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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. Explain why the sample gives some evidence for the alternative hypothesis.
The sample result gives some evidence for Ha: p>0.13 because p^ = 17/100 = 0.17, which is greater than 0.13.
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. Calculate the standardized test statistic and P -value.
z = (0.17 - 0.13) / sqrt((0.13)(0.87)/100) = 1.19; P-value = 0.1170
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. Based on a test with a P-value of 0.1170, what conclusion would you make?
Because the P-value of 0.1170 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of all students at this elementary school who typically walk to school is greater than 0.13.
A test of H0: p = 0.5 versus Ha: p > 0.5 based on a sample of size 200 yields the standardized test statistic z = 2.19. Assume that the conditions for performing inference are met. Find and interpret the P -value.
The P-value is P(z≥ 2.19) = 0.0143. Assuming that the true population proportion is 0.5, there is a 0.0143 probability of getting a sample proportion as large as or larger than the one observed just by chance in a random sample of size n=200.
A test of H0: p = 0.5 versus Ha: p > 0.5 based on a sample of size 200 yields the standardized test statistic z = 2.19. Assume that the conditions for performing inference are met. What conclusion would you make at the α=0.01? Would your conclusion change if you used α=0.05 instead? Explain your reasoning.
α = 0.01: Because the P-value of 0.0143 < α = 0.01, we fail to reject H0. There is not convincing evidence that p < 0.5. α = 0.05: Yes! The P-value of 0.0143 < α = 0.05, so this time we will reject H0. There is convincing evidence that p > 0.5.
A test of H0: p = 0.5 versus Ha: p > 0.5 based on a sample of size 200 yields the standardized test statistic z = 2.19. Assume that the conditions for performing inference are met. Determine the value of p^= the sample proportion of successes.
Solve for p^: 2.19 = (p^ - 0.5) sqrt((0.5(0.5)) / 200); p^ = 0.5774
A test of H0: p = 0.65 versus Ha: p <0.65 based on a sample of size 400 yields the standardized test statistic z =− 1.78. Find and interpret the P -value.
P-value = 0.0375; if the true population proportion is 0.65, a 0.0375 probability exists of getting a sample proportion as small as or smaller than the one observed by chance in a random sample of size n= 400.
A test of H0: p = 0.65 versus Ha: p <0.65 based on a sample of size 400 yields the standardized test statistic z =− 1.78. What conclusion would you make at the α=0.10? Would your conclusion change if you used α=0.05 instead? Explain your reasoning.
When α = 0.10 and α = 0.05, reject H0.
A test of H0: p = 0.65 versus Ha: p <0.65 based on a sample of size 400 yields the standardized test statistic z =− 1.78. Determine the value of p^= the sample proportion of successes.
p^ = 0.6075
A media report claims that more than 75% of middle school students engage in bullying behavior. A University of Illinois study on aggressive behavior surveyed a random sample of 558 middle school students. When asked to describe their significance level? Would your conclusion change if you used behavior in the last 30 days, 445 students admitted that they had engaged in physical aggression, social ridicule, teasing, name-calling, and issuing threats—all of which would be classified as bullying. Do these data provide convincing evidence at the α=0.05 significance level that the media report’s claim is correct?
S: H0: p = 0.75, Ha: p > 0.75, where p= the true proportion of all middle school students who engage in bullying behavior using α = 0.05. P: One-sample z test for p. Random: Random sample. 10%: 558 < 10% of all middle school students. Large Counts: 418.5 ≥ 10 and 139.5 ≥ 10. D: z= 2.59; P-value = 0.0048. C: Because the P-value of 0.0048 < α = 0.05, we reject H0. We have convincing evidence that the true proportion of all middle school students who engage in bullying behavior is > 0.75.
The germination rate of seeds is defined as the proportion of seeds that sprout and grow when properly planted and watered. A certain variety of grass seed usually has a germination rate of 0.80. A company wants to see if spraying the seeds with a chemical that is known to increase germination rates in other species will increase the germination rate of this variety of grass. The company researchers spray a random sample of 400 grass seeds with the chemical, and 339 of the seeds germinate. Do these data provide convincing evidence at the α=0.05 significance level that the chemical is effective for this variety of grass? Explain why the sample result gives some evidence for the alternative hypothesis.
S: H0: p= 0.80, Ha: p> 0.80 P: Large Counts: 320 ≥ 10 and 80 ≥ 10. D: z= 2.38; P-value = 0.0087. C: Reject H0.
A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of 200 from the more than 2500 students at the school. In all, 83 students say that they approve of the new parking arrangement. The principal cites this as evidence that the change was effective. Describe a Type I error and a Type II error in this setting, and give a possible consequence of each.
Type I: Finding convincing evidence that more than 37% of students were satisfied with the new arrangement, when in reality only 37% were satisfied. Consequence: The principal believes students are satisfied and takes no further action. Type II: Failing to find convincing evidence that more than 37% are satisfied with the new arrangement, when in reality more than 37% are satisfied. Consequence: The principal takes further action when none is needed.
A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of 200 from the more than 2500 students at the school. In all, 83 students say that they approve of the new parking arrangement. The principal cites this as evidence that the change was effective. Is there convincing evidence that the principal’s claim is true?
S: H0: p= 0.37, Ha: p> 0.37, where p= the true proportion of all students who are satisfied with the parking situation after the change using α = 0.05. P: One-sample z test for p. Random: Random sample. 10%: 200 < 10% of 2500. Large Counts: 74 ≥ 10 and 126 ≥ 10. D: z= 1.32; P-value = 0.0934. C: Because the P-value of 0.0934 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of all students who are satisfied with the parking situation after the change is greater than 0.37.
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, council members survey a random sample of 300 city residents. In the sample, 158 residents say that they are in favor of the sales tax increase. Describe a Type I error and a Type II error in this setting, and give a possible consequence of each.
Type I error: Finding convincing evidence that a majority of city residents support a 1% increase in the sales tax to fund road repairs, when in reality at most 50% would. Consequence: The sales tax will be increased by 1%, potentially upsetting most city residents. Type II error: Failing to find convincing evidence that a majority of city residents support a 1% increase in the sales tax to fund road repairs, when in reality a majority would. Consequence: The sales tax will not increase and the road repairs will not take place, despite the support of the city residents.
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, council members survey a random sample of 300 city residents. In the sample, 158 residents say that they are in favor of the sales tax increase. Do these data provide convincing evidence that a majority of city residents support the tax increase?
S: H0: p= 0.5, Ha: p> 0.5. P: Large Counts: 150 ≥ 10 and 150 ≥ 10. D: z= 0.92; P-value = 0.1788. C: Fail to reject H0.
Cell-phone passwords A consumer organization suspects that less than half of parents know their child’s cell-phone password. The Pew Research Center asked a random sample of parents if they knew their child’s cell-phone password. Of the 1060 parents surveyed, 551 reported that they knew the password. Explain why it isn’t necessary to carry out a significance test in this setting.
Because p^ > 0.5, there is no evidence for Ha: p< 0.50.
A political organization wants to determine if there is convincing evidence that a majority of registered voters in a large city favor Proposition X. In an SRS of 1000 registered voters, 482 favor the proposition. Explain why it isn’t necessary to carry out a significance test in this setting.
Because p^ < 0.5, there is no evidence for Ha: p> 0.50.
Gregor Mendel (1822–1884), an Austrian monk, is considered the father of genetics. Mendel studied the inheritance of various traits in pea plants. One such trait is whether the pea is smooth or wrinkled. Mendel predicted a ratio of 3 smooth peas for every 1 wrinkled pea. In one experiment, he observed 423 smooth and 133 wrinkled peas. Assume that the conditions for inference are met. State appropriate hypotheses for testing Mendel’s claim about the true proportion of smooth peas.
H0: p= 0.75, Ha: p≠ 0.75, where p5 the true proportion of peas that will be smooth using α = 0.05.
Gregor Mendel (1822–1884), an Austrian monk, is considered the father of genetics. Mendel studied the inheritance of various traits in pea plants. One such trait is whether the pea is smooth or wrinkled. Mendel predicted a ratio of 3 smooth peas for every 1 wrinkled pea. In one experiment, he observed 423 smooth and 133 wrinkled peas. Assume that the conditions for inference are met. Calculate the standardized test statistic and P -value.
z= 0.59; P-value = 0.5552
Gregor Mendel (1822–1884), an Austrian monk, is considered the father of genetics. Mendel studied the inheritance of various traits in pea plants. One such trait is whether the pea is smooth or wrinkled. Mendel predicted a ratio of 3 smooth peas for every 1 wrinkled pea. In one experiment, he observed 423 smooth and 133 wrinkled peas. Assume that the conditions for inference are met. A test was done and a P-value was found to be 0.5552. Interpret the P -value. What conclusion would you make?
Assuming the true proportion of smooth peas is 0.75, there is a 0.5552 probability of getting a sample proportion as different from 0.75 as 0.761 (in either direction) by chance in a random sample of 556 peas. Because the P-value of 0.5552 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of peas that are smooth differs from 0.75.
When a fair coin is flipped, we all know that the probability the coin lands on heads is 0.50. However, what if a coin is spun? According to the article “Euro Coin Accused of Unfair Flipping” in the New Scientist, two Polish math professors and their students spun a Belgian euro coin 250 times. It landed heads 140 times. One of the professors concluded that the coin was minted asymmetrically. A representative from the Belgian mint indicated the result was just chance. Assume that the conditions for inference are met. State appropriate hypotheses for testing these competing claims about the true proportion of spins that will land on heads.
H0: p= 0.50, Ha: p≠ 0.50
When a fair coin is flipped, we all know that the probability the coin lands on heads is 0.50. However, what if a coin is spun? According to the article “Euro Coin Accused of Unfair Flipping” in the New Scientist, two Polish math professors and their students spun a Belgian euro coin 250 times. It landed heads 140 times. One of the professors concluded that the coin was minted asymmetrically. A representative from the Belgian mint indicated the result was just chance. Assume that the conditions for inference are met. Calculate the standardized test statistic and P -value.
z= 1.90; P-value = 0.0574
When a fair coin is flipped, we all know that the probability the coin lands on heads is 0.50. However, what if a coin is spun? According to the article “Euro Coin Accused of Unfair Flipping” in the New Scientist, two Polish math professors and their students spun a Belgian euro coin 250 times. It landed heads 140 times. One of the professors concluded that the coin was minted asymmetrically. A representative from the Belgian mint indicated the result was just chance. Assume that the conditions for inference are met. A test was done and a P-value was found to be 0.074. Interpret the P -value. What conclusion would you make?
If the true proportion of heads is 0.50, a 0.0574 probability exists of getting a sample proportion as different from 0.50 as 0.56 (in either direction) just by chance in a random sample of 250 spun Belgian euros. Fail to reject H0.
A state’s Division of Motor Vehicles (DMV) claims that 60% of all teens pass their driving test on the first attempt. An investigative reporter examines an SRS of the DMV records for 125 teens; 86 of them passed the test on their first try. Is there convincing evidence at the α =0.05 significance level that the DMV’s claim is incorrect?
S: H0: p= 0.60, Ha: p≠ 0.60, where p= the true proportion of teens who pass on the first attempt using α = 0.05. P: One-sample z test for p. Random: Random sample. 10%: 125 < 10% of population. Large Counts: 75 ≥ 10 and 50 $≥ 10. D: z= 2.01; P-value = 0.0444. C: Reject H0. There is convincing evidence that the true proportion of teens who pass on their first attempt differs from 0.60.
In a recent year, 73% of first-year college students responding to a national survey identified “being very well-off financially” as an important personal goal. A state university finds that 132 of an SRS of 200 of its first-year students say that this goal is important. Is there convincing evidence at the α=0.05 significance level that the proportion of all first-year students at this university who think being very well-off is important differs from the national value of 73%?
S: H0: p= 0.73, Ha: p≠ 0.73 P: Large Counts: 146 ≥ 10 and 54 $≥ 10. D: z=-2.23; P-value = 0.0258. C: Reject H0.
A state’s Division of Motor Vehicles (DMV) claims that 60% of all teens pass their driving test on the first attempt. An investigative reporter examines an SRS of the DMV records for 125 teens; 86 of them passed the test on their first try. Construct and interpret a 95% confidence interval for the true proportion p of all teens in the state who passed their driving test on the first attempt. Assume that the conditions for inference are met.
S: p= the true proportion of teens who pass on the first attempt. P: One-sample z interval for p. D: (0.607,0.769) C: We are 95% confident that the interval from 0.607 to 0.769 captures the true proportion of teens who pass on the first attempt.
A state’s Division of Motor Vehicles (DMV) claims that 60% of all teens pass their driving test on the first attempt. An investigative reporter examines an SRS of the DMV records for 125 teens; 86 of them passed the test on their first try. Explain why calculating this as an interval provides more information than a significance test.
The confidence interval gives the values of p that are plausible based on the sample data and not just a reject or fail to reject decision about H0.
In a recent year, 73% of first-year college students responding to a national survey identified “being very well-off financially” as an important personal goal. A state university finds that 132 of an SRS of 200 of its first-year students say that this goal is important. Construct and interpret a 95% confidence interval for the true proportion p of all first-year students at the university who would identify being very well-off as an important personal goal. Assume that the conditions for inference are met.
S: p= the true proportion of first-year students who think being very well-off financially is an important personal goal. P: One-sample z interval for p. D: (0.594, 0.726) C: We are 95% confident that the interval from 0.594 to 0.726 captures the true proportion of first-year students who identify being very well-off as an important personal goal.
In a recent year, 73% of first-year college students responding to a national survey identified “being very well-off financially” as an important personal goal. A state university finds that 132 of an SRS of 200 of its first-year students say that this goal is important. Explain why calculating this as an interval provides more information than a significance test.
The confidence interval gives the values of p that are plausible based on the sample data and not just a reject or fail to reject decision about H0.
The Pew Internet and American Life Project asked a random sample of U.S. adults, “Do you ever . . . use Twitter or another service to share updates about yourself or to see updates about others?” According to Pew, the resulting 95% confidence interval is (0.123, 0.177). Based on the confidence interval, is there convincing evidence that the true proportion of U.S. adults who would say they use Twitter or another service to share updates differs from 0.17? Explain your reasoning.
No; because the value 0.17 is included in the confidence interval, it is a plausible value for the true proportion of U.S. adults who would say they use Twitter.
A Gallup poll found that 59% of the people in its sample said “Yes” when asked, “Would you like to lose weight?” Gallup announced: “For results based on the total sample of national adults, one can say with 95% confidence that the margin of (sampling) error is 3± percentage points.” Based on the confidence interval, is there convincing evidence that the true proportion of U.S. adults who would say they want to lose weight differs from 0.55? Explain your reasoning.
Yes; because 0.55 is not included in the confidence interval from 0.56 to 0.62, it is not a plausible value for the true proportion of U.S. adults who would say that they want to lose weight.
A company that makes potato chips requires each shipment of potatoes to meet certain quality standards. If the company finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent back to the supplier to get another load of potatoes. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. He will then perform a test of H0 p =: 0.08 versus Ha: p > 0.08, where p is the true proportion of potatoes with blemishes in a given truckload. The power of the test to detect that p = 0.11, based on a random sample of 500 potatoes and significance level α= 0.05 is 0.764. Interpret this value.
If the true proportion of potatoes with blemishes in a given truckload is p = 0.11, there is a 0.764 probability that the company will find convincing evidence for Ha: p > 0.08.
You are thinking about opening a restaurant and are searching for a good location. From the 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 site. Based on the mean income of this sample, you will perform a test at the α= 0.05 significance level of H0: μ= $85,000 versus Ha: μ > $85,000, where μ is the true mean income in the population of people who live near the restaurant. The power of the test to detect that μ= $85,000 is 0.64. Interpret this value.
If the true mean income in the population of people who live near the restaurant is µ = $86,000, there is a 0.64 probability that I will find convincing evidence for Ha: µ > $85,000.
A company that makes potato chips requires each shipment of potatoes to meet certain quality standards. If the company finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent back to the supplier to get another load of potatoes. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. He will then perform a test of H0 p =: 0.08 versus Ha: p > 0.08, where p is the true proportion of potatoes with blemishes in a given truckload. The power of the test to detect that p = 0.11, based on a random sample of 500 potatoes and significance level α= 0.05 is 0.764. Determine if the following changes would increase or decrease the power of the test. Explain your answers. Change the significance level to α= 0.10.
Increase; using a larger significance level makes it easier to reject H0 when Ha is true.
A company that makes potato chips requires each shipment of potatoes to meet certain quality standards. If the company finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent back to the supplier to get another load of potatoes. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. He will then perform a test of H0 p =: 0.08 versus Ha: p > 0.08, where p is the true proportion of potatoes with blemishes in a given truckload. The power of the test to detect that p = 0.11, based on a random sample of 500 potatoes and significance level α= 0.05 is 0.764. Determine if the following changes would increase or decrease the power of the test. Explain your answers. Take a random sample of 250 potatoes instead of 500 potatoes.
Decrease; a smaller sample size gives less information about the true proportion p.
A company that makes potato chips requires each shipment of potatoes to meet certain quality standards. If the company finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent back to the supplier to get another load of potatoes. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. He will then perform a test of H0 p =: 0.08 versus Ha: p > 0.08, where p is the true proportion of potatoes with blemishes in a given truckload. The power of the test to detect that p = 0.11, based on a random sample of 500 potatoes and significance level α= 0.05 is 0.764. Determine if the following changes would increase or decrease the power of the test. Explain your answers. The true proportion is p = 0.10 instead of p = 0.11.
Decrease; it is harder to detect a smaller difference between the null and alternative parameter value.
You are thinking about opening a restaurant and are searching for a good location. From the 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 site. Based on the mean income of this sample, you will perform a test at the α= 0.05 significance level of H0: μ= $85,000 versus Ha: μ > $85,000, where μ is the true mean income in the population of people who live near the restaurant. The power of the test to detect that μ= $85,000 is 0.64. Determine if the following change would increase or decrease the power of the test. Explain your answers. Use a random sample of 30 people instead of 50 people.
Decrease; a smaller sample size gives less information about the true mean µ.
You are thinking about opening a restaurant and are searching for a good location. From the 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 site. Based on the mean income of this sample, you will perform a test at the α= 0.05 significance level of H0: μ= $85,000 versus Ha: μ > $85,000, where μ is the true mean income in the population of people who live near the restaurant. The power of the test to detect that μ= $85,000 is 0.64. Determine if the following change would increase or decrease the power of the test. Explain your answers. Try to detect that μ= $85,500 instead of μ= $86,000.
Decrease; it is harder to detect a smaller difference between the null and alternative parameter value.
You are thinking about opening a restaurant and are searching for a good location. From the 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 site. Based on the mean income of this sample, you will perform a test at the α= 0.05 significance level of H0: μ= $85,000 versus Ha: μ > $85,000, where μ is the true mean income in the population of people who live near the restaurant. The power of the test to detect that μ= $85,000 is 0.64. Determine if the following change would increase or decrease the power of the test. Explain your answers. Change the significance level to α 0.10
Increase; using a larger significance level makes it easier to reject H0 when Ha is true.
A company that makes potato chips requires each shipment of potatoes to meet certain quality standards. If the company finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent back to the supplier to get another load of potatoes. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. He will then perform a test of H0 p =: 0.08 versus Ha: p > 0.08, where p is the true proportion of potatoes with blemishes in a given truckload. The power of the test to detect that p = 0.11, based on a random sample of 500 potatoes and significance level α= 0.05 is 0.764. Explain one disadvantage of using α= 0.10 α= 0.05 when performing the test.
The larger significance level will increase the probability of a Type I error.
A company that makes potato chips requires each shipment of potatoes to meet certain quality standards. If the company finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent back to the supplier to get another load of potatoes. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. He will then perform a test of H0 p =: 0.08 versus Ha: p > 0.08, where p is the true proportion of potatoes with blemishes in a given truckload. The power of the test to detect that p = 0.11, based on a random sample of 500 potatoes and significance level α= 0.05 is 0.764. Explain one disadvantage of taking a random sample of 500 potatoes instead of 250 potatoes from the shipment.
The larger sample size would require more time and money.
You are thinking about opening a restaurant and are searching for a good location. From the 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 site. Based on the mean income of this sample, you will perform a test at the α= 0.05 significance level of H0: μ= $85,000 versus Ha: μ > $85,000, where μ is the true mean income in the population of people who live near the restaurant. The power of the test to detect that μ= $85,000 is 0.64. Explain one disadvantage of using α= 0.10 instead of α= 0.05 when performing the test.
The larger significance level will increase the probability of a Type I error.
You are thinking about opening a restaurant and are searching for a good location. From the 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 site. Based on the mean income of this sample, you will perform a test at the α= 0.05 significance level of H0: μ= $85,000 versus Ha: μ > $85,000, where μ is the true mean income in the population of people who live near the restaurant. The power of the test to detect that μ= $85,000 is 0.64. Explain one disadvantage of taking a random sample of 50 people instead of 30 people.
The larger sample size would require more time and money.
A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of students at the school. She would like to perform a test of H0 p =: 0.37 verses Ha: p >: 0.37 where p is the true proportion of students at school who are satisfied with the parking situation after the change. The power of the test to detect that p = 0.45 based on a random sample of 200 students and a significance level of α= 0.05 is 0.75. Interpret this value.
If the true proportion of students at the school who are satisfied with the parking situation after the change is p= 0.45 there is a 0.75 probability that the principal will find convincing evidence for Ha: p > 0.37.
A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of students at the school. She would like to perform a test of H0 p =: 0.37 verses Ha: p >: 0.37 where p is the true proportion of students at school who are satisfied with the parking situation after the change. The power of the test to detect that p = 0.45 based on a random sample of 200 students and a significance level of α= 0.05 is 0.75. Find the probability of a Type I error and the probability of a Type II error for the test.
P(Type I error) 5 α = 0.05 and P(Type II error) = 1 - Power = 1 - 0.75 = 0.25.
A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of students at the school. She would like to perform a test of H0 p =: 0.37 verses Ha: p >: 0.37 where p is the true proportion of students at school who are satisfied with the parking situation after the change. The power of the test to detect that p = 0.45 based on a random sample of 200 students and a significance level of α= 0.05 is 0.75. Describe two ways to increase the power of the test.
The power would increase by increasing the sample size or using a larger significance level.
A company that manufactures classroom chairs for high school students claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of H0: μ = 300 verses Ha: μ < 300 where μ is the true mean breaking strength of this company’s classroom chairs. The power of the test to detect that μ= 294 based on a random sample of 30 chairs and a significance level of α= 0.05 is 0.71. Interpret this value.
If the true mean breaking strength of this company’s classroom chairs is µ = 294 pounds, there is a 0.71 probability that I will find convincing evidence for Ha: µ > 300.
A company that manufactures classroom chairs for high school students claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of H0: μ = 300 verses Ha: μ < 300 where μ is the true mean breaking strength of this company’s classroom chairs. The power of the test to detect that μ= 294 based on a random sample of 30 chairs and a significance level of α= 0.05 is 0.71. Find the probability of a Type I error and the probability of a Type II error for the test.
P(Type I error) = α = 0.05 and P(Type II error) = 1 - 0.71 = 0.29
A company that manufactures classroom chairs for high school students claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of H0: μ = 300 verses Ha: μ < 300 where μ is the true mean breaking strength of this company’s classroom chairs. The power of the test to detect that μ= 294 based on a random sample of 30 chairs and a significance level of α= 0.05 is 0.71. Describe two ways to increase the power of the test.
The power would increase by increasing the sample size or using a larger significance level.
You read that a significance test at the α= 0.01 significance level has probability 0.14 of making a Type II error when a specific alternative is true. What is the power of the test against this alternative?
(a) Power = 1 - P (Type II error) = 1 - 0.14 = 0.86
You read that a significance test at the α= 0.01 significance level has probability 0.14 of making a Type II error when a specific alternative is true. What’s the probability of making a Type I error?
P (Type I error) = α = 0.01
A scientist calculates that a test at the α= 0.05 significance level has probability 0.23 of making a Type II error when a specific alternative is true. What is the power of the test against this alternative?
Power = 1 - P (Type II error) = 1 - 0.23 = 0.77
A scientist calculates that a test at the α= 0.05 significance level has probability 0.23 of making a Type II error when a specific alternative is true. What’s the probability of making a Type I error?
P (Type I error) = α = 0.05
After once again losing a football game to the archrival, a college’s alumni association conducted a survey to see if alumni were in favor of firing the coach. An SRS of 100 alumni from the population of all living alumni was taken, and 64 of the alumni in the sample were in favor of firing the coach. Suppose you wish to see if a majority of all living alumni is in favor of firing the coach. The appropriate standardized test statistic is (a) z = (0.64 – 0.5) / sqrt((0.64(0.36))/100) (b) z = (0.5 – 0.64) / sqrt((0.64(0.36))/100) (c) z = (0.64 – 0.5) / sqrt((0.5(0.5))/100) (d) z = (0.64 – 0.5) / sqrt((0.64(0.36))/64) (e) z = (0.5 – 0.64) / sqrt((0.5(0.5))/100)
c
Which of choices (a) through (d) is not a condition for performing a significance test about a population proportion p? (a) The data should come from a random sample from the population of interest. (b) Both np0 and (1 – np0) should be at least 10. (c) If you are sampling without replacement from a finite population, then you should sample less than 10% of the population. (d) The population distribution should be approximately Normal, unless the sample size is large. (e) All of the above are conditions for performing a significance test about a population proportion.
d
The standardized test statistic for a test of H0: p = 0.4 versus Ha: p ≠ 0.4 is z = 2.43. This test is (a) not significant at either α= 0.05 or α= 0.01. (b) significant at α= 0.05, but not at α= 0.01. (c) significant at α= 0.01, but not at α= 0.05. (d) significant at both α= 0.05 and α= 0.01. (e) inconclusive because we don’t know the value of p^.
b
Which of the following 95% confidence intervals would lead us to reject H0: p =0.30 in favor of Ha: p ≠ 0.30 at the 5% significance level? (a) (0.19, 0.27) (b) (0.24, 0.30) (c) (0.27, 0.31) (d) (0.29, 0.38) (e) None of these
a
A researcher plans to conduct a significance test at the α= 0.01 significance level. She designs her study to have a power of 0.90 at a particular alternative value of the parameter of interest. The probability that the researcher will commit a Type II error for the particular alternative value of the parameter she used is (a) 0.01. (b) 0.10. (c) 0.89. (d) 0.90. (e) 0.99.
b
A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance. Researchers want to know if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids. State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest.
H0: p1 - p2 = 0, Ha: p1 - p2 ≠ 0, where p1 = the true proportion of high school freshmen in Illinois who have used anabolic steroids and p2 = the true proportion of high school seniors in Illinois who have used anabolic steroids.
A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance. Researchers want to know if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids. Check if the conditions for performing a significance test are met.
Random: Independent random samples. 10%: 1679 < 10% of all high school freshmen in Illinois and 1366 < 10% of all high school seniors in Illinois. Large Counts: p^C= (34 + 24) / (1679 + 1366) = 0.019; 31.9, 1647.1, 25.95, 1340.05 are ≥ 10.
Fire is a serious threat to shrubs in dry climates. Some shrubs can resprout from their roots after their tops are destroyed. Researchers wondered if fire would help with resprouting. One study of resprouting took place in a dry area of Mexico. 21 The researchers randomly assigned shrubs to treatment and control groups. They clipped the tops of all the shrubs. They then applied a propane torch to the stumps of the treatment group to simulate a fire. All 12 of the shrubs in the treatment group resprouted. Only 8 of the 12 shrubs in the control group resprouted. State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest.
H0: p1 - p2 = 0, Ha: p1 - p2 > 0, where p1 = true proportion of shrubs that would resprout after being clipped and burned and p2 = . . . after being clipped.
Fire is a serious threat to shrubs in dry climates. Some shrubs can resprout from their roots after their tops are destroyed. Researchers wondered if fire would help with resprouting. One study of resprouting took place in a dry area of Mexico. 21 The researchers randomly assigned shrubs to treatment and control groups. They clipped the tops of all the shrubs. They then applied a propane torch to the stumps of the treatment group to simulate a fire. All 12 of the shrubs in the treatment group resprouted. Only 8 of the 12 shrubs in the control group resprouted. Check if the conditions for performing a significance test are met.
Random: Random assignment. Large Counts: Not met because all are < 10
Lyme disease is spread in the northeastern United States by infected ticks. The ticks are infected mainly by feeding on mice, so more mice result in more infected ticks. The mouse population, in turn, rises and falls with the abundance of acorns, their favored food. Experimenters studied two similar forest areas in a year when the acorn crop failed. To see if mice are more likely to breed when there are more acorns, the researchers added hundreds of thousands of acorns to one area to imitate an abundant acorn crop, while leaving the other area untouched. The next spring, 54 of the 72 mice trapped in the first area were in breeding condition, versus 10 of the 17 mice trapped in the second area. State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest.
H0: p1 - p2 = 0, Ha: p1 - p2 > 0, where p1 = true proportion of mice that are in breeding condition in the area of abundant acorn crop and p2 = . . . in the untouched area.
Lyme disease is spread in the northeastern United States by infected ticks. The ticks are infected mainly by feeding on mice, so more mice result in more infected ticks. The mouse population, in turn, rises and falls with the abundance of acorns, their favored food. Experimenters studied two similar forest areas in a year when the acorn crop failed. To see if mice are more likely to breed when there are more acorns, the researchers added hundreds of thousands of acorns to one area to imitate an abundant acorn crop, while leaving the other area untouched. The next spring, 54 of the 72 mice trapped in the first area were in breeding condition, versus 10 of the 17 mice trapped in the second area. Check if the conditions for performing a significance test are met.
Random: Unknown. Large Counts: Not met because 4.777 < 10.
A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance. Researchers want to know if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids. Explain why the sample results give some evidence for the alternative hypothesis.
p^1 - p^2 = 0.0027 ≠ 0
A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance. Researchers want to know if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids. Calculate the standardized test statistic and P -value.
z= 0.54, P-value = 0.5904
A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance. Researchers want to know if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids. A test was done and a P-value was found to be 0.5904. What conclusion would you make?
Because the P-value of 0.5904 > α= 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of high school freshmen in Illinois who have used anabolic steroids differs from the true proportion of high school seniors in Illinois who have.
Phoebe has a hunch that older students at her very large high school are more likely to bring a bag lunch than younger students because they have grown tired of cafeteria food. She takes a simple random sample of 80 sophomores and finds that 52 of them bring a bag lunch. A simple random sample of 104 seniors reveals that 78 of them bring a bag lunch. Do these data give convincing evidence to support Phoebe’s hunch at the α= 0.05 significance level?
S: H0: p1 - p2 = 0, Ha: p1 - p2 < 0, where p1 = true proportion of sophomores who bring a bag lunch and p2 = . . . seniors . . . P: Two-sample z test for p1 - p2. Random: Independent random samples. 10%: 80 < 10% of all sophomores and 104 < 10% of all seniors. Large Counts: 56.56, 23.44, 73.528, 30.472 are all ≥ 10. D: z=-1.48, P-value = 0.0699. C: Because the P-value of 0.0699 > α= 0.05, we fail to reject H0. There is not convincing evidence that the true proportion of sophomores who bring a bag lunch is less than the true proportion of seniors who do.
Phoebe has a hunch that older students at her very large high school are more likely to bring a bag lunch than younger students because they have grown tired of cafeteria food. She takes a simple random sample of 80 sophomores and finds that 52 of them bring a bag lunch. A simple random sample of 104 seniors reveals that 78 of them bring a bag lunch. Interpret the P -value of 0.0699 in the context of this study.
If there is no difference in the true proportion of sophomores and seniors who bring a bag lunch, there is a 0.0699 probability of getting a difference in the proportions as large as or larger than the one observed (0.65 - 0.75 =-0.10) by chance alone.
In a study of 3000 randomly selected teenagers in 1990, 450 showed some hearing loss. In a similar study of 1800 teenagers reported in 2010, 351 showed some hearing loss. Do these data give convincing evidence that the proportion of all teens with hearing loss has increased at the α= 0.01 significance level?
S: H0: p1 - p2 = 0, Ha: p1 - p2 < 0, where p1 = true proportion of teenagers in 1990 with some hearing loss and p2 = . . . in 2010 . . . P: Two-sample z test for p1 - p2. Random: Independent random samples. 10%: 3000 < 10% of all teenagers in 1990 and 1800 < 10% of all teenagers in 2010. Large Counts: 501, 2499, 300.6, 1499.4 ≥ 10. D: z=-4.05, P-value ~ 0. C: Because the P-value of approximately 0 < α= 0.05, we reject H0. There is convincing evidence that the true proportion of teenagers in 1990 with some hearing loss is less than the true proportion of teenagers in 2010 with some hearing loss.
In a study of 3000 randomly selected teenagers in 1990, 450 showed some hearing loss. In a similar study of 1800 teenagers reported in 2010, 351 showed some hearing loss. Interpret the P -value of ~0 in the context of this study.
If there is no difference in the true proportion of teenagers in 1990 and 2010 with some hearing loss, there is about a 0 probability of getting a difference in the proportions
A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. Does this study provide convincing evidence of a difference at the α= 0.05 significance level in the development of peanut allergies in infants like the ones in this study who consume or avoid peanut butter?
S: H0: p1 - p2 = 0, Ha: p1 - p2 ≠ 0, where p1 = true proportion of children like the ones in this study who are exposed to peanut butter as infants who are allergic to peanuts at age 5 and p2 = . . . not exposed . . . P: Two-sample z test for p1 - p2. Random: Random assignment. Large Counts: 31.928, 275.072, 33.384, 287.616 ≥ 10. D: z=-5.71, P-value ~0. C: Because the P-value of approximately 0 < α= 0.05, we reject H0. There is convincing evidence that the true proportion of children like the ones in this study who are exposed to peanut butter as infants who are allergic to peanuts at age 5 differs from the true proportion of children like the ones in this study who are not exposed to peanut butter as infants who are allergic to peanuts at age 5.
A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. Based on the conclusion of rejecting the H0, which mistake—a Type I error or a Type II error—could you have made? Explain your answer.
Type I error
A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. Should you generalize the result of a significance test to all infants? Why or why not?
No; when subjects are not randomly selected, we should not generalize the results of an experiment to some larger population.
A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. A 95% confidence interval for p1 – p2 is (0.185, 0.093). Explain how the confidence interval provides more information than a significance test.
The confidence interval does not include 0 as a plausible value for p1 - p2, which is consistent with the decision to reject H0: p1 - p2 = 0 in part (a). The confidence interval tells us that any value between 20.185 and 20.093 is plausible for p1 - p2 based on the sample data.
Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. Does this study provide convincing evidence at the α= 0.05 significance level of a difference in the effectiveness of Lipitor and Pravachol for people like the ones in this study?
S: H0: p1-p2 =0, Ha: p1 - p2 ≠ 0, where p1 = true proportion of subjects like these who would die, have a heart attack, or suffer other serious consequences within 2 years if they took Pravachol and p2 = true proportion who would suffer serious consequences if they took Lipitor; α= 0.05. P: Two-sample z test for p1 - p2. Random: Volunteers were randomly assigned to the treatments. Large Counts: 510.057, 1588.943, 501.309, 1561.691 are ≥ 10. D: z=-2.95; P-value =0.0031. C: Because the P-value of 0.0031 < α= 0.05, we reject H0. There is convincing evidence the true proportion of subjects like these who would die, have a heart attack, or suffer other serious consequences within 2 years if they took Pravachol differs from the true proportion of subjects like these who would die, have a heart attack, or suffer other serious consequences within 2 years if they took Lipitor.
Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. Based on the conclusion of rejecting the H0, which mistake—a Type I error or a Type II error—could you have made? Explain your answer.
Because we rejected H0, it is possible we made a Type I error.
Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. Should you generalize the result of a significance test to all people with heart disease? Why or why not?
No; this experiment recruited volunteers as subjects. We should not generalize to some larger population of interest.
Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. A 95% confidence interval for pPR – pL is (0.013, 0.065). Explain how the confidence interval provides more information than a significance test.
The confidence interval does not include 0 as a plausible value for p1 - p2, which is consistent with the decision to reject H0: p1 - p2 = 0 in part (a). The confidence interval tells us that any value between 0.013 and 0.065 is plausible for p1 - p2 based on the sample data.
A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. Explain how the following change to the design of the experiment would affect the power of the test. Then give a drawback of making that change. Suppose that researchers had recruited twice as many infants for the LEAP trial.
Increasing the sample size will increase the power of the test. Increasing the sample size decreases the variability of both the null and alternative distributions, making it easier to reject the null hypothesis when it is false. A drawback is that an experiment that uses twice as many infants will be more expensive and will require a lot more work in following up with the parents of these infants when they are 5 years old.
A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. Explain how the following change to the design of the experiment would affect the power of the test. Then give a drawback of making that change. Suppose that researchers had used α= 0.10 instead of α= 0.05.
Using α = 0.10 instead of α = 0.05 will increase the power of the test. When α is larger, it is easier to reject the null hypothesis because the P-value doesn’t need to be as small. A drawback to increasing a is that doing so increases the probability of making a Type I error.
A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. Explain how the following change to the design of the experiment would affect the power of the test. Then give a drawback of making that change. Suppose that researchers had used 628 male subjects but no female subjects in the study.
Using male infants only would increase the power of the test by eliminating a source of variability, making it easier to reject the null hypothesis when it is false. A drawback is that this also limits the scope of inference to males only.
Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. Explain how the following change to the design of the experiment would affect the power of the test. Then give a drawback of making that change. Suppose that researchers had recruited twice as many subjects for the PROVE-IT Study.
Increasing the sample size will increase the power of the test. Increasing the sample size decreases the variability of both the null and alternative distributions, making it easier to reject the null hypothesis when it is false. A drawback is that an experiment that uses twice as many subjects will be more expensive and will require a lot more work in following up with the subjects at the end of the 2-year long experiment.
Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. Explain how the following change to the design of the experiment would affect the power of the test. Then give a drawback of making that change. Suppose that researchers had used α= 0.10 instead of α= 0.05.
Using α = 0.10 instead of α = 0.05 will increase the power of the test. When α is larger, it is easier to reject the null hypothesis because the P-value doesn’t need to be as small. A drawback to increasing a is that doing so increases the probability of making a Type I error.
Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. Explain how the following change to the design of the experiment would affect the power of the test. Then give a drawback of making that change. Suppose that researchers had used 4162 subjects under age 60 but no older subjects in the study.
Using only subjects under the age of 60 would increase the power of the test by eliminating a source of variability, making it easier to reject the null hypothesis when it is false. A drawback is that this also limits the scope of inference to those under the age of 60.
Some women would like to have children but cannot do so for medical reasons. One option for these women is a procedure called in vitro fertilization (IVF), which involves fertilizing an egg outside the woman’s body and implanting it in her uterus. Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to either a treatment group or a control group. Several people (called intercessors) prayed intentionally for the women in the treatment group, although they did not know the women, a process known as intercessory prayer. Prayers continued for 3 weeks following IVF. The intercessors did not pray for the women in the control group. Here are the results: 44 of the 88 women in the treatment group got pregnant, compared to 21 out of 81 in the control group. Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of H0: p1 = p2 verses Ha: p1> p2 where p1 and p2 are the actual pregnancy rates for women like those in the study who do and don’t receive intercessory prayer, respectively. Name the appropriate test and check that the conditions for carrying out this test are met.
Two-sample z test for p1 - p2. Random: Two groups in a randomized experiment. Large Counts: 33.88, 54.12, 31.185, 49.815 are ≥ 10.
Some women would like to have children but cannot do so for medical reasons. One option for these women is a procedure called in vitro fertilization (IVF), which involves fertilizing an egg outside the woman’s body and implanting it in her uterus. Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to either a treatment group or a control group. Several people (called intercessors) prayed intentionally for the women in the treatment group, although they did not know the women, a process known as intercessory prayer. Prayers continued for 3 weeks following IVF. The intercessors did not pray for the women in the control group. Here are the results: 44 of the 88 women in the treatment group got pregnant, compared to 21 out of 81 in the control group. Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of H0: p1 = p2 verses Ha: p1> p2 where p1 and p2 are the actual pregnancy rates for women like those in the study who do and don’t receive intercessory prayer, respectively. The appropriate test yields a P-value of 0.0007. Interpret this P-value in context.
If there is no difference in the true pregnancy rates of women prayed for and those who are not, there is a 0.0007 probability of getting a difference in pregnancy rates as large as or larger than the one observed (0.500 - 0.259 = 0.241) by chance alone.
Some women would like to have children but cannot do so for medical reasons. One option for these women is a procedure called in vitro fertilization (IVF), which involves fertilizing an egg outside the woman’s body and implanting it in her uterus. Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to either a treatment group or a control group. Several people (called intercessors) prayed intentionally for the women in the treatment group, although they did not know the women, a process known as intercessory prayer. Prayers continued for 3 weeks following IVF. The intercessors did not pray for the women in the control group. Here are the results: 44 of the 88 women in the treatment group got pregnant, compared to 21 out of 81 in the control group. Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of H0: p1 = p2 verses Ha: p1> p2 where p1 and p2 are the actual pregnancy rates for women like those in the study who do and don’t receive intercessory prayer, respectively. What conclusion should researchers draw at the α 0.05= significance level based on a P-level of 0.007?
Because the P-value of 0.0007 is less than α = 0.05, we reject H0. There is convincing evidence the pregnancy rates among women like these who are prayed for is higher than the rates for those not prayed for.
Some women would like to have children but cannot do so for medical reasons. One option for these women is a procedure called in vitro fertilization (IVF), which involves fertilizing an egg outside the woman’s body and implanting it in her uterus. Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to either a treatment group or a control group. Several people (called intercessors) prayed intentionally for the women in the treatment group, although they did not know the women, a process known as intercessory prayer. Prayers continued for 3 weeks following IVF. The intercessors did not pray for the women in the control group. Here are the results: 44 of the 88 women in the treatment group got pregnant, compared to 21 out of 81 in the control group. Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of H0: p1 = p2 verses Ha: p1> p2 where p1 and p2 are the actual pregnancy rates for women like those in the study who do and don’t receive intercessory prayer, respectively. The women in the study did not know whether they were being prayed for. Explain why this is important.
Knowing might have affected their behavior in some way (even unconsciously) that would have affected if they became pregnant. Then we wouldn’t know if the prayer or other behaviors caused the higher pregnancy rate.
A sample survey interviews SRSs of 500 female college students and 550 male college students. Researchers want to determine whether there is a difference in the proportion of male and female college students who worked for pay last summer. In all, 410 of the females and 484 of the males say they worked for pay last summer. Let pM and pF be the proportions of all college males and females who worked last summer. The hypotheses to be tested are (a) H0: pM – pF = 0 versus: Ha: pM – pF ≠ 0 (b) H0: pM – pF = 0 versus: Ha: pM – pF > 0 (c) H0: pM – pF = 0 versus: Ha: pM – pF < 0 (d) H0: pM – pF > 0 versus: Ha: pM – pF = 0 (e) H0: pM – pF ≠ 0 versus: Ha: pM – pF = 0
a
A sample survey interviews SRSs of 500 female college students and 550 male college students. Researchers want to determine whether there is a difference in the proportion of male and female college students who worked for pay last summer. In all, 410 of the females and 484 of the males say they worked for pay last summer. The researchers report that the results were statistically significant at the 1% level. Which of the following is the most appropriate conclusion? (a) Because the P-value is less than 1%, fail to reject H0. There is not convincing evidence that the proportion of male college students in the study who worked for pay last summer is different from the proportion of female college students in the study who worked for pay last summer. (b) Because the P-value is less than 1%, fail to reject H0. There is not convincing evidence that the proportion of all male college students who worked for pay last summer is different from the proportion of all female college students who worked for pay last summer. (c) Because the P-value is less than 1%, reject H0. There is convincing evidence that the proportion of all male college students who worked for pay last summer is the same as the proportion of all female college students who worked for pay last summer. (d) Because the P-value is less than 1%, reject H0. There is convincing evidence that the proportion of all male college students in the study who worked for pay last summer is different from the proportion of all female college students in the study who worked for pay last summer. (e) Because the P-value is less than 1%, reject H0. There is convincing evidence that the proportion of all male college students who worked for pay last summer is different from the proportion of all female college students who worked for pay last summer.
e
A sample survey interviews SRSs of 500 female college students and 550 male college students. Researchers want to determine whether there is a difference in the proportion of male and female college students who worked for pay last summer. In all, 410 of the females and 484 of the males say they worked for pay last summer. Let pM and pF be the proportions of all college males and females who worked last summer. The hypotheses to be tested are H0: pM – pF = 0 versus: Ha: pM – pF ≠ 0. Which of the following is the correct standard error for the test? (a) sqrt((0.851(0.149))/1050) (b) sqrt(((0.851(0.149))/550) + ((0.851(0.149))/500)) (c) sqrt(((0.880(0.120))/550) + ((0.820(0.180))/500)) (d) sqrt(((0.851(0.149))/1050) + ((0.851(0.149))/1050)) (e) sqrt(((0.880(0.120))/1050) + ((0.820(0.180))/1050))
b
In an experiment to learn whether substance M can help restore memory, the brains of 20 rats were treated to damage their memories. First, the rats were trained to run a maze. After a day, 10 rats (determined at random) were given substance M and 7 of them succeeded in the maze. Only 2 of the 10 control rats were successful. The two-sample z test for the difference in the true proportions (a) gives z = 2.25, P < 0.02. (b) gives z = 2.60, P < 0.005. (c) gives z = 2.25, P < 0.04 but not <0.02. (d) should not be used because the Random condition is violated. (e) should not be used because the Large Counts condition is violated.
e
Rob once read that one-quarter of all people have played/danced in the rain at some point in their lives. His friend Justin thinks that the proportion is higher than 0.25 for their high school. To settle their dispute, they ask a random sample of 80 students in their school and find out that 28 have played/danced in the rain. State the appropriate null and alternative hypotheses.
H0: p= 0.25; Ha: p> 0.25, where p = the true proportion of all students at this school who have played/ danced in the rain.
Rob once read that one-quarter of all people have played/danced in the rain at some point in their lives. His friend Justin thinks that the proportion is higher than 0.25 for their high school. To settle their dispute, they ask a random sample of 80 students in their school and find out that 28 have played/danced in the rain. Explain why the sample data give some evidence for Ha.
The sample data give some evidence for Ha because p^ = 28/80 = 0.35 > 0.25.
Rob once read that one-quarter of all people have played/danced in the rain at some point in their lives. His friend Justin thinks that the proportion is higher than 0.25 for their high school. To settle their dispute, they ask a random sample of 80 students in their school and find out that 28 have played/danced in the rain. Identify the appropriate test to perform and show that the conditions for carrying out the test are met.
Appropriate test: One-sample z test for p. Random: We have a random sample of 80 students from this school. 10%: Assume the sample size (80) is less than 10% of all students at this school. Normal/Large Sample: np0 = 80(0.25) = 20 ≥ 10 and n(1 - p0) = 80(0.75) = 60 ≥ 10.
Rob once read that one-quarter of all people have played/danced in the rain at some point in their lives. His friend Justin thinks that the proportion is higher than 0.25 for their high school. To settle their dispute, they ask a random sample of 80 students in their school and find out that 28 have played/danced in the rain. Find the standardized test statistic and P-value, and make an appropriate conclusion.
z= 2.07; P-value = 0.0188 Conclusion: We will use α = 0.05. Because the P-value of 0.0118 > α = 0.05, we reject H0. We have convincing evidence that the true proportion of all students at this school who have played/ danced in the rain in their lifetime is greater than 0.25
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. State appropriate hypotheses for testing the company’s claim. Be sure to define your parameter.
H0: p= 0.05, Ha: p< 0.05, where p is the true proportion of adults who will get the flu after using the vaccine. We will use α = 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. Describe a Type I error and a Type II error in this setting, and give the consequences of each.
Type I: Finding convincing evidence that less than 5% of patients would get the flu, when in reality at least 5% would. Consequence: The company might get sued for false advertising. Type II: Failing to find convincing evidence that less than 5% would get the flu, when in reality less than 5% would. Consequence: Loss of potential income.
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. Would you recommend a significance level of 0.01, 0.05, or 0.10 for this test? Justify your choice.
Because a Type I error is more serious in this case, and P(Type I error) = α, I would recommend a significance level of α = 0.01 to minimize the possibility of making this type of error.