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A drug company has developed a new vaccine for preventing the flu. The company claims that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults. The power of the test to detect the fact that only 3% of adults who use this vaccine would develop flu using α= 0.05 is 0.9437. Interpret this value.
If the true proportion of adults who will get the flu after using the vaccine is p= 0.03, there is a 0.9437 probability that the researchers will find convincing evidence for Ha: p< 0.05.
A drug company has developed a new vaccine for preventing the flu. The company claims that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults. The power of the test to detect the fact that only 3% of adults who use this vaccine would develop flu using α= 0.05 is 0.9437. Explain two ways that you could increase the power of the test.
The power could be increased by increasing the sample size or increasing the significance level α.
A drug company has developed a new vaccine for preventing the flu. The company claims that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults. Of the 1000 adults who were given the vaccine, 43 got the flu. Do these data provide convincing evidence to support the company’s claim?
S: H0: p= 0.05, Ha: p< 0.05, where p = the true proportion of adults who will get the flu after using the vaccine. We will use α = 0.05. P: One-sample z test for p. Random: We have a random sample of 1000 adults. 10%: The sample size (1000) is less than 10% of the population of adults. Large Counts: np0 = 1000(0.05) = 50 ≥ 10 and n(1 - p0) = 1000(0.95) = 950 ≥ 10. D: z=-1.02; P-value = 0.1539. C: Because the P-value of 0.1539 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of adults who will get the flu after using the vaccine is less than 0.05.
An American roulette wheel has 18 red slots among its 38 slots. To test if a particular roulette wheel is fair, you spin the wheel 50 times and the ball lands in a red slot 31 times. The resulting P -value is 0.0384. Interpret the P -value.
Assuming the roulette wheel is fair, there is a 0.0384 probability that we would get a sample proportion of reds (p^ = 31/50) at least this different from the expected proportion of reds (18/38) by chance alone.
An American roulette wheel has 18 red slots among its 38 slots. To test if a particular roulette wheel is fair, you spin the wheel 50 times and the ball lands in a red slot 31 times. The resulting P -value is 0.0384. What conclusion would you make at the α= 0.05 level?
Because the P-value of 0.0384 < α = 0.05, we reject H0. We have convincing evidence that the true proportion of reds is not equal to 18/38 = 0.474.
An American roulette wheel has 18 red slots among its 38 slots. To test if a particular roulette wheel is fair, you spin the wheel 50 times and the ball lands in a red slot 31 times. The resulting P -value is 0.0384. The casino manager uses your data to produce a 99% confidence interval for p and gets (0.44, 0.80). He says that this interval provides convincing evidence that the wheel is fair. How do you respond?
Because 18/38 = 0.474 is one of the plausible values in the interval, this interval does not provide convincing evidence that the wheel is unfair. But it also does not prove that the wheel is fair; there are many plausible values in the interval that are not equal to 18/38. This conclusion is inconsistent with the conclusion in part (b) because the manager used a 99% CI, which is equivalent to a test using α = 0.01. If the manager had used a 95% CI, 18/38 would not be considered a plausible value.
AZT was the first drug that seemed effective in delaying the onset of AIDS. Evidence of AZT’s effectiveness came from a large randomized comparative experiment. The subjects were 870 volunteers who were infected with HIV (the virus that causes AIDS), but who did not yet have AIDS. The study assigned 435 of the subjects at random to take 500 milligrams of AZT each day and another 435 sub-jects to take a placebo. At the end of the study, 38 of the placebo subjects and 17 of the AZT subjects had developed AIDS. If the results of the study are statistically significant, is it reasonable to conclude that AZT is the cause of the decrease in the proportion of people like these who will develop AIDS?
Yes, if the results are statistically significant, we have reason to conclude that AZT is the cause of the decrease in the proportion of people like these who will develop AIDS because the subjects were randomly assigned to the treatment and control groups in this randomized comparative experiment.
AZT was the first drug that seemed effective in delaying the onset of AIDS. Evidence of AZT’s effectiveness came from a large randomized comparative experiment. The subjects were 870 volunteers who were infected with HIV (the virus that causes AIDS), but who did not yet have AIDS. The study assigned 435 of the subjects at random to take 500 milligrams of AZT each day and another 435 sub-jects to take a placebo. At the end of the study, 38 of the placebo subjects and 17 of the AZT subjects had developed AIDS. Do the data provide convincing evidence at the α= 0.05 level that taking AZT lowers the proportion of infected people like the ones in this study who will develop AIDS in a given period of time?
S: H0: p1 - p2 = 0, Ha: p1 - p2 < 0, where p1 = true proportion of patients like these who take AZT and develop AIDS and p2 = true proportion of patients like these who take placebo and develop AIDS; α = 0.05. P: Two-sample z test for p1 - p2. Random: The subjects were assigned at random to take AZT or a placebo. Large Counts: 27.405, 407.595, 27.405, 407.595 are ≥ 10. D: z = -2.91; P-value = 0.0018. C: Because the P-value of 0.0018 < α = 0.05, we reject H0 . We have convincing evidence that taking AZT lowers the proportion of patients like these who develop AIDS compared to a placebo.
An opinion poll asks a random sample of adults whether they favor banning ownership of handguns by private citizens. A commentator believes that more than half of all adults favor such a ban. The null and alternative hypotheses you would use to test this claim are (a) H0: p^ = 0.5; Ha: p^ > 0.5. (b) H0: p = 0.5; Ha: p > 0.5. (c) H0: p = 0.5; Ha: p < 0.5. (d) H0: p = 0.5; Ha: p ≠ 0.5. (e) H0: p > 0.5; Ha: p = 0.5.
b
The power takeoff driveline on tractors used in agriculture can be a serious hazard to operators of farm equipment. The driveline is covered by a shield in new tractors, but the shield is often missing on older tractors. Two types of shields are the bolt-on and the flip-up. It was believed that the bolt-on shield was perceived as a nuisance by the operators and deliberately removed, but the flip-up shield is easily lifted for inspection and maintenance and may be left in place. In a study by the U.S. National Safety Council, random samples of older tractors with both types of shields were taken to see what proportion of shields were removed. Of 183 tractors designed to have bolt-on shields, 35 had been removed. Of the 136 tractors with flip-up shields, 15 were removed. We wish to perform a test of H0: pB = pF versus Ha: pB > pF where pB and pF are the proportions of all tractors with the bolt-on and flip-up shields removed, respectively. Which of the following is not a condition for performing the significance test? (a) Both populations are Normally distributed. (b) The data come from two independent samples. (c) Both samples were chosen at random. (d) The expected counts of successes and failures are large enough to use Normal calculations. (e) Both populations are more than 10 times the corresponding sample sizes.
a
A significance test allows you to reject a null hypothesis H0 in favor of an alternative hypothesis Ha at the 5% significance level. What can you say about significance at the 1% level? (a) H0 can be rejected at the 1% significance level. (b) There is insufficient evidence to reject H0 at the 1% significance level. (c) There is sufficient evidence to accept H0 at the 1% significance level. (d) Ha can be rejected at the 1% significance level. (e) The answer can’t be determined from the information given.
e
A random sample of 100 likely voters in a small city produced 59 voters in favor of Candidate A. The value of the standardized test statistic for performing a test of H0: p = 0.5 versus Ha: p > 0.5 following? (a) z = (0.59-0.5) / sqrt((0.59(0.41))/100) (b) z = (0.59-0.5) / sqrt((0.5(0.5))/100) (c) z = (0.5-0.59) / sqrt((0.59(0.41))/100) (d) z = (0.5-0.59) / sqrt((0.5(0.5))/100) (e) z = (0.59-0.5) / sqrt(100)
b
A researcher claims to have found a drug that causes people to grow taller. The coach of the basketball team at Brandon University has expressed interest but demands evidence. Over 1000 Brandon students volunteer to participate in an experiment to test this new drug. Fifty of the volunteers are randomly selected, their heights are measured, and they are given the drug. Two weeks later, their heights are measured again. The power of the test to detect an average increase in height of 1 inch could be increased by (a) using only the 12 members of the basketball team in the experiment. (b) using α 0.01= instead ofα= 0.05. (c) using α= 0.05 instead of α= 0.01. (d) giving the drug to 25 randomly selected students instead of 50. (e) using a two-sided test instead of a one-sided test.
c
A 95% confidence interval for the proportion of viewers of a certain reality television show who are over 30 years old is (0.26, 0.35). Suppose the show’s producers want to test the hypothesis H0: p = 0.25 against Ha: p ≠ 0.25. Which of the following is an appropriate conclusion for them to draw at the α= 0.05 significance level? (a) Fail to reject H0; there is convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old equals 0.25. (b) Fail to reject H0; there is not convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old differs from 0.25. (c) Reject H0; there is not convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old differs from 0.25. (d) Reject H0; there is convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old is greater than 0.25. (e) Reject H0; there is convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old differs from 0.25.
e
In a test of H0: p = 0.4 against Ha: p ≠ 0.4, a random sample of size 100 yields a standardized test statistic of z = 1.28. Which of the following is closest to the P-value for this test? (a) 0.90 (b) 0.40 (c) 0.05 (d) 0.20 (e) 0.10
d
Bags of a certain brand of tortilla chips claim to have a net weight of 14 ounces. A representative of a consumer advocacy group wishes to see if there is convincing evidence that the mean net weight is less than advertised and so intends to test the hypotheses H0: μ =14; Ha: μ < 14. A Type I error in this situation would mean concluding that the bags (a) are being underfilled when they really aren’t. (b) are being underfilled when they really are. (c) are not being underfilled when they really are. (d) are not being underfilled when they really aren’t. (e) are being overfilled when they are really underfilled.
a
Conference organizers wondered whether posting a sign that says “Please take only one cookie” would reduce the proportion of conference attendees who take multiple cookies from the snack table during a break. To find out, the organizers randomly assigned 212 attendees to take their break in a room where the snack table had the sign posted, and 189 attendees to take their break in a room where the snack table did not have a sign posted. In the room without the sign posted, 24.3% of attendees took multiple cookies. In the room with the sign posted, 17.0% of attendees took multiple cookies. Is this decrease in proportions statistically significant at the α = 0.05 level? (a) No. The P-value is 0.034. (b) No. The P-value is 0.068. (c) Yes. The P-value is 0.034. (d) Yes. The P-value is 0.068. (e) Cannot be determined from the information given.
c
A software company is trying to decide whether to pro-duce an upgrade of one of its programs. Customers would have to pay $100 for the upgrade. For the upgrade to be profitable, the company must sell it to more than 20% of their customers. The company will survey a random sample of 60 customers about this issue. Which would be a more serious mistake in this setting—a Type I error or a Type II error? Justify your answer.
Type I: Finding convincing evidence that more than 20% of customers would pay for the upgrade, when in reality they would not. Type II: Finding convincing evidence that more than 20% of customers would pay for the upgrade, when in reality more than 20% would. For the company, a Type I error is worse because they would go ahead with the upgrade and lose money.
A software company is trying to decide whether to pro-duce an upgrade of one of its programs. Customers would have to pay $100 for the upgrade. For the upgrade to be profitable, the company must sell it to more than 20% of their customers. The company will survey a random sample of 60 customers about this issue. In the company’s survey, 16 customers are willing to pay $100 each for the upgrade. Do the sample data give convincing evidence that more than 20% of the company’s customers are willing to purchase the upgrade? Carry out an appropriate test at the significance level.
S: H0: p= 0.20, Ha: p≠ 0.20, where p = the true proportion of customers who would pay $100 for the upgrade using α = 0.05. P: One-sample z test for p. Random: We have a random sample of 60 customers. 10%: The sample size (60) is less than 10% of this company’s customers. Large Counts: np0 = 60(0.20) = 12 ≥ 10 and n(1 - p0) = 60(0.8) = 48 ≥ 10. D: z= 1.29; P-value = 0.0985. C: Because the P-value of 0.0985 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of customers who would pay $100 for the upgrade is greater than 0.20.
A random sample of 100 of last year’s model of a certain popular car found that 20 had a specific minor defect in the brakes. The automaker adjusted the production process to try to reduce the proportion of cars with the brake problem. A random sample of 350 of this year’s model found that 50 had the minor brake defect. Was the company’s adjustment successful? Carry out an appropriate test using α= 0.05 to support your answer.
STATE: H0: p1 - p2 = 0, Ha: p1 - p2 > 0, where p1 = true proportion of cars that have the brake defect in last year’s model and p2 = true proportion of cars with the brake defect in this year’s model; α = 0.05. PLAN: Two-sample z test for p1 - p2. Random: Independent random samples. 10%: n1 = 100 < 10% of last year’s model and n2 = 350 < 10% of this year’s model. Large Counts: 15.6, 84.4, 54.6, and 295.4 are ≥ 10. DO: z= 1.39; P-value = 0.0822. CONCLUDE: Because the P-value of 0.0822 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of brake defects is smaller in this year’s model compared to last year’s model.
A random sample of 100 of last year’s model of a certain popular car found that 20 had a specific minor defect in the brakes. The automaker adjusted the production process to try to reduce the proportion of cars with the brake problem. A random sample of 350 of this year’s model found that 50 had the minor brake defect. Suppose that the proportion of cars with the defect was reduced by 0.10 from last year to this year. The power of the test to detect this decrease is 0.72. Interpret this value.
If the true proportion of cars with the defect has decreased by 0.10 from last year to this year, there is a 0.72 probability that the automaker will find convincing evidence for Ha: p1 - p2 > 0.
A random sample of 100 of last year’s model of a certain popular car found that 20 had a specific minor defect in the brakes. The automaker adjusted the production process to try to reduce the proportion of cars with the brake problem. A random sample of 350 of this year’s model found that 50 had the minor brake defect. Other than increasing the sample sizes, identify one way of increasing the power of the test.
Besides increasing the sample sizes, the power of the test can be increased by using a larger value for a, the significance level of the test.
What critical value *t should be used for a confidence interval for the population mean for the following situation? A 95% confidence interval based on n =10 randomly selected observations
df = 9, t* = 2.262; Tech: invT(area: 0.025, df: 9) 522.262, so t*= 2.262.
What critical value *t should be used for a confidence interval for the population mean for the following situation? A 99% confidence interval from an SRS of 20 observations
df = 19, t* = 2.861; Tech: invT(area: 0.005, df: 19) 522.861, so t* = 2.861.
What critical value *t should be used for a confidence interval for the population mean for the following situation? A 90% confidence interval based on a random sample of 77 individuals
df = 60, t* = 1.671; Tech: invT(area: 0.05, df: 76) 521.665, so t* = 1.665.
What critical value *t should be used for a confidence interval for the population mean for the following situation? A 90% confidence interval based on n =12 randomly selected observations
df = 11, t* = 1.796; Tech: invT(area: 0.05, df: 11) 521.796, so t* = 1.796.
What critical value *t should be used for a confidence interval for the population mean for the following situation? A 95% confidence interval from an SRS of 30 observations
df = 29, t* = 2.045; Tech: invT(area: 0.025, df: 29) 522.045, so t*= 2.045.
What critical value *t should be used for a confidence interval for the population mean for the following situation? A 99% confidence interval based on a random sample of size 58
df = 50, t* = 2.678’ Tech: invT(area: 0.005, df: 57) 522.665, so t* = 2.665.
Determine if the conditions are met for constructing a confidence interval for the population mean in the following setting: How much time do students at your school spend on the Internet? You collect data from the 32 members of your AP® Statistics class and calculate the mean amount of time that these students spent on the Internet yesterday.
Random: No; the AP® Statistics students are not a random sample of all students. 10%: 32 < 10% of all students in the school. (Condition met if the number of students at the school is at least 320). Normal/Large Sample: Yes; n= 32 ≥ 30.
Determine if the conditions are met for constructing a confidence interval for the population mean in the following setting: We want to estimate the average age at which U.S. presidents have died. So we obtain a list of all U.S. presidents who have died and their ages at death.
Random: No; a list of all U.S. presidents is not a random sample. 10%: No; the n presidents who have died is not less than 10% of all presidents that have died. Normal/Large Sample: Yes; n≥ 30.
Blood pressure A medical study finds that x= 114.9 and sx = 9.3 for the seated systolic blood pressure of 27 randomly selected adults. What is the standard error of the mean? Interpret this value in context.
SEx = sx / sqrt(n) = 9.3 / sqrt(27) = 1.7898. In many random samples of size 27, the sample mean blood pressure will typically vary by about 1.7898 from the population mean blood pressure.
A study of commuting times reports the travel times to work of a random sample of 20 employed adults in New York State. The mean is x = 31.25 minutes and the standard deviation is sx = 21.88 minutes. What is the standard error of the mean? Interpret this value in context.
SEx = sx / sqrt(n) = 21.88 / sqrt(20) = 4.8925. In many random samples of size 20, the sample mean commute time will typically vary by about 4.8925 from the population mean commute time.
Breast-feeding mothers secrete calcium into their milk. Some of the calcium may come from their bones, so mothers may lose bone mineral. Researchers measured the percent change in bone mineral content (BMC) of the spines of 47 randomly selected mothers during three months of breast-feeding. The mean change in BMC was −3.587% and the standard deviation was 2.506%. Construct and interpret a 99% confidence interval to estimate the mean percent change in BMC in the population of breast-feeding mothers.
S: µ = the true mean percent change in BMC for breast-feeding mothers. P: One-sample t interval. Random: The mothers were randomly selected. 10%: 47 is less than 10% of all breast-feeding mothers. Normal/Large Sample: n= 47 ≥ 30. D: df = 40 (24.575, 22.599); Tech: (24.569, 22.605) with df = 40. C: We are 99% confident that the interval from 24.569 to 22.605 captures µ = the true mean percent change in BMC for breast-feeding mothers.
Breast-feeding mothers secrete calcium into their milk. Some of the calcium may come from their bones, so mothers may lose bone mineral. Researchers measured the percent change in bone mineral content (BMC) of the spines of 47 randomly selected mothers during three months of breast-feeding. The mean change in BMC was −3.587% and the standard deviation was 2.506%. A 99% confidence interval was found to be (-4.569, -2.599). Do these data give convincing evidence that nursing mothers lose bone mineral, on average? Explain your answer.
Because all the plausible values in the interval are negative (indicating bone loss), the data give convincing evidence.
The Trial Urban District Assessment (TUDA) is a government-sponsored study of student achievement in large urban school districts. TUDA gives a reading test scored from 0 to 500. A score of 243 is a “basic” reading level and a score of 281 is “proficient.” Scores for a random sample of 1470 eighth-graders in Atlanta had a mean of 240 with standard deviation of 42.17. Construct and interpret a 99% confidence interval for the mean reading test score of all Atlanta eighth-graders.
S: µ = the true mean reading test score for all Atlanta eighth-graders. P: One-sample t interval. Random: Students were randomly selected. 10%: 1470 is less than 10% of eighth-graders in Atlanta. Normal/Large Sample: n= 1470 ≥ 30. D: df = 1469 (237.16, 242.84); Tech: (237.16, 242.84) with df = 1469. C: We are 99% confident that the interval from 237.16 to 242.84 captures µ = the true mean reading test score for all Atlanta eighth-graders.
The Trial Urban District Assessment (TUDA) is a government-sponsored study of student achievement in large urban school districts. TUDA gives a reading test scored from 0 to 500. A score of 243 is a “basic” reading level and a score of 281 is “proficient.” Scores for a random sample of 1470 eighth-graders in Atlanta had a mean of 240 with standard deviation of 42.17. A 99% confidence interval was found to be (237.16, 242.84). Is there convincing evidence that the mean reading test score for all Atlanta eighth-graders is less than the basic level? Explain your answer.
Because all the plausible values in the interval are less than 243, there is convincing evidence.
Ann and Tori wanted to estimate the average weight of an Oreo cookie to determine if it was less than advertised (34 grams for 3 cookies). They selected a random sample of 36 cookies and found the weight of each cookie (in grams). The mean weight was x¯ =11.3921 grams with a standard deviation of sx = 0.0817 grams. Construct and interpret a 90% confidence interval for the true mean weight of an Oreo cookie.
S: µ = the true mean weight of an Oreo cookie. P: One-sample t interval. Random: The cookies were randomly selected. 10%: 36 is less than 10% of all Oreo cookies. Normal/Large Sample: n= 36 ≥ 30. D: df = 30 (11.369, 11.4152); Tech: (11.369, 11.415) with df = 30. C: We are 90% confident that the interval from 11.369 to 11.4152 captures µ = the true mean weight for all Oreo cookies.
Ann and Tori wanted to estimate the average weight of an Oreo cookie to determine if it was less than advertised (34 grams for 3 cookies). They selected a random sample of 36 cookies and found the weight of each cookie (in grams). The mean weight was x¯ =11.3921 grams with a standard deviation of sx = 0.0817 grams. A 90% confidence interval was found to be (11.369, 11.4152). Interpret the confidence level.
If we were to select many random samples of size 36 from the population of all Oreo cookies and construct a 90% confidence interval using each sample, about 90% of the intervals would capture the true mean weight of an Oreo cookie.
Fruit flies are used frequently in genetic research because of their quick reproductive cycle. The length of the thorax (in millimeters) was measured for each fly in a random sample of 49 male fruit flies. The mean length was x¯ = 0.8004 mm, with a standard deviation of sx = 0.0782 mm. Construct and interpret a 90% confidence interval for the true mean thorax length of a male fruit fly.
S: µ = the true mean thorax length of a male fruit fly. P: One-sample t interval. Random: Flies were randomly selected. 10%: 49 is less than 10% of all male fruit flies. Normal/ Large Sample: n= 49 ≥ 30. D: df = 48 (0.7816, 0.8192); Tech: (0.7817, 0.8191) with df = 48. C: We are 90% confident that the interval from 0.7817 to 0.8191 captures µ = the true mean thorax length for all male fruit flies.
Fruit flies are used frequently in genetic research because of their quick reproductive cycle. The length of the thorax (in millimeters) was measured for each fly in a random sample of 49 male fruit flies. The mean length was x¯ = 0.8004 mm, with a standard deviation of sx = 0.0782 mm. A 90% confidence interval was found to be (0.7816, 0.8192). Interpret the confidence level.
If we were to select many random samples of size 49 from the population of all male fruit flies and construct a 90% confidence interval using each sample, about 90% of the intervals would capture the true mean thorax length.
Melissa and Madeline love pepperoni pizza, but sometimes they are disappointed with the small number of pepperonis on their pizza. To investigate, they went to their favorite pizza restaurant at 10 random times during the week and ordered a large pepperoni pizza. 8 Here are the number of pepperonis on each pizza: 47 36 25 37 46 36 49 32 32 34 Construct and interpret a 95% confidence interval for the true mean number of pepperonis on a large pizza at this restaurant.
S: m 5 the true mean number of pepperonis on a large pizza at this restaurant. P: One-sample t interval. Random: Pizzas were randomly selected. 10%: 10 is less than 10% of all pepperoni pizzas made at this restaurant. Normal/ Large Sample: The dotplot doesn’t show any outliers or strong skewness. D: x¯= 37.4, sx= 7.662, and n= 10; df = 9 (31.919, 42.881); Tech: (31.919, 42.881) with df = 9. C: We are 95% confident that the interval from 31.919 to 42.881 captures µ = the true mean.
Carly and Maysem plan to be preschool teachers after they graduate from college. To prepare for snack time, they want to know the mean number of goldfish crackers in a bag of original-flavored goldfish. To estimate this value, they randomly selected 12 bags of original-flavored goldfish and counted the number of crackers in each bag. 9 Here are their data: 317 330 325 323 332 337 324 342 330 349 335 333 Construct and interpret a 95% confidence interval for the true mean number of crackers in a bag of original-flavored goldfish.
S: µ = the true mean number of crackers in a bag of original goldfish. P: One-sample t interval. Random: Bags of goldfish were randomly selected. 10%: 12 is less than 10% of all bags of original-flavored goldfish. Normal/Large Sample: The dotplot doesn’t show any outliers or strong skewness. D: x¯= 331.417, sx= 8.775, and n= 12; df = 11 (325.842, 336.992); Tech: (325.84, 336.99) with df = 11. C: We are 95% confident that the interval from 325.84 to 336.99 captures µ = the true mean.
Melissa and Madeline love pepperoni pizza, but sometimes they are disappointed with the small number of pepperonis on their pizza. To investigate, they went to their favorite pizza restaurant at 10 random times during the week and ordered a large pepperoni pizza. 8 Here are the number of pepperonis on each pizza: 47 36 25 37 46 36 49 32 32 34. A 95% confidence interval was found to be (31.919, 42.881). Explain why it was necessary to inspect a graph of the sample data when checking the Normal/Large Sample condition.
It was necessary because the sample size was not large (<30). When the sample size is less than 30, we must assume the population is Normally distributed.
Melissa and Madeline love pepperoni pizza, but sometimes they are disappointed with the small number of pepperonis on their pizza. To investigate, they went to their favorite pizza restaurant at 10 random times during the week and ordered a large pepperoni pizza. 8 Here are the number of pepperonis on each pizza: 47 36 25 37 46 36 49 32 32 34. A 95% confidence interval was found to be (31.919, 42.881). According to the manager of the restaurant, there should be an average of 40 pepperonis on a large pizza. Based on the interval, is there convincing evidence that the average number of pepperonis is less than 40? Explain your answer.
The value 40 is a plausible value found within the confidence interval, so we do not have convincing evidence that the average number of pepperonis is less than 40.
Melissa and Madeline love pepperoni pizza, but sometimes they are disappointed with the small number of pepperonis on their pizza. To investigate, they went to their favorite pizza restaurant at 10 random times during the week and ordered a large pepperoni pizza. 8 Here are the number of pepperonis on each pizza: 47 36 25 37 46 36 49 32 32 34. A 95% confidence interval was found to be (31.919, 42.881). Explain two ways that Melissa and Madeline could reduce the margin of error of their estimate. Why might they object to these changes?
Melissa and Madeline could increase the sample size or decrease the confidence level. They may not want to increase the sample size because it would be expensive and they also may not want to eat that much pizza. They may not want to decrease the confidence level because it is desirable to have a high degree of confidence that the population parameter has been captured by the interval.
Carly and Maysem plan to be preschool teachers after they graduate from college. To prepare for snack time, they want to know the mean number of goldfish crackers in a bag of original-flavored goldfish. To estimate this value, they randomly selected 12 bags of original-flavored goldfish and counted the number of crackers in each bag. 9 Here are their data: 317 330 325 323 332 337 324 342 330 349 335 333. A 95% confidence interval was found to be (325.842, 336.992). Explain why it was necessary to inspect a graph of the sample data when checking the Normal/Large Sample condition.
It was necessary because the sample size was not large (<30). When the sample size is less than 30, we must assume the population is Normally distributed.
Carly and Maysem plan to be preschool teachers after they graduate from college. To prepare for snack time, they want to know the mean number of goldfish crackers in a bag of original-flavored goldfish. To estimate this value, they randomly selected 12 bags of original-flavored goldfish and counted the number of crackers in each bag. 9 Here are their data: 317 330 325 323 332 337 324 342 330 349 335 333. A 95% confidence interval was found to be (325.842, 336.992). According to the packaging, there are supposed to be 330 goldfish in each bag of crackers. Based on the interval, is there convincing evidence that the average number of goldfish is less than 330? Explain your answer.
The value 330 is a plausible value found within the confidence interval, so we do not have convincing evidence that the average number of goldfish is less than 330.
Carly and Maysem plan to be preschool teachers after they graduate from college. To prepare for snack time, they want to know the mean number of goldfish crackers in a bag of original-flavored goldfish. To estimate this value, they randomly selected 12 bags of original-flavored goldfish and counted the number of crackers in each bag. 9 Here are their data: 317 330 325 323 332 337 324 342 330 349 335 333. A 95% confidence interval was found to be (325.842, 336.992). Explain two ways that Carly and Maysem could reduce the margin of error of their estimate. Why might they object to these changes?
Carly and Maysem could increase the sample size or decrease the confidence level. They may not want to increase the sample size because that would become expensive. They may not want to decrease the confidence level because it is desirable to have a high degree of confidence that the population parameter has been captured by the interval.
The body mass index (BMI) of all American young women is believed to follow a Normal distribution with a standard deviation of about 7.5. How large a sample would be needed to estimate the mean BMI μ in this population to within ±1 with 99% confidence?
Solving n≥ (2.576(7.5) / 1)^2 = 373.26. Select an SRS of 374 women.
The SAT again High school students who take the SAT Math exam a second time generally score higher than on their first try. Past data suggest that the score increase has a standard deviation of about 50 points. How large a sample of high school students would be needed to estimate the mean change in SAT score to within 2 points with 95% confidence?
Solving n≥ (1.96(50) / 2)^2 = 2401. Select a random sample of 2401 students.
Writers in some fields summarize data by giving x and its standard error rather than x¯ and sx. Biologists studying willow plants in Yellowstone National Park reported their results in a table with columns labeled x¯ ± SE. The table entry for the heights of willow plants (in centimeters) in one region of the park was 61.55 ± 19.03. The researchers measured a total of 23 plants. Find the sample standard deviation sx for these measurements.
SEx= 19.03 = sx/sqrt(n) = sx/sqrt(23), so sx= 19.03sqrt(23) = 91.26.
Writers in some fields summarize data by giving x and its standard error rather than x¯ and sx. Biologists studying willow plants in Yellowstone National Park reported their results in a table with columns labeled x¯ ± SE. The table entry for the heights of willow plants (in centimeters) in one region of the park was 61.55 ± 19.03. The researchers measured a total of 23 plants. A hasty reader believes that the interval given in the table is a 95% confidence interval for the mean height of willow plants in this region of the park. Find the actual confidence level for the given interval.
Because the researchers are using a critical value of t* = 1. With df = 23 - 1 = 22, the area between t=-1 and t= 1 is approximately tcdf(lower: 21, upper: 1, df: 22) = 0.67. So, the confidence level is 67%.
When two lights that are close together blink alternately, we “see” one light moving back and forth if the time between blinks is short. What is the longest interval of time between blinks that preserves the illusion of motion? Ask subjects to turn a knob that slows the blinking until they “see” two lights rather than one light moving. A report gives the results in the form “mean plus or minus the standard error of the mean.” Data for 12 subjects are summarized as 251 ± 45 (in milliseconds). Find the sample standard deviation sx for these measurements.
SEx= 15 = sx/sqrt(n) = sx/sqrt(12), so sx= 45sqrt(12) = 155.88.
When two lights that are close together blink alternately, we “see” one light moving back and forth if the time between blinks is short. What is the longest interval of time between blinks that preserves the illusion of motion? Ask subjects to turn a knob that slows the blinking until they “see” two lights rather than one light moving. A report gives the results in the form “mean plus or minus the standard error of the mean.” Data for 12 subjects are summarized as 251 ± 45 (in milliseconds). A hasty reader believes that the interval given in the report is a 95% confidence interval for the population mean. Find the actual confidence level for the given interval.
Because the researchers are using a critical value of t* = 1. With df = 12 - 1 = 11, the area between t=-1 and t= 1 is approximately tcdf(lower: 21, upper: 1, df: 11) = 0.66. So, the confidence level is 66%.
One reason for using a t distribution instead of the standard Normal distribution to find critical values when calculating a level C confidence interval for a population mean is that (a) z can be used only for large samples. (b) z requires that you know the population standard deviation σ. (c) z requires that you can regard your data as an SRS from the population. (d) z requires that the sample size is less than 10% of the population size. (e) a z critical value will lead to a wider interval than a t critical value.
b
You have an SRS of 23 observations from a large population. The distribution of sample values is roughly symmetric with no outliers. What critical value would you use to obtain a 98% confidence interval for the mean of the population? (a) 2.177 (b) 2.183 (c) 2.326 (d) 2.500 (e) 2.508
e
A quality control inspector will measure the salt content (in milligrams) in a random sample of bags of potato chips from an hour of production. Which of the following would result in the smallest margin of error in estimating the mean salt content μ? (a) 90% confidence; n = 25 (b) 90% confidence; n =50 (c) 95% confidence; n = 25 (d) 95% confidence; n =50 (e) n =100 at any confidence level
b
Scientists collect data on the blood cholesterol levels (milligrams per deciliter of blood) of a random sample of 24 laboratory rats. A 95% confidence interval for the mean blood cholesterol level μ is 80.2 to 89.8. Which of the following would cause the most worry about the validity of this interval? (a) There is a clear outlier in the data. (b) A stemplot of the data shows a mild right skew. (c) You do not know the population standard deviation σ. (d) The population distribution is not exactly Normal. (e) None of these are a problem when using a t interval.
a
How many pairs of shoes do teenagers have? To find out, a group of AP® Statistics students conducted a survey. They selected a random sample of 20 female students and a separate random sample of 20 male students from their school. Then they recorded the number of pairs of shoes that each student reported having. Here are their data: Males 14 7 6 5 12 38 8 7 10 10 10 11 4 5 22 7 5 10 35 7 Females 50 26 26 31 57 19 24 22 23 38 13 50 13 34 23 30 49 13 15 51. Let μ1 = the true mean number of pairs of shoes that male students at the school have and μ2 = the true mean number of pairs of shoes that female students at the school have. Check if the conditions for calculating a confidence interval for μ1- μ2 are met.
Random: Met because these are two independent random samples. 10%: Met because 20 < 10% of all males at the school and 20 < 10% of all females at the school. Normal/Large Sample: Not met; there are fewer than 30 observations in each group and a dotplot for males shows several outliers.
For their final project, a group of AP® Statistics students wanted to compare the texting habits of males and females. They asked a random sample of students from their school to record the number of text messages sent and received over a 2-day period. Here are their data: Males 127 44 28 83 0 6 78 6 5 213 73 20 214 28 11 Females 112 203 102 54 379 305 179 24 127 65 41 27 298 6 130 0 Let μ1 = the true mean number of texts sent by male students at the school and μ2 = the true mean number of texts sent by female students at the school. Check if the conditions for calculating a confidence interval for μ1-μ2 are met.
Random: Met; even though the data came from a single random sample, it is reasonable to consider the two samples independent because knowing the response of a male shouldn’t predict the response of a female. 10%: Met because 15 < 10% of all males at the school and 16 < 10% of all females at the school. Normal/Large Sample: Not met; there are fewer than 30 observations in each group and a dotplot for males shows several outliers
College financial aid offices expect students to use summer earnings to help pay for college. But how large are these earnings? One large university studied this question by asking a random sample of 1296 students who had summer jobs how much they earned. The financial aid office separated the responses into two groups based on gender, so these can be viewed as independent samples. Here are the data in summary form: Males: n=675, x¯=$1884.52, sx=1360.39 Females n=621, x¯=$1368.37, sx= $1037.46. How can you tell from the summary statistics that the distribution of earnings in each group is strongly skewed to the right? The use of two-sample t procedures is still justified. Why?
Skewed to the right because the earnings cannot be negative, yet the standard deviation is almost as large as the distance between the mean and 0. The use of the two-sample t procedures is justified because the sample sizes are both very large (675 ≥ 30 and 621 ≥ 30).
College financial aid offices expect students to use summer earnings to help pay for college. But how large are these earnings? One large university studied this question by asking a random sample of 1296 students who had summer jobs how much they earned. The financial aid office separated the responses into two groups based on gender, so these can be viewed as independent samples. Here are the data in summary form: Males: n=675, x¯=$1884.52, sx=1360.39 Females n=621, x¯=$1368.37, sx= $1037.46. Construct and interpret a 90% confidence interval for the difference between the true mean summer earnings of male and female students at this university.
S: µ1 = true mean summer earnings of male students and µ2 = true mean for female students. P: Two-sample t interval for µ1 - µ2. Random: It is reasonable to consider the two samples independent because knowing the response of a male shouldn’t predict the response of a female. 10%: n1 = 675 < 10% of male students and n2 = 621 < 10% of female students. Normal/Large Sample: n1 = 675 ≥ 30 and n2 = 621 ≥ 30. D: Using df = 1249.21, (413.62, 634.64); using df = 620, (413.52, 634.72). C: We are 90% confident the interval from $413.62 to $634.64 captures µ1 - µ2 = true difference in mean summer earnings of male and female students at this university.
College financial aid offices expect students to use summer earnings to help pay for college. But how large are these earnings? One large university studied this question by asking a random sample of 1296 students who had summer jobs how much they earned. The financial aid office separated the responses into two groups based on gender, so these can be viewed as independent samples. Here are the data in summary form: Males: n=675, x¯=$1884.52, sx=1360.39 Females n=621, x¯=$1368.37, sx= $1037.46. Interpret the 90% confidence level in the context of this study.
If we took many random samples of 675 males and 621 females from this university and each time constructed a 90% confidence interval in this same way, about 90% of the resulting intervals would capture the true difference in mean earnings for males and females.
In a study of heart surgery, one issue was the effect of drugs called beta blockers on the pulse rate of patients during surgery. The available subjects were randomly assigned into two groups. One group received a beta blocker; the other group received a placebo. The pulse rate of each patient at a critical point during the operation was recorded. Here are the data in summary form: Beta blocker: n=30, x¯=65.2, sx=7.8; Placebo: n=30, x¯=70.3, sx=8.3. The distribution of pulse rate in each group is not Normal. The use of two-sample t procedures is still justified. Why?
Because the sample sizes are both large (30 ≥ 30 and 30 ≥ 30).
In a study of heart surgery, one issue was the effect of drugs called beta blockers on the pulse rate of patients during surgery. The available subjects were randomly assigned into two groups. One group received a beta blocker; the other group received a placebo. The pulse rate of each patient at a critical point during the operation was recorded. Here are the data in summary form: Beta blocker: n=30, x¯=65.2, sx=7.8; Placebo: n=30, x¯=70.3, sx=8.3. Construct and interpret a 99% confidence interval for the difference in mean pulse rates for patients like these who receive a beta blocker or a placebo.
S: µ1 = true mean pulse rate of patients like those in the experiment who take beta-blockers and µ2 = mean pulse rate for those who do not. P: Two-sample t interval for µ1 - µ2. Random: Two groups in a randomized experiment. Normal/Large Sample: n1 = 30 ≥ 30 and n2 = 30 ≥ 30. D: Using df = 57.78, (210.64, 0.44); using df = 29, (210.83, 0.63). C: We are 99% confident the interval from 210.64 to 0.44 captures µ1 - µ2 = true difference in the mean pulse rate of patients receiving a beta blocker during surgery and those who take a placebo.
In a study of heart surgery, one issue was the effect of drugs called beta blockers on the pulse rate of patients during surgery. The available subjects were randomly assigned into two groups. One group received a beta blocker; the other group received a placebo. The pulse rate of each patient at a critical point during the operation was recorded. Here are the data in summary form: Beta blocker: n=30, x¯=65.2, sx=7.8; Placebo: n=30, x¯=70.3, sx=8.3. Interpret the 99% confidence level in the context of this study.
If we took 60 patients and repeated many random assignments of them to the treatments of a beta blocker and a placebo and each time constructed a 99% confidence interval in this same way, about 99% of the resulting intervals would capture the true difference in mean pulse rates for patients given a beta blocker or placebo.
Catherine and Ana wanted to know if student athletes (students on at least one varsity team) have faster reaction times than non-athletes. They took separate random samples of 33 athletes and 30 non-athletes from their school and tested their reaction time using an online reaction test, which measured the time (in seconds) between when a green light went on and the subject pressed a key on the computer keyboard. A 95% confidence interval for the difference (Non-athlete − Athlete) in the mean reaction time was 0.018 ±0.034 seconds. Does the interval provide convincing evidence of a difference in the true mean reaction time of athletes and non-athletes? Explain your answer.
No, because our interval includes a difference of 0 (no difference) as a plausible value.
Catherine and Ana wanted to know if student athletes (students on at least one varsity team) have faster reaction times than non-athletes. They took separate random samples of 33 athletes and 30 non-athletes from their school and tested their reaction time using an online reaction test, which measured the time (in seconds) between when a green light went on and the subject pressed a key on the computer keyboard. A 95% confidence interval for the difference (Non-athlete − Athlete) in the mean reaction time was 0.018 ±0.034 seconds. Does the interval provide convincing evidence that the true mean reaction time of athletes and non-athletes is the same? Explain your answer.
No; instead, we don’t have convincing evidence the mean reaction times differ. Zero is a plausible value for the difference in means, but many other plausible values besides 0 are in the confidence interval.
Catherine and Ana wanted to know if student athletes (students on at least one varsity team) have faster reaction times than non-athletes. They took separate random samples of 33 athletes and 30 non-athletes from their school and tested their reaction time using an online reaction test, which measured the time (in seconds) between when a green light went on and the subject pressed a key on the computer keyboard. A 95% confidence interval for the difference (Non-athlete − Athlete) in the mean reaction time was 0.018 ±0.034 seconds. Identify two ways Catherine and Ana could reduce the width of their interval. Describe any drawbacks to these actions.
They could increase the sample sizes or decrease the confidence level. Drawbacks: Increasing the sample sizes would require more work; decreasing the confidence level would give them less certainty that they captured the difference in the true mean reaction time of athletes and non-athletes.
A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random samples of nests: one of small nests and the other of large nests. Then she weighs one egg (chosen at random if there is more than one egg) from each nest. A 95% confidence interval for the difference (Large Small) between the mean mass (in grams) of eggs in small and large nests is 1.6±2.0. Does the interval provide convincing evidence of a difference in the true mean egg mass of birds with small nests and birds with large nests? Explain your answer.
No, because our interval includes a difference of 0 (no difference) as a plausible value.
A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random samples of nests: one of small nests and the other of large nests. Then she weighs one egg (chosen at random if there is more than one egg) from each nest. A 95% confidence interval for the difference (Large Small) between the mean mass (in grams) of eggs in small and large nests is 1.6±2.0. Does the interval provide convincing evidence that the true mean egg mass of birds with small nests and birds with large nests is the same? Explain your answer.
No; instead, we don't have convincing evidence the mean reaction times differ. Zero is a plausible value for the difference in means, but many other plausible values besides 0 are in the confidence interval.
A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random samples of nests: one of small nests and the other of large nests. Then she weighs one egg (chosen at random if there is more than one egg) from each nest. A 95% confidence interval for the difference (Large Small) between the mean mass (in grams) of eggs in small and large nests is 1.6±2.0. Identify two ways the researcher could reduce the width of her interval. Describe any drawbacks to these actions.
The researcher could increase the sample sizes or decrease the confidence level. Drawbacks: Increasing the sample sizes could make it difficult to find additional nests with eggs; decreasing the confidence level decreases the certainty of capturing the difference in the true mean egg mass of birds with small nests and birds with large nests.
Do piano lessons improve the spatial-temporal reasoning of preschool children? A study designed to investigate this question measured the spatial-temporal reasoning of a random sample of 34 preschool children before and after 6 months of piano lessons. The difference (After - Before) in the reasoning scores for each student has mean 3.618 and standard deviation 3.055. Construct and interpret a 90% confidence interval for the true mean difference.
S: µdiff 5 true mean difference (After − Before) in reasoning scores for all preschool students who take 6 months of piano lessons. P: One-sample t interval for µdiff. Random: Random sample of 34 preschool children. 10%: 34 < 10% of all preschool children. Normal/Large Sample: ndiff = 34 ≥ 30. D: x¯diff = 3.618, sdiff = 3.055, and ndiff = 34; df = 33; (2.731, 4.505). C: We are 90% confident the interval from 2.731 to 4.505 captures the true mean difference (After − Before) in reasoning scores for all preschool students who take 6 months of piano lessons.
Do piano lessons improve the spatial-temporal reasoning of preschool children? A study designed to investigate this question measured the spatial-temporal reasoning of a random sample of 34 preschool children before and after 6 months of piano lessons. The difference (After - Before) in the reasoning scores for each student has mean 3.618 and standard deviation 3.055. An 90% interval was calculated to be (2.731, 4.505). Can you conclude that taking 6 months of piano lessons would cause an increase in preschool students’ average reasoning scores? Why or why not?
No; a randomized experiment is needed to show causation.
A bank wonders if omitting the annual credit card fee for customers who charge at least $2400 in a year will increase the amount charged on its credit cards. The bank makes this offer to an SRS of 200 of its credit card customers. It then compares how much these customers charge this year with the amount that they charged last year. The mean increase in the sample is $332, and the standard deviation is $108. Construct and interpret a 99% confidence interval for the true mean increase.
S: µdiff 5 true mean increase in the amount this bank’s credit card customers would spend with no annual fee. P: One-sample t interval for µdiff. Random: SRS of 200 credit card customers. 10%: 200 < 10% of the population of all credit card customers. Normal/Large Sample: ndiff = 200 ≥ 30. D: x¯diff = 332, sdiff = 108, and ndiff = 200; df = 199; (312.14, 351.86). C: We are 99% confident that the interval from $312.14 to $351.86 captures the true mean increase in the amount this bank’s credit card customers would spend with no annual fee.
A bank wonders if omitting the annual credit card fee for customers who charge at least $2400 in a year will increase the amount charged on its credit cards. The bank makes this offer to an SRS of 200 of its credit card customers. It then compares how much these customers charge this year with the amount that they charged last year. The mean increase in the sample is $332, and the standard deviation is $108. A 99% confidence interval was calculated to be (312.14, 351.86). Can you conclude that dropping the annual fee would cause an increase in the average amount spent by this bank’s credit card customers? Why or why not?
No; there is no control group to compare results.
After hearing that students can improve short-term memory by chewing the same flavor of gum while studying for and taking a test, Leila and Valerie designed an experiment to investigate Using 30 volunteers, they randomly assigned 15 to chew gum while they studied a list of 40 words for 90 seconds. Immediately after the 90-second study period—and while chewing the same gum—the subjects wrote down as many words as they could remember. The remaining 15 followed the same procedure without chewing gum. Two weeks later, each of the 30 subjects did the oppo-site treatment. The number of words correctly remembered for each test was recorded for each subject. Explain why these are paired data.
They result from recording two values for each individual.
After hearing that students can improve short-term memory by chewing the same flavor of gum while studying for and taking a test, Leila and Valerie designed an experiment to investigate Using 30 volunteers, they randomly assigned 15 to chew gum while they studied a list of 40 words for 90 seconds. Immediately after the 90-second study period—and while chewing the same gum—the subjects wrote down as many words as they could remember. The remaining 15 followed the same procedure without chewing gum. Two weeks later, each of the 30 subjects did the oppo-site treatment. The number of words correctly remembered for each test was recorded for each subject. Verify that the conditions for constructing a one-sample t interval for a mean difference are satisfied.
PLAN: One-sample t interval for µdiff. Random: The students were randomly assigned a treatment order. Normal/ Large Sample: ndiff = 30 ≥ 30.
After hearing that students can improve short-term memory by chewing the same flavor of gum while studying for and taking a test, Leila and Valerie designed an experiment to investigate Using 30 volunteers, they randomly assigned 15 to chew gum while they studied a list of 40 words for 90 seconds. Immediately after the 90-second study period—and while chewing the same gum—the subjects wrote down as many words as they could remember. The remaining 15 followed the same procedure without chewing gum. Two weeks later, each of the 30 subjects did the oppo-site treatment. The number of words correctly remembered for each test was recorded for each subject. The 95% confidence interval for the true mean difference (Gum – No gum) in number of words remembered is –0.67 to 1.54. Interpret the confidence level.
If they repeated this experiment many times with these 30 volunteers and constructed a 95% CI using the results of each experiment, about 95% of the intervals would capture the true mean difference (Gum – No gum) in the number of words remembered for students like these.
After hearing that students can improve short-term memory by chewing the same flavor of gum while studying for and taking a test, Leila and Valerie designed an experiment to investigate Using 30 volunteers, they randomly assigned 15 to chew gum while they studied a list of 40 words for 90 seconds. Immediately after the 90-second study period—and while chewing the same gum—the subjects wrote down as many words as they could remember. The remaining 15 followed the same procedure without chewing gum. Two weeks later, each of the 30 subjects did the oppo-site treatment. The number of words correctly remembered for each test was recorded for each subject. The 95% confidence interval for the true mean difference (Gum – No gum) in number of words remembered is –0.67 to 1.54. Is there convincing evidence that chewing gum helps subjects like these with short-term memory? Explain your answer.
Because 0 is included in the CI, there is not convincing evidence that chewing gum helps subjects like these with short-term memory.
Do people get stressed out when other people watch them work? To find out, Sean and Shelby recruited 30 volunteers to take part in an experiment.25 Fifteen of the subjects were randomly assigned to complete a word search puzzle while Sean and Shelby stood close by and visibly took notes. The remaining 15 were assigned to complete a word search puzzle while Sean and Shelby stood at a distance. After each subject completed the word search, they completed a second word search under the opposite treatment. The amount of time required to complete each puzzle was recorded for each subject. Explain why these are paired data.
They result from recording two values for each individual.
Do people get stressed out when other people watch them work? To find out, Sean and Shelby recruited 30 volunteers to take part in an experiment.25 Fifteen of the subjects were randomly assigned to complete a word search puzzle while Sean and Shelby stood close by and visibly took notes. The remaining 15 were assigned to complete a word search puzzle while Sean and Shelby stood at a distance. After each subject completed the word search, they completed a second word search under the opposite treatment. The amount of time required to complete each puzzle was recorded for each subject. Verify that the conditions for constructing a one-sample t interval for a mean difference are satisfied.
PLAN: One-sample t interval for µdiff. Random: The students were randomly assigned a treatment order. Normal/Large Sample: ndiff = 30 ≥ 30.
Do people get stressed out when other people watch them work? To find out, Sean and Shelby recruited 30 volunteers to take part in an experiment.25 Fifteen of the subjects were randomly assigned to complete a word search puzzle while Sean and Shelby stood close by and visibly took notes. The remaining 15 were assigned to complete a word search puzzle while Sean and Shelby stood at a distance. After each subject completed the word search, they completed a second word search under the opposite treatment. The amount of time required to complete each puzzle was recorded for each subject. The 95% confidence interval for the true mean difference (Close by – At a distance) in amount of time needed to complete the puzzle is –12.7 seconds to 119.4 seconds. Interpret the confidence level.
If they conducted this experiment many times with these 30 volunteers and constructed a 95% CI using the results of each experiment, about 95% of the intervals would capture the true mean difference (Close by - At a distance) in the amount of time needed to complete the puzzle for subjects like these.
Do people get stressed out when other people watch them work? To find out, Sean and Shelby recruited 30 volunteers to take part in an experiment.25 Fifteen of the subjects were randomly assigned to complete a word search puzzle while Sean and Shelby stood close by and visibly took notes. The remaining 15 were assigned to complete a word search puzzle while Sean and Shelby stood at a distance. After each subject completed the word search, they completed a second word search under the opposite treatment. The amount of time required to complete each puzzle was recorded for each subject. The 95% confidence interval for the true mean difference (Close by – At a distance) in amount of time needed to complete the puzzle is –12.7 seconds to 119.4 seconds. Is there convincing evidence that standing close by causes subjects like these to take longer to complete a word search? Explain your answer.
Because 0 is included in the CI, there is not convincing evidence that standing close by causes subjects like these to take longer to complete a word search.
Researchers suspect that Variety A tomato plants have a different average yield than Variety B tomato plants. To find out, researchers randomly select 10 Variety A and 10 Variety B tomato plants. Then the researchers divide in half each of 10 small plots of land in different locations. For each plot, a coin toss deter-mines which half of the plot gets a Variety A plant; a Variety B plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The 10 differences (Variety A - Variety B) in yield are recorded. A graph of the differences looks roughly symmetric and unimodal with no outliers. The mean difference is x¯(A-B) = 0.34 and the standard deviation of the differences is s(A-B) = 0.83. Let µ(A-B) = the true mean difference (Variety A -Variety B) in yield for tomato plants of these two varieties. Which of the following is the best reason to use a one-sample t interval for a mean difference rather than a two-sample t interval for a difference in means to analyze these data? (a) The number of plots is the same for Variety A and Variety B plants. (b) The response variable, yield of tomatoes, is quantitative. (c) This is an experiment with randomly assigned treatments. (d) Each plot is given both varieties of tomato plant. (e) The sample size is less than 30 for both treatments.
d
Researchers suspect that Variety A tomato plants have a different average yield than Variety B tomato plants. To find out, researchers randomly select 10 Variety A and 10 Variety B tomato plants. Then the researchers divide in half each of 10 small plots of land in different locations. For each plot, a coin toss deter-mines which half of the plot gets a Variety A plant; a Variety B plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The 10 differences (Variety A - Variety B) in yield are recorded. A graph of the differences looks roughly symmetric and unimodal with no outliers. The mean difference is x¯(A-B) = 0.34 and the standard deviation of the differences is s(A-B) = 0.83. Let µ(A-B) = the true mean difference (Variety A -Variety B) in yield for tomato plants of these two varieties. A 95% confidence interval for μA − µB is given by (a) 0.34 ± 1.96(0.83) (b) 0.34 ± 1.96(0.83/sqrt(10)) (c) 0.34 ± 1.812(0.83/sqrt(10)) (d) 0.34 ± 2.262(0.83) (e) 0.34 ± 2.262(0.83/sqrt(10))
e
Jordan wondered if the bean burritos at Restaurant A tend to be heavier than the bean burritos at Restaurant B. To investigate, she visited each restaurant at 10 randomly selected times, ordered a bean burrito, and weighed the burrito. The 95% confidence interval for the difference (A – B) in the mean weight of burrito is 0.06 ounce ± 0.20 ounce. Based on the confidence interval, which conclusion is most appropriate? (a) Because 0 is included in the interval, there is convincing evidence that the mean weight is the same at both restaurants. (b) Because 0 is included in the interval, there isn’t convincing evidence that the mean weight is different at the two restaurants. (c) Because 0.06 is included in the interval, there is convincing evidence that the mean weight is greater at Restaurant A than Restaurant B. (d) Because 0.06 is included in the interval, there isn’t convincing evidence that the mean weight is greater at Restaurant A than Restaurant B. (e) Because there are more positive values in the interval than negative values, there is convincing evidence that the mean weight is greater at Restaurant A than Restaurant B.
b
A random sample of 30 words from Jane Austen’s Pride and Prejudice had a mean length of 4.08 letters with a standard deviation of 2.40. A random sample of 30 words from Henry James’s What Maisie Knew had a mean length of 3.85 letters with a standard deviation of 2.26. Which of the following is a correct expression for a 95% confidence interval for the difference in mean word length for these two novels? (a) (4.08 – 3.85) ± 2.576((2.40/sqrt(30)) + (2.26/sqrt(30)) (b) (4.08 – 3.85) ± 2.045((2.40/sqrt(30)) + (2.26/sqrt(30)) (c) (4.08 – 3.85) ± 2.045sqrt((2.40^2/29) + (2.26^2/29) (d) (4.08 – 3.85) ± 2.045sqrt((2.40^2/30) + (2.26^2/30) (e) (4.08 – 3.85) ± 2.576sqrt((2.40^2/30) + (2.26^2/30)
d
Find the appropriate critical value for constructing a confidence interval in the following setting: Estimating a population mean μ at a 95% confidence level based on an SRS of size 12.
df = 11, t* = 2.201
Find the appropriate critical value for constructing a confidence interval in the following setting: Estimating a true mean difference μdiff at a 99% confidence level based on paired data set with 75 differences.
df = 74, using Table B and df = 60, t* = 2.660. Using technology and df = 74, t* = 2.644.
A company that produces AA batteries tests the lifetime of a random sample of 30 batteries using a special device designed to imitate real-world use. Based on the testing, the company makes the following statement: “Our AA batteries last an average of 430 to 470 minutes, and our confidence in that interval is 95%.” Determine the point estimate, margin of error, standard error, and sample standard deviation.
Point estimate = (430 1 470)/2 = 450 minutes; margin of error = 470 – 450 = 20. SE = sx / sqrt(30) = 9.780 because 20 = 2.045 sx / sqrt(30). Finally, sx = 53.57 because sx / sqrt(30) = 9.780.
A company that produces AA batteries tests the lifetime of a random sample of 30 batteries using a special device designed to imitate real-world use. Based on the testing, the company makes the following statement: “Our AA batteries last an average of 430 to 470 minutes, and our confidence in that interval is 95%.” Explain the phrase “our confidence in that interval is 95%.”
If we were to select many samples of 30 batteries from this population and compute a 95% confidence interval for the mean lifetime from each sample, about 95% of these intervals will capture the true mean lifetime of the batteries.
A company that produces AA batteries tests the lifetime of a random sample of 30 batteries using a special device designed to imitate real-world use. Based on the testing, the company makes the following statement: “Our AA batteries last an average of 430 to 470 minutes, and our confidence in that interval is 95%.” Explain two ways the company could reduce the margin of error.
The company could reduce the margin of error by increasing the sample size or decreasing the confidence level.
A random sample of 16 of the more than 200 auto engine crankshafts produced in one day was selected. Here are measurements (in millimeters) of a critical component on these crankshafts: 224.120 224.001 224.017 223.982 223.989 223.961 223.960 224.089 223.987 223.976 223.902 223.980 224.098 224.057 223.913 223.999 Construct and interpret a 95% confidence interval for the mean length of this component on all the crankshafts produced on that day.
S: m 5 the true mean measurement of the critical dimension for the engine crankshafts produced in one day. P: One-sample t interval. Random: Random sample. 10%: 16 < 10% of all crankshafts produced in one day. Normal/ Large Sample: The histogram shows no strong skewness or outliers. D: x = 224.002, sx= 0.0618, and n= 16. Thus, df = 15 and t* = 2.131. (223.969, 224.035). C: We are 95% confident that the interval from 223.969 to 224.035 mm captures m 5 the true mean measurement of the critical dimension for engine crankshafts produced on this day.
A random sample of 16 of the more than 200 auto engine crankshafts produced in one day was selected. Here are measurements (in millimeters) of a critical component on these crankshafts: 224.120 224.001 224.017 223.982 223.989 223.961 223.960 224.089 223.987 223.976 223.902 223.980 224.098 224.057 223.913 223.999 The mean length is supposed to be μ = 224 mm but can drift away from this target during production. Does the interval (223.969, 224.035) provide convincing evidence that the mean has drifted from 224 mm? Explain your answer.
Because 224 is in this interval, it is a plausible value for the true mean. We don’t have convincing evidence that the process mean has drifted.
The National Assessment of Educational Progress (NAEP) Young Adult Literacy Assessment Survey interviewed separate random samples of 840 men and 1077 women aged 21 to 25 years. 29 The mean and standard deviation of scores on the NAEP’s test of quantitative skills were x¯1 = 272.401 and s1 = 59.21 for the men in the sample. For the women, the results were x¯2 = 274.732 and s2 = 57.52. Construct and interpret a 90% confidence interval for the difference in mean score for male and female young adults.
S: µ1 = true mean NAEP quantitative skills test score for young men and µ2 = true mean NAEP quantitative skills test score for young women. P: Two-sample t interval for µ1 - µ2. Random: It is reasonable to consider the two samples independent because knowing the response of a male shouldn’t predict the response of a female. 10%: n1 = 840 < 10% of all young men and n2 = 10.77 < 10% of all young women. Normal/Large Sample: n1 = 840 ≥ 30 and n2 = 1077 ≥ 30. D: x¯1 = 272.40, s1 = 59.2, n1 = 840, x¯2 = 274.73, s2 = 57.5, n2 =1077. Using df = 1777.52, (26.76, 2.10); using df = 839, (26.76, 2.10). C: We are 90% confident that the interval from 26.76 to 2.10 captures m1 2 m2 5 true difference in the mean NAEP quantitative skills test score for young men and the mean test score for young women.
Anne claims that a store-brand fertilizer works better than homemade compost as a soil enhancement when growing tomatoes. To test her theory, she plants two tomato plants in each of five planters. One plant in each planter is grown in soil with store-brand fertilizer and the other plant is grown in soil with homemade compost, with the choice of soil determined at random. In three months, she will harvest and weigh the tomatoes from each plant. Which of the following is the correct confidence interval Anne should use to analyze these data? (a) Two-sample z interval for μ1 – μ2 (b) Paired z interval for µdiff (c) Two-sample t interval for μ μ−12 (d) Paired t interval for µdiff (e) The correct interval cannot be determined without the data.
d
The weights (in pounds) of three adult males are 160, 215, and 195. What is the standard error of the mean for these data? (a) 190 (b) 27.84 (c) 22.73 (d) 16.07 (e) 13.13
d
You want to compute a 90% confidence interval for the mean difference in height for mothers and their adult daughters using a random sample of 30 mothers who have an adult daughter. What critical value should you use for this interval? (a) 1.645 (b) 1.671 (c) 1.697 (d) 1.699 (e) 1.761
d
We want to construct a one-sample t interval for a population mean using data from a population with unknown shape. In which of the following circumstances would it be inappropriate to construct the interval based on an SRS of size 14 from the population? (a) A stemplot of the data is roughly bell-shaped. (b) A histogram of the data shows slight skewness. (c) A boxplot shows that the values above the median are much more variable than the values below the median. (d) The sample standard deviation is large. (e) The sample standard deviation is small.
c