increase. (b) finds convincing evidence that a majority of residents supports the tax increase, when in reality at most 50% of city residents support the increase. (c) finds convincing evidence that a majority of residents supports the tax increase, when in reality more than 50% of city residents do support the increase. (d) does not find convincing evidence that a majority of residents supports the tax increase, when in reality more than 50% of city residents do support the increase. (e) does not find convincing evidence that a majority of residents supports the tax increase, when in reality at most 50% of city residents do support the increase.
b
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: p = 0.50; Ha: p > 0.50 where p is the proportion of all city residents who support a 1% increase in the sales tax to fund road repairs. In the sample, p^= 158 / 300 = 0.527. The resulting P-value is 0.18. What is the correct interpretation of this P-value? (a) Only 18% of the city residents support the tax increase. (b) There is an 18% chance that the majority of residents supports the tax increase. (c) Assuming that 50% of residents support the tax increase, there is an 18% probability that the sample proportion would be 0.527 or greater by chance alone. (d) Assuming that more than 50% of residents support the tax increase, there is an 18% probability that the sample proportion would be 0.527 or greater by chance alone. (e) Assuming that 50% of residents support the tax increase, there is an 18% chance that the null hypothesis is true by chance alone.
c
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: p =0.50; Ha: p > 0.50 where p is the proportion of all city residents who support a 1% increase in the sales tax to fund road repairs. Based on the P-value of 0.18, which of the following would be the most appropriate conclusion? (a) Because the P-value is large, we reject H0. There is sufficient evidence that more than 50% of city residents support the tax increase. (b) Because the P-value is large, we fail to reject H0. There is sufficient evidence that more than 50% of city residents support the tax increase. (c) Because the P-value is large, we reject H0. There is sufficient evidence that at most 50% of city residents support the tax increase. (d) Because the P-value is large, we fail to reject H0. There is sufficient evidence that at most 50% of city residents support the tax increase. (e) Because the P-value is large, we fail to reject H0. There is not sufficient evidence that more than 50% of city residents support the tax increase.
e
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: p = 0.80 versus Ha: p < 0.80, where p = the true proportion of all students at Jason’s high school who have a computer at home. Check if the conditions for performing the significance test are met.
Random: We have an SRS of 60 students from a large rural high school. 10%: The sample size (60) is less than 10% of all students at this large high school. Large Counts: np0 = 60(0.80) = 48 ≥ 10 and n(1 - p0) = 60(0.20) = 12 ≥ 10.
A recent report claimed that 13% of students typically walk to school. DeAnna thinks that the proportion is higher than 0.13 at her large elementary school. She surveys a random sample of 100 students and finds that 17 typically walk to school. DeAnna would like to carry out a test at the α= 0.05 significance level of H0: p = 0.13 versus Ha: p > 0.13, where p = the true proportion of all students at her elementary school who typically walk to school. Check if the conditions for performing the significance test are met.
Random: We have a random sample of 100 students from a large elementary school. 10%: The sample size (100) is less than 10% of the students at this large elementary school. Large Counts: np0 = 100(0.13) = 13 ≥ 10 and n(1 - p0) = 100(0.87) = 87 ≥ 10.
The chips project Zenon decided to investigate whether students at his school prefer name-brand potato chips to generic potato chips. He randomly selected 50 of the 400 students at his school and asked each student to try both types of chips, in a random order. Overall, 41 of the 50 students preferred the name-brand chips. Zenon wants to perform a test of H0: p = 0.5 versus Ha: p > 0.5, where p = the true proportion of students at his school who prefer name-brand chips. Can the observations in the sample be viewed as independent? Justify your answer.
No; the students (n = 50) are selected from a finite population (N = 400) without replacement. Because n = 50 is not less than 10% of the population size (400), the observations are not independent.
The chips project Zenon decided to investigate whether students at his school prefer name-brand potato chips to generic potato chips. He randomly selected 50 of the 400 students at his school and asked each student to try both types of chips, in a random order. Overall, 41 of the 50 students preferred the name-brand chips. Zenon wants to perform a test of H0: p = 0.5 versus Ha: p > 0.5, where p = the true proportion of students at his school who prefer name-brand chips. Is the sampling distribution of p^ approximately Normal? Explain your reasoning.
Yes; assuming p = 0.5 is true, the sampling distribution of p^ will be approximately Normal because np0 = 50(0.5) = 25 and n(1 - p0) = 50(1 - 0.5) = 25 are both ≥ 10.
On TV shows that feature singing competitions, contestants often wonder if there is an advantage in performing last. To investigate, researchers selected a random sample of 100 students from their large university and showed each student the audition video of 12 different singers with similar vocal skills. Each student viewed the videos in a random order. We would expect approximately 1/12 of the students to prefer the last singer seen, assuming order doesn’t matter. In this study, 11 of the 100 students preferred the last singer they viewed. The researchers want to perform a test of H0: p =1/12 versus Ha: p > 1/12, where p = the true proportion of students at this university who prefer the singer they last see. Can the observations in the sample be viewed as independent? Justify your answer.
Yes; although the students (n = 100) are selected from a finite population (a large university) without replacement, n = 100, 10% of the population size (all students at the large university).
On TV shows that feature singing competitions, contestants often wonder if there is an advantage in performing last. To investigate, researchers selected a random sample of 100 students from their large university and showed each student the audition video of 12 different singers with similar vocal skills. Each student viewed the videos in a random order. We would expect approximately 1/12 of the students to prefer the last singer seen, assuming order doesn’t matter. In this study, 11 of the 100 students preferred the last singer they viewed. The researchers want to perform a test of H0: p =1/12 versus Ha: p > 1/12, where p = the true proportion of students at this university who prefer the singer they last see. Is the sampling distribution of p^ approximately Normal? Explain your reasoning.
No; assuming that p = 1/12 is true, the sampling distribution of p^ will not be approximately Normal because np0 = 100(1/12) = 8.33 and n(1 - p0) = 100(1 - 1/12) = 91.67 are not both ≥ 10.
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: p = 0.80 versus Ha: p < 0.80, where p = the true proportion of all students at Jason’s high school who have a computer at home. Explain why the sample result gives some evidence for the alternative hypothesis.
The sample result gives some evidence for Ha: p < 0.80 because p^ = 41/60 = 0.683, which is less than 0.80.
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: p = 0.80 versus Ha: p < 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 P -value.
z = (0.683 - 0.80) / sqrt((0.80(0.20) / 60) = -2.27; P-value = 0.0116.
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: p = 0.80 versus Ha: p < 0.80, where p = the true proportion of all students at Jason’s high school who have a computer at home. Based on a test with a P-value of 0.0116. 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.