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A study of road rage asked random samples of 596 men and 523 women about their behavior while driving. Based on their answers, each person was assigned a road rage score on a scale of 0 to 20. The participants were chosen by random digit dialing of phone numbers. The researchers performed a test of the following hypotheses: H0: µM = µF versus Ha: µM ≠µF. The P-value for the stated hypotheses is 0.002. Interpret this value in the context of this study. (a) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting a difference in sample means equal to the one observed in this study. (b) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting a difference in sample means at least as large in either direction as the one observed in this study. (c) Assuming that the true mean road rage score is different for males and females, there is a 0.002 probability of getting a difference in sample means at least as large in either direction as the one observed in this study. (d) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the null hypothesis is true. (e) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the alternative hypothesis is true.
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b
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A study of road rage asked random samples of 596 men and 523 women about their behavior while driving. Based on their answers, each person was assigned a road rage score on a scale of 0 to 20. The participants were chosen by random digit dialing of phone numbers. The researchers performed a test of the following hypotheses: H0: µM = µF versus Ha: µM ≠µF. Based on the P-value in Exercise 66, which of the following must be true? (a) A 90% confidence interval for µM - µF will contain 0. (b) A 95% confidence interval for µM - µF will contain 0. (c) A 99% confidence interval for µM - µF will contain 0. (d) A 99.9% confidence interval for µM - µF will contain 0. (e) It is impossible to determine whether any of these statements is true based only on the P-value.
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d
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The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from a large local high school differs from the national average. You measure an SRS of 48 female graduates and find that x¯ = 63.5 inches and sx = 3.7 inches. State the appropriate null and alternative hypotheses for performing a significance test.
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H0: µ = 64.2, Ha: µ ≠ 64.2, where µ = the true mean height of all female graduates from the large local high school this year.
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The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from a large local high school differs from the national average. You measure an SRS of 48 female graduates and find that x¯ = 63.5 inches and sx = 3.7 inches. Explain why the sample data give some evidence for Ha.
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The observed sample mean is x= 63.5 inches, which does not equal 64.2 inches.
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The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from a large local high school differs from the national average. You measure an SRS of 48 female graduates and find that x¯ = 63.5 inches and sx = 3.7 inches. Identify the appropriate test and show that the conditions for carrying out the test are met.
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One-sample t test for a population mean. Random: Random sample of 48 female graduates. 10%: n = 48 < 10% of all female graduates from this large high school. Normal/ Large Sample: n = 48 ≥ 30
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The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from a large local high school differs from the national average. You measure an SRS of 48 female graduates and find that x¯ = 63.5 inches and sx = 3.7 inches. Find the standardized test statistic and P-value.
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t = -1.31, df = 47, P-value = 0.1966
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The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from a large local high school differs from the national average. You measure an SRS of 48 female graduates and find that x¯ = 63.5 inches and sx = 3.7 inches. Interpret the P-value. What conclusion would you make?
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P-value interpretation: Assuming that the true mean height of all female graduates is 64.2 inches, there is a 0.1966 probability of getting a sample mean as different as or more different than 63.5 inches (in either direction) by chance alone. Conclusion: Because the P-value of 0.1966 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true mean height of all female graduates from this high school is different from the national average of 64.2 inches.
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Does the use of fancy type fonts slow down the reading of text on a computer screen? Adults can read four paragraphs of a certain text in the common Times New Roman font in an average of 22 seconds. Researchers asked a random sample of 24 adults to read this text in the ornate font named Gigi. Here are their times (in seconds). 23.2 21.2 28.9 27.7 29.1 27.3 16.1 22.6 25.6 34.2 23.9 26.8 20.5 34.3 21.4 32.6 26.2 34.1 31.5 24.6 23.0 28.6 24.4 28.1 Do these data provide convincing evidence that it takes adults longer than 22 seconds, on average, to read these four paragraphs in Gigi font?
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S: H0: µ = 22, Ha: µ > 22, where µ = the true mean amount of time (in seconds) it takes adults to read four paragraphs of text in the ornate font Gigi. We will use α = 0.05. P: One-sample t test for µ. Random: We have a random sample of 24 adults. 10%: The sample size (24) is less than 10% of the population of adults. Normal/ Large Sample: Because the sample size is small, we need to graph the sample data. The histogram shows that the distribution is roughly symmetric with no outliers, so using a t procedure is appropriate. D: x¯= 26.496, sx= 4.728, t= 4.66, df = 23, and P-value = 0.000054. C: Because the P-value of 0.000054 < α = 0.05, we reject H0 . There is convincing evidence that the true mean amount of time it takes adults to read four paragraphs of text in the ornate font Gigi is greater than 22 seconds.
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In the children’s game Don’t Break the Ice, small plastic ice cubes are squeezed into a square frame. Each child takes turns tapping out a cube of “ice” with a plastic hammer, hoping that the remaining cubes don’t collapse. For the game to work correctly, the cubes must be big enough so that they hold each other in place in the plastic frame, but not so big that they are too difficult to tap out. The machine that produces the plastic cubes is designed to make cubes that are 25.4 millimeters (mm) wide, but the width varies a little. To ensure that the machine is working well, a supervisor inspects a random sample of 50 cubes every hour and measures their width. If the sample provides convincing evidence at the α= 0.05 significance level that the true mean width μ of the cubes produced that hour differs from 25.4 mm, the supervisor will discard all of the cubes. Describe a Type II error in this setting.
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Type II error: Failing to find convincing evidence that the true mean width of the cubes differs from 25.4 mm, when the true mean width of the cubes differs from 25.4 mm.
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In the children’s game Don’t Break the Ice, small plastic ice cubes are squeezed into a square frame. Each child takes turns tapping out a cube of “ice” with a plastic hammer, hoping that the remaining cubes don’t collapse. For the game to work correctly, the cubes must be big enough so that they hold each other in place in the plastic frame, but not so big that they are too difficult to tap out. The machine that produces the plastic cubes is designed to make cubes that are 25.4 millimeters (mm) wide, but the width varies a little. To ensure that the machine is working well, a supervisor inspects a random sample of 50 cubes every hour and measures their width. If the sample provides convincing evidence at the α= 0.05 significance level that the true mean width μ of the cubes produced that hour differs from 25.4 mm, the supervisor will discard all of the cubes. Identify one benefit and one drawback of changing the significance level to α= 0.10.
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This change will reduce the probability of making a Type II error. A drawback to this change is an increased chance of rejecting batches of cubes that are of the appropriate width.
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In the children’s game Don’t Break the Ice, small plastic ice cubes are squeezed into a square frame. Each child takes turns tapping out a cube of “ice” with a plastic hammer, hoping that the remaining cubes don’t collapse. For the game to work correctly, the cubes must be big enough so that they hold each other in place in the plastic frame, but not so big that they are too difficult to tap out. The machine that produces the plastic cubes is designed to make cubes that are 25.4 millimeters (mm) wide, but the width varies a little. To ensure that the machine is working well, a supervisor inspects a random sample of 50 cubes every hour and measures their width. If the sample provides convincing evidence at the α= 0.05 significance level that the true mean width μ of the cubes produced that hour differs from 25.4 mm, the supervisor will discard all of the cubes. The data from a sample taken during 1 hour result in a 95% confidence interval of (25.39, 25.441). What conclusion should the supervisor make?
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Because m 5 25.4 mm is one of the plausible values in this 95% confidence interval, the supervisor should not reject this batch as being significantly different from the intended mean width of 25.4 mm.
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Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. The mean time is 18 seconds for one particular maze. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures how long each of 10 mice takes with a noise stimulus. What are the null and alternative hypotheses for the appropriate significance test? (a) H0: μ = 18 vs. Ha: μ ≠ 18 (b) H0: μ = 18 vs. Ha: μ< 18 (c) H0: μ = 18 vs. Ha: μ> 18 (d) H0: μ< 18 vs. Ha: μ = 18 (e) H0: μ≠ 18 vs. Ha: μ = 18
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b
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You are thinking of conducting a one-sample t test about a population mean μ using a 0.05 significance level. Which of the following statements is correct? (a) You should not carry out the test if the sample does not have a Normal distribution. (b) You can safely carry out the test if there are no outliers, regardless of the sample size. (c) You can carry out the test if a graph of the data shows no strong skewness, regardless of the sample size. (d) You can carry out the test only if the population standard deviation is known. (e) You can safely carry out the test if your sample size is at least 30.
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e
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A 95% confidence interval for μ based on n= 15 observations from a Normal population is (–0.73, 1.92). If we use this confidence interval to test the hypothesis H0: μ= 0 against Ha: μ ≠ 0, which of the following is the most appropriate conclusion? (a) Reject H0 at the α= 0.05 level of significance. (b) Fail to reject H0 at the α= 0.05 level of significance. (c) Reject H0 at the α= 0.10 level of significance. (d) Fail to reject H0 at the α= 0.10 level of significance.(e) We cannot perform the required test since we do not know the value of the standardized test statistic.
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b
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Which of the following has the smallest probability? (a)P(t>2) if t has 5 degrees of freedom (b) P(t>2) if t has 2 degrees of freedom. (c) P(z>2) if z is a standard Normal random variable. (d) P(t<2) if t has 5 degrees of freedom. (e) P(z<2) if z is a standard Normal random variable.
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c
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A significance test was performed to test H0: μ = 2 versus the alternative Ha: μ ≠ 2. A sample of size 28 produced a standardized test statistic of t= 2.051. Assuming all conditions for inference were met, which of the following intervals contains the P-value for this test? (a) 0.01 < P < 0.02 (b) 0.02 < P <0.025 (c) 0.025 < P < 0.05 (d) 0.05 < P < 0.10 (e) P > 0.10
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d
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A study of road rage asked separate random samples of 596 men and 523 women about their behavior while driving. Based on their answers, each respondent was assigned a road rage score on a scale of 0 to 20. Are the conditions for performing a two-sample t test satisfied? (a) Maybe; we have independent random samples, but we should look at the data to check Normality. (b) No; road rage scores on a scale from 0 to 20 can’t be Normal. (c) No; we don’t know the population standard deviations. (d) Yes; the large sample sizes guarantee that the corresponding population distributions will be Normal. (e) Yes; we have two independent random samples and large sample sizes.
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e
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A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with the average time spent by students in a large city school district. The researcher obtained an SRS of 60 high school students in a large suburban school district and found the mean time spent in extracurricular activities per week to be 6 hours with a standard deviation of 3 hours. The researcher also obtained an independent SRS of 40 high school students in a large city school district and found the mean time spent in extracurricular activities per week to be 5 hours with a standard deviation of 2 hours. Suppose that the researcher decides to carry out a significance test of H0: µsuburban = µcity versus a two-sided alternative. Which is the correct standardized test statistic? (a) z = ((6-5)-0) / sqrt((3/60) + (2/40)) (b) z = ((6-5)-0) / sqrt((3^2/60) + (2^2/40)) (c) t = ((6-5)-0) / (3 / sqrt(60) + 2 / sqrt(40)) (d) t = ((6-5)-0) / sqrt((3/60) + (2/40)) (e) t = ((6-5)-0) / sqrt((3^2/60) + (2^2/40))
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e
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A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with the average time spent by students in a large city school district. The researcher obtained an SRS of 60 high school students in a large suburban school district and found the mean time spent in extracurricular activities per week to be 6 hours with a standard deviation of 3 hours. The researcher also obtained an independent SRS of 40 high school students in a large city school district and found the mean time spent in extracurricular activities per week to be 5 hours with a standard deviation of 2 hours. Suppose that the researcher decides to carry out a significance test of H0: µsuburban = µcity versus a two-sided alternative. The P-value for the test is 0.048. A correct conclusion is to (a) fail to reject H0 because 0.048<α= 0.05. There is convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts. (b) fail to reject H0 because 0.048<α= 0.05. There is not convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts. (c) fail to reject H0 because 0.048<α= 0.05. There is convincing evidence that the average time spent on extracurricular activities by students in the suburban and city school districts is the same. (d) reject H0 because 0.048<α= 0.05. There is not convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts. (e) reject H0 because 0.048<α= 0.05. There is convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts.
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e
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Are TV commercials louder than their surrounding programs? To find out, researchers collected data on 50 randomly selected commercials in a given week. With the television’s volume at a fixed setting, they measured the maximum loudness of each commercial and the maximum loudness in the first 30 seconds of regular programming that followed. Assuming conditions for inference are met, the most appropriate method for answering the question of interest is (a) a two-sample t test for a difference in means. (b) a two-sample t interval for a difference in means. (c) a paired t test for a mean difference. (d) a paired t interval for a mean difference. (e) a two-sample z test for a difference in proportions.
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c
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Researchers want to evaluate the effect of a natural product on reducing blood pressure. They plan to carry out a randomized experiment to compare the mean reduction in blood pressure of a treatment (natural product) group and a placebo group. Then they will use the data to perform a test of H0: µT - µP = 0 versus Ha: µT - µP > 0, where μT = the true mean reduction in blood pressure when taking the natural product and μP = the true mean reduction in blood pressure when taking a placebo for subjects like the ones in the experiment. The researchers would like to detect whether the natural product reduces blood pressure by at least 7 points more, on average, than the placebo. If groups of size 50 are used in the experiment, a two-sample t test using α= 0.01 will have a power of 80% to detect a 7-point difference in mean blood pressure reduction. If the researchers want to be able to detect a 5-point difference instead, then the power of the test(a) would be less than 80%. (b) would be greater than 80%. (c) would still be 80%. (d) could be either less than or greater than 80%. (e) would vary depending on the values of μT and μP.
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a
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A government report says that the average amount of money spent per U.S. household per week on food is about $158. A random sample of 50 households in a small city is selected, and their weekly spending on food is recorded. The sample data have a mean of $165 and a standard deviation of $32. Is there convincing evidence that the mean weekly spending on food in this city differs from the national figure of $158? State appropriate hypotheses for performing a significance test in this setting. Be sure to define the parameter of interest.
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We want to test H0: µ = $158, Ha: µ ≠ $158, where µ = the true mean amount spent on food by households in this city. We will perform the test at the α = 0.05 significance level.
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A government report says that the average amount of money spent per U.S. household per week on food is about $158. A random sample of 50 households in a small city is selected, and their weekly spending on food is recorded. The sample data have a mean of $165 and a standard deviation of $32. Is there convincing evidence that the mean weekly spending on food in this city differs from the national figure of $158? The distribution of household spending in this small city is heavily skewed to the right. Explain why the Normal/Large Sample condition is met in this case.
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The Normal/Large sample condition is met because n = 50 ≥ 30.
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A government report says that the average amount of money spent per U.S. household per week on food is about $158. A random sample of 50 households in a small city is selected, and their weekly spending on food is recorded. The sample data have a mean of $165 and a standard deviation of $32. Is there convincing evidence that the mean weekly spending on food in this city differs from the national figure of $158? The P-value of the test is 0.128. Interpret this value. What conclusion would you make?
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Assuming that the true mean amount of money spent on food per household in this city is $158, there is a 0.1283 probability of getting a sample mean as different as or more different than $165 (in either direction) by chance alone. Conclusion: Because the P-value of 0.128 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true mean amount spent on food per household in this city is different from the national average of $158.
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For estimating a population characteristic, why is an unbiased statistic with a small standard error preferred over an unbiased statistic with a larger standard error?
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An unbiased statistic with a smaller standard error is preferred because it is likely to result in an estimate that is closer to the actual value of the population characteristic than an unbiased statistic that has a larger standard error.
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A researcher wants to estimate the proportion of students enrolled at a university who eat fast food more than three times in a typical week. Would the standard error of the sample proportion p^ be smaller for random samples of size n = 50 or random samples of size n = 200?
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n = 200
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Use the formula for the standard error of p^ to explain why the standard error is greater when the value of the population proportion p is near 0.5 than when it is near 1. 459.
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The formula for the standard error of p^ is σp^ = sqrt((p(1 - p))/n). The quantity p(1-p) reaches a maximum value when p = 0.5.
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Use the formula for the standard error of p^ to explain why the standard error of p^ is the same when the value of the population proportion is p = 0.2 as it is when p =0.8.
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The standard error of p^ is the same when p = 0.2 as when p = 0.8 because when p = 0.2, (1 - p) = (1 – 0.2) = 0.8. Similarly, when p = 0.8, (1 - p) = (1 – 0.8) = 0.2. So, the quantity p(1-p) is the same in both cases.
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A random sample will be selected from the population of all adult residents of a particular city. The sample proportion p^ will be used to estimate p, the proportion of all adult residents who are employed full time. For which of the following situations will the estimate tend to be closest to the actual value of p? i. n = 500, p = 0.6 ii. n = 450, p = 0.7 iii. n = 400, p = 0.8
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n = 400 and p = 0.8
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If two statistics are available for estimating a population characteristic, under what circumstances might you choose a biased statistic over an unbiased statistic?
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A biased statistic might be chosen over an unbiased statistic if the bias is not too large, and the standard error of the biased statistic is much smaller than the standard error of the unbiased statistic. In this case, the observed value of the biased statistic might be closer to the actual value than the value of an unbiased statistic.
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A researcher wants to estimate the proportion of city residents who favor spending city funds to promote tourism. Would the standard error of the sample proportion be smaller for random samples of size n = 100 or random samples of size n = 200?
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n = 200
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A large online retailer is interested in learning about the proportion of customers making a purchase during a particular month who were satisfied with the online ordering process. A random sample of 600 of these customers included 492 who indicated they were satisfied. For the following statement, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement: It is unlikely that the estimate p^=0.82 differs from the value of the actual population proportion by more than 0.0157.
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Incorrect, because the value 0.0157 is the standard error of p^, and therefore approximately 32% of all possible values of p^ would differ from the value of the actual population proportion by more than 0.0157 (using properties of the normal distribution).
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A large online retailer is interested in learning about the proportion of customers making a purchase during a particular month who were satisfied with the online ordering process. A random sample of 600 of these customers included 492 who indicated they were satisfied. For the following statement, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement: It is unlikely that the estimate p^=0.82 differs from the value of the actual population proportion by more than 0.0307.
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Correct
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A large online retailer is interested in learning about the proportion of customers making a purchase during a particular month who were satisfied with the online ordering process. A random sample of 600 of these customers included 492 who indicated they were satisfied. For the following statement, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement: The estimate p^=0.82 will never differ from the value of the actual population proportion by more than 0.0307.
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Incorrect, because the phrase “will never differ from the value of the actual population proportion” is wrong. The value 0.0307 is the margin of error and indicates that in about 95% of all possible random samples, the estimation error will be less than the margin of error. In about 5% of the random samples, the estimation error will be greater than the margin of error.
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Consider taking a random sample from a population with p = 0.40. What is the standard error of p^ for random samples of size 100?
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0.049
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Consider taking a random sample from a population with p = 0.40. Would the standard error of p^ be larger for samples of size 100 or samples of size 200?
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n = 100
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Consider taking a random sample from a population with p = 0.40. If the sample size were doubled from 100 to 200, by what factor would the standard error of p^ decrease?
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1 / sqrt(2) = 0.707
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The report “2005 Electronic Monitoring & Surveillance Survey: Many Companies Monitoring, Recording, Videotaping—and Firing—Employees” (American Management Association, 2005) summarized a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet. Assume that this sample is representative of businesses in the United States. Estimate the proportion of all businesses in the U.S. that have fired workers for misuse of the Internet. What statistic did you use?
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p^ = 0.260
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The report “2005 Electronic Monitoring & Surveillance Survey: Many Companies Monitoring, Recording, Videotaping—and Firing—Employees” (American Management Association, 2005) summarized a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet. Assume that this sample is representative of businesses in the United States. Use the sample data to estimate the standard error of p^.
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σp^ = 0.019
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The report “2005 Electronic Monitoring & Surveillance Survey: Many Companies Monitoring, Recording, Videotaping—and Firing—Employees” (American Management Association, 2005) summarized a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet. Assume that this sample is representative of businesses in the United States. Compute and interpret the margin of error.
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margin of error = 0.037. The estimate of the proportion of all businesses that have fired workers for misuse of the Internet is unlikely to differ from the actual population proportion by more than 0.037.
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The use of the formula for margin of error requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for use of this formula to be appropriate: n = 50 and p^ = 0.30.
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yes
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The use of the formula for margin of error requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for use of this formula to be appropriate: n = 50 and p^ = 0.05.
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no
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The use of the formula for margin of error requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for use of this formula to be appropriate: n= 15 and p^ = 0.45.
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no
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The use of the formula for margin of error requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for use of this formula to be appropriate: n = 100 and p^ = 0.01.
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no
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Most American college students make use of the Internet for both academic and social purposes. What proportion of students use it for more than 3 hours a day? The authors of the paper “U.S. College Students’ Internet Use: Race, Gender and Digital Divides” (Journal of Computer-Mediated Communication [2009]: 244–264) describe a survey of 7,421 students at 40 colleges and universities. The sample was selected to reflect general demographics of U.S. college students. Of the students surveyed, 2,998 reported Internet use of more than 3 hours per day. Use the given information to estimate the proportion of college students who use the Internet more than 3 hours per day.
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p^ = 0.404
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Most American college students make use of the Internet for both academic and social purposes. What proportion of students use it for more than 3 hours a day? The authors of the paper “U.S. College Students’ Internet Use: Race, Gender and Digital Divides” (Journal of Computer-Mediated Communication [2009]: 244–264) describe a survey of 7,421 students at 40 colleges and universities. The sample was selected to reflect general demographics of U.S. college students. Of the students surveyed, 2,998 reported Internet use of more than 3 hours per day. Verify that the conditions needed in order for the margin of error formula to be appropriate are met.
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The sample was selected in such a way that makes it representative of the population of U.S. college students. Additionally, there are 2,998 successes and 4,423 failures in the sample, which are both at least 10.
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Most American college students make use of the Internet for both academic and social purposes. What proportion of students use it for more than 3 hours a day? The authors of the paper “U.S. College Students’ Internet Use: Race, Gender and Digital Divides” (Journal of Computer-Mediated Communication [2009]: 244–264) describe a survey of 7,421 students at 40 colleges and universities. The sample was selected to reflect general demographics of U.S. college students. Of the students surveyed, 2,998 reported Internet use of more than 3 hours per day. Compute the margin of error.
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margin of error = 0.011
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Most American college students make use of the Internet for both academic and social purposes. What proportion of students use it for more than 3 hours a day? The authors of the paper “U.S. College Students’ Internet Use: Race, Gender and Digital Divides” (Journal of Computer-Mediated Communication [2009]: 244–264) describe a survey of 7,421 students at 40 colleges and universities. The sample was selected to reflect general demographics of U.S. college students. Of the students surveyed, 2,998 reported Internet use of more than 3 hours per day. Interpret the margin of error in the context of this problem.
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It is unlikely that the estimated proportion of U.S. college students who use the Internet more than 3 hours per day (p^ = 0.404) will differ from the actual population proportion by more than 0.011 (or 1.1%).
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Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article “Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children” (Pediatrics [2004]: 112–118) reported that 1,720 of those in a random sample of 6,212 American children indicated that they ate fast food on a typical day. Use the given information to estimate the proportion of American children who eat fast food on a typical day.
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p^ = 0.277
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Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article “Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children” (Pediatrics [2004]: 112–118) reported that 1,720 of those in a random sample of 6,212 American children indicated that they ate fast food on a typical day. Verify that the conditions needed in order for the margin of error formula to be appropriate are met.
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The sample is a random sample from the population of American children. Additionally, there are 1,720 successes and 4,492 failures in the sample, which are both greater than 10.
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Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article “Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children” (Pediatrics [2004]: 112–118) reported that 1,720 of those in a random sample of 6,212 American children indicated that they ate fast food on a typical day. Compute the margin of error.
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margin of error = 0.011
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Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article “Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children” (Pediatrics [2004]: 112–118) reported that 1,720 of those in a random sample of 6,212 American children indicated that they ate fast food on a typical day. Interpret the margin of error in the context of this problem.
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It is unlikely that the estimated proportion of American children who indicated that they eat fast food on a typical day (p^ = 0.277) will differ from the actual population proportion by more than 0.011.
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The article “Viewers Speak Out Against Reality TV” (Associated Press, September 12, 2005) included the following statement: “Few people believe there’s much reality in reality TV: a total of 82% said the shows are either ‘totally made up’ or ‘mostly distorted.’” This statement was based on a survey of 1,002 randomly selected adults. Compute and interpret the margin of error for the reported percentage.
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margin of error = 0.024. It is unlikely that the estimated proportion of adults who believe that the shows are mostly or totally made up (p^ = 0.82) will differ from the actual population proportion by more than 0.024.
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The Gallup Organization conducts and annual survey on crime. It was reported that 25% of all households experienced some sort of crime during the past year. This estimate was based on a sample of 1,002 selected adults. The report states, “One can say with 95% confidence that the margin of sampling error is ±3 percentage points.” Explain how this statement is justified.
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In this case, margin of error = 0.027, which rounds to 3%.
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An article in the Chicago Tribue (August 29, 1999) reported that in a poll of residents of the Chicago suburbs, 43% felt that their financial situation had improved during the past year. The following state is from the article: “The findings of this Tribune poll are based on interviews with 930 randomly selected suburban residents. The sample included suburban Cook County plus DuPage, Kane, Lake, McHenry, and Will Counties. In a sample of this size, one can say with 95% certainty that results will differ by no more than 3% from results obtained if all residents had been included in the poll.” Give a statistical argument to justify the claim that the estimate of 43% is within 3% of the actual percentage of all residents who feel that their financial situation has improved.
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In this case, margin of error = 0.032, which rounds to 3%. The margin of error tells you that it is unlikely that the estimate will differ from the actual population proportion by more than 0.03.
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Suppose that a city planning commission wants to know the proportion of city residents who support installing streetlights in the downtown area. Two different people independently selected random samples of city residents and used their sample data to construct the following confidence intervals for the population proportion: Interval 1: (0.28, 0.34) Interval 2: (0.31, 0.33). Explain how it is possible that the two confidence intervals are not centered in the same place.
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The confidence intervals are centered at p^. In this case, the intervals are not centered in the same place because two different samples were taken, each yielding a different value of p^.
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Suppose that a city planning commission wants to know the proportion of city residents who support installing streetlights in the downtown area. Two different people independently selected random samples of city residents and used their sample data to construct the following confidence intervals for the population proportion: Interval 1: (0.28, 0.34) Interval 2: (0.31, 0.33). Which of the two intervals conveys more precise information about the value of the population proportion?
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Interval 2 conveys more precise information about the value of the population proportion because Interval 2 is narrower than Interval 1.
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Suppose that a city planning commission wants to know the proportion of city residents who support installing streetlights in the downtown area. Two different people independently selected random samples of city residents and used their sample data to construct the following confidence intervals for the population proportion: Interval 1: (0.28, 0.34) Interval 2: (0.31, 0.33). If both confidence intervals have a 95% confidence level, which confidence interval was based on the smaller sample size? How can you tell?
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A smaller sample size produces a larger margin of error. In this case, Interval 1 (being wider than Interval 2) was based on the smaller sample size.
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Suppose that a city planning commission wants to know the proportion of city residents who support installing streetlights in the downtown area. Two different people independently selected random samples of city residents and used their sample data to construct the following confidence intervals for the population proportion: Interval 1: (0.28, 0.34) Interval 2: (0.31, 0.33). If both confidence intervals were based on the same sample size, which interval has the higher confidence level? How can you tell?
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Interval 1 would have the higher confidence level, because the z critical value for higher confidence is larger, resulting in a wider confidence interval.
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Explain which one would result in a wider large-sample confidence interval for p: 90% confidence level or 95% confidence level
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0.95
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Explain which one would result in a wider large-sample confidence interval for p: n = 100 or n = 400
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n = 100
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Based on data from a survey of 1,200 randomly selected Facebook users (USA Today, March 24, 2010), a 95% confidence interval for the proportion of all Facebook users who say it is OK for someone to “friend” his or her boss is (0.41, 0.47). What is the meaning of the confidence level of 95% that is associated with this interval?
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The method used to construct this interval estimate is successful in capturing the actual value of the population proportion about 95% of the time.
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Appropriate use of the interval p^ ± (z critical values) sqrt((p^(1-p^))/n) requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for this interval to be appropriate: n = 50 and p^ = 0.30
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yes
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Appropriate use of the interval p^ ± (z critical values) sqrt((p^(1-p^))/n) requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for this interval to be appropriate: n = 50 and p^ = 0.05
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no
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Appropriate use of the interval p^ ± (z critical values) sqrt((p^(1-p^))/n) requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for this interval to be appropriate: n = 15 and p^ = 0.45
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no
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Appropriate use of the interval p^ ± (z critical values) sqrt((p^(1-p^))/n) requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for this interval to be appropriate: n = 100 and p^ = 0.01
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no
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The formula used to compute a large-sample confidence interval for p is p^ ± (z critical values) sqrt((p^(1-p^))/n). What is the appropriate z critical value for a 90% confidence level?
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1.1645
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The formula used to compute a large-sample confidence interval for p is p^ ± (z critical values) sqrt((p^(1-p^))/n). What is the appropriate z critical value for a 99% confidence level?
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2.58
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The formula used to compute a large-sample confidence interval for p is p^ ± (z critical values) sqrt((p^(1-p^))/n). What is the appropriate z critical value for a 80% confidence level?
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1.28
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The article “Career Expert Provides DOs and DON'Ts for Job Seekers on Social Networking” (CareerBuilder.com, August 19, 2009) included data from a survey of 2,667 hiring managers and human resource professionals. The article noted that more employers are now using social networks to screen job applicants. Of the 2,667 people who participated in the survey, 1,200 indicated that they use social networking sites such as Facebook, MySpace, and LinkedIn to research job applicants. Assume that the sample is representative of hiring managers and human resource professionals. Answer the four key questions (QSTN) to confirm that the suggested method in this situation is a confidence interval for a population proportion.
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Question type (Q): Estimation; Study type (S): Sample data; Type of data (T): One categorical variable; Number of samples or treatments (N): One sample.
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The article “Career Expert Provides DOs and DON'Ts for Job Seekers on Social Networking” (CareerBuilder.com, August 19, 2009) included data from a survey of 2,667 hiring managers and human resource professionals. The article noted that more employers are now using social networks to screen job applicants. Of the 2,667 people who participated in the survey, 1,200 indicated that they use social networking sites such as Facebook, MySpace, and LinkedIn to research job applicants. Assume that the sample is representative of hiring managers and human resource professionals. Use the five-step process for estimation problems (EMC) to construct and interpret a 95% confidence interval for the proportion of hiring managers and human resource professionals who use social networking sites to research job applicants. Identify each of the five steps in your solution.
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Estimate (E): The proportion of hiring managers and human resources professionals who use social networking sites to research job applicants, p, will be estimated. Method (M): Because the answers to the four key questions are estimation, sample data, one categorical variable, and one sample, consider a 95% confidence interval for the proportion of hiring managers and human resource professionals who use social networking sites to research job applicants. Check (C): The sample is representative of hiring managers and human resource professionals. In addition, the sample includes 1,200 successes and 1,467 failures, which are both greater than 10. The two required conditions are satisfied. Calculations (C): (0.4311, 0.4689) Communicate Results (C): Interpret confidence interval: You can be 95% confident that the actual proportion of hiring managers and human resources professionals is somewhere between 0.4311 and 0.4689. Interpret confidence level: The method used to construct this interval estimate is successful in capturing the actual value of the population proportion about 95% of the time.
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The article “Students Increasingly Turn to Credit Cards” (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college freshmen and 48% of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1,000 college freshmen and 1,000 college seniors. Construct and interpret a 90% confidence interval for the proportion of college freshmen who carry a credit card balance from month to month.
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(0.3449, 0.3951). You can be 90% confident that the actual proportion of college freshmen who carry a credit card balance is somewhere between 0.3449 and 0.3951.
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The article “Students Increasingly Turn to Credit Cards” (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college freshmen and 48% of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1,000 college freshmen and 1,000 college seniors. Construct and interpret a 90% confidence interval for the proportion of college seniors who carry a credit card balance from month to month.
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(0.454, 0.506). You can be 90% confident that the actual proportion of college seniors who carry a credit card balance is somewhere between 0.454 and 0.506.
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The article “Students Increasingly Turn to Credit Cards” (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college freshmen and 48% of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1,000 college freshmen and 1,000 college seniors. Explain why the 90% confidence intervals for freshman and seniors are not the same width.
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The two confidence intervals do not have the same width because the standard errors, and hence the margins of error, are based on two different values for p^.
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In a survey of 1,000 randomly selected adults in the United States, participants were asked what their most favorite and least favorite subjects were when they were in school (Associated Press, August 17, 2005). In what might seem like a contradiction, math was chosen more often than any other subject in both categories. Math was chosen by 230 of the 1,000 as their most favorite subject and chosen by 370 of the 1,000 as their least favorite subject. Construct and interpret a 95% confidence interval for the proportion of U.S. adults for whom math was their most favorite subject.
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(0.2039, 0.2561). You can be 95% confident that the actual proportion of U.S. adults for whom math was their most favorite subject is somewhere between 0.2039 and 0.2561.
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In a survey of 1,000 randomly selected adults in the United States, participants were asked what their most favorite and least favorite subjects were when they were in school (Associated Press, August 17, 2005). In what might seem like a contradiction, math was chosen more often than any other subject in both categories. Math was chosen by 230 of the 1,000 as their most favorite subject and chosen by 370 of the 1,000 as their least favorite subject. Construct and interpret a 95% confidence interval for the proportion of U.S. adults for whom math was their least favorite subject.
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(0.3401, 0.3999). You can be 95% confident that the actual proportion of U.S. adults for whom math was their least favorite subject is somewhere between 0.3401 and 0.3999.
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It probably wouldn’t surprise you to know that Valentine’s Day means big business for florists, jewelry stores, and restaurants. But did you know that it is also a big day for pet stores? In January 2010, the National Retail Federation conducted a survey of consumers in a representative sample of adult Americans (“This Valentine's Day, Couples Cut Back on Gifts to Each Other, According to NRF Survey,” www.nrf.com). One of the questions in the survey asked if the respondent planned to spend money on a Valentine’s Day gift for his or her pet. The proportion who responded that they did plan to purchase a gift for their pet was 0.173. Suppose that the sample size for this survey was n = 200. Construct and interpret a 95% confidence interval for the proportion of all adult Americans who planned to purchase a Valentine’s Day gift for their pet.
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(0.1206, 0.2254). You can be 95% confident that the actual proportion of all adult Americans who planned to purchase a Valentine’s Day gift for their pet is somewhere between 0.1206 and 0.2254.
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It probably wouldn’t surprise you to know that Valentine’s Day means big business for florists, jewelry stores, and restaurants. But did you know that it is also a big day for pet stores? In January 2010, the National Retail Federation conducted a survey of consumers in a representative sample of adult Americans (“This Valentine's Day, Couples Cut Back on Gifts to Each Other, According to NRF Survey,” www.nrf.com). One of the questions in the survey asked if the respondent planned to spend money on a Valentine’s Day gift for his or her pet. A confidence interval was calculated using n = 200. The actual sample size for the survey was much larger than 200. Would a 95% confidence interval computed using the actual sample size have been narrower or wider than the confidence interval computed?
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The 95% confidence interval computed for the actual sample size would have been narrower than the 95% confidence interval computed with n = 200 because the standard error, and hence the margin of error, would have been smaller.
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One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked “Do you think it is sometimes justified to lie, or do you think lying is never justified?” 52% responded that lying was never justified. When asked about lying to avoid hurting someone’s feelings, 650 responded that this was often or sometimes OK. Construct and interpret a 90% confidence interval for the proportion of adult Americans who would say that lying is never justified.
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(0.494, 0.546). You can be 90% confident that the actual proportion of adult Americans who would say that lying is never justified is somewhere between 0.494 and 0.546.
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One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked “Do you think it is sometimes justified to lie, or do you think lying is never justified?” 52% responded that lying was never justified. When asked about lying to avoid hurting someone’s feelings, 650 responded that this was often or sometimes OK. Construct and interpret a 90% confidence interval for the proportion of adult Americans who think that it is often or sometimes OK to lie to avoid hurting someone’s feelings.
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(0.6252, 0.6748). You can be 90% confident that the actual proportion of adult Americans who would say that it is often or sometimes OK to lie to avoid hurting someone’s feelings is somewhere between 0.6252 and 0.6748.
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One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked “Do you think it is sometimes justified to lie, or do you think lying is never justified?” 52% responded that lying was never justified. When asked about lying to avoid hurting someone’s feelings, 650 responded that this was often or sometimes OK. Using the 90% confidence intervals for the proportion of adult Americans who think lying is never justified and who think it is often or sometimes OK to lie, comment on the apparent inconsistency in the responses.
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The confidence interval for the proportion of adult Americans who would say that lying is never justified indicates that it is plausible that at least 50% of adult Americans would say that lying is never justified, and the confidence interval for proportion of adult Americans who would say that it is often or sometimes OK to lie indicates that it is also plausible that well over 50% of adult Americans would say that it is often or sometimes OK to lie to avoid hurting someone’s feelings. These are contradictory responses.
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In a study of 1,710 schoolchildren in Australia (Herald Sun, October 27, 1994), 1,060 children indicated that they normally watch TV before school in the morning. (Interestingly, only 35% of the parents said their children watched TV before school.) Construct and interpret a 95% confidence interval for the proportion of all Australian children who say they watch TV before school. In order for the method used to construct the interval to be valid, what assumption about the sample must be reasonable?
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(0.597, 0.643). You can be 95% confident that the actual proportion of all Australian children who would say that they watch TV before school is between 0.597 and 0.643. For the method used to construct the interval to be valid, the sample must have either been randomly selected from the population of interest or the sample must have been selected in such a way that it should result in a sample that is representative of the population.
|
An Associated Press article on potential violent behavior reported the results of a survey of 750 workers who were employed full time (San Luis Obispo Tribune, September 7, 1999). Of those surveyed, 125 indicated that they were so angered by a coworker during the past year that they felt like hitting the coworker (but didn’t). Assuming that it is reasonable to regard this sample as representative of the population of full-time workers, use this information to construct and interpret a 90% confidence interval estimate of p, the proportion of all full-time workers so angered in the last year that they wanted to hit a coworker.
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(0.1446, 0.1894). You can be 90% confident that the actual proportion of all full-time workers so angered in the past year that they wanted to hit a co-worker is between 0.1446 and 0.1894.
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A discussion of digital ethics appears in the article “Academic Cheating, Aided by Cell Phones or Web, Shown to be Common” (Los Angeles Times, June 17, 2009). One question posed in the article is: What proportion of college students have used cell phones to cheat on an exam? Suppose you have been asked to estimate this proportion for students enrolled at a large college. How many students should you include in your sample if you want to estimate this proportion with a margin of error of 0.02?
|
Assuming a 95% confidence level, and using a conservative estimate of p = 0.5, n = 2,401.
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The article “Consumers Show Increased Liking for Diesel Autos” (USA Today, January 29, 2003) reported that 27% of U.S. consumers would opt for a diesel car if it ran as cleanly and performed as well as a car with a gas engine. Suppose that you suspect that the proportion might be different in your area. You decide to conduct a survey to estimate this proportion for the adult residents of your city. What is the required sample size if you want to estimate this proportion with a margin of error of 0.05? Compute the required sample size first using 0.27 as a preliminary estimate of p and then using the conservative value of 0.5. How do the two sample sizes compare? What sample size would you recommend for this study?
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Assuming a 95% confidence level, and using the preliminary estimate of p = 0.27, n = 303. Using the conservative estimate of p = 0.5, n = 385.
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A manufacturer of small appliances purchases plastic handles for coffeepots from an outside vendor. If a handle is cracked, it is considered defective and can’t be used. A large shipment of plastic handles is received. How many handles from the shipment should be inspected in order to estimate p, the proportion of defective handles in the shipment, with a margin of error of 0.1?
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Assuming a 95% confidence level, and using a conservative estimate of p = 0.5, n = 97.
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Data from a representative sample were used to estimate that 32% of all computer users in 2011 had tried to get on a Wi-Fi network that was not their own in order to save money (USA Today, May 16, 2011). You decide to conduct a survey to estimate this proportion for the current year. What is the required sample size if you want to estimate this proportion with a margin of error of 0.05? Compute the required sample size first using 0.32 as a preliminary estimate of p and then using the conservative value of 0.5. How do the two sample sizes compare? What sample size would you recommend for this study?
| null |
The article “Hospitals Dispute Medtronic Data on Wires” (The Wall Street Journal, February 4, 2010) describes several studies of the failure rate of defibrillators used in the treatment of heart problems. In one study conducted by the Mayo Clinic, it was reported that failures within the first 2 years were experienced by 18 of 89 patients under 50 years old and 13 of 362 patients age 50 and older. Assume that these two samples are representative of patients who receive this type of defibrillator in the two age groups. Construct and interpret a 95% confidence interval for the proportion of patients under 50 years old who experience a failure within the first 2 years.
| null |
The article “Hospitals Dispute Medtronic Data on Wires” (The Wall Street Journal, February 4, 2010) describes several studies of the failure rate of defibrillators used in the treatment of heart problems. In one study conducted by the Mayo Clinic, it was reported that failures within the first 2 years were experienced by 18 of 89 patients under 50 years old and 13 of 362 patients age 50 and older. Assume that these two samples are representative of patients who receive this type of defibrillator in the two age groups. Construct and interpret a 99% confidence interval for the proportion of patients age 50 and older who experience a failure within the first 2 years.
| null |
The article “Hospitals Dispute Medtronic Data on Wires” (The Wall Street Journal, February 4, 2010) describes several studies of the failure rate of defibrillators used in the treatment of heart problems. In one study conducted by the Mayo Clinic, it was reported that failures within the first 2 years were experienced by 18 of 89 patients under 50 years old and 13 of 362 patients age 50 and older. Assume that these two samples are representative of patients who receive this type of defibrillator in the two age groups. Suppose that the researchers wanted to estimate the proportion of patients under 50 years old who experience this type of failure with a margin of error of 0.03. How large a sample should be used? Use the given study results to obtain a preliminary estimate of the population proportion.
| null |
Will p^ from a random sample of size 400 tend to be closer to the actual value of the population proportion when p = 0.4 or when p = 0.7? Provide an explanation for your choice.
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p^ from a random sample of size 400 tends to be closer to the actual value when p = 0.7.
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In response to budget cuts, county officials are interested in learning about the proportion of county residents who favor closure of a county park rather than closure of a county library. In a random sample of 500 county residents, 198 favored closure of a county park. For the statement below, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement: It is unlikely that the estimate p^ = 0.396 differs from the value of the actual population proportion by more than 0.0429.
|
correct
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In response to budget cuts, county officials are interested in learning about the proportion of county residents who favor closure of a county park rather than closure of a county library. In a random sample of 500 county residents, 198 favored closure of a county park. For the statement below, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement: The estimate p^ = 0.396 will never differ from the value of the actual population proportion by more than 0.0429.
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Incorrect. About 95% of all possible sample proportions computed from random samples of size 500 will be within 0.0429 of the actual population proportion (this is the margin of error for a 95% confidence level). About 5% of the sample proportions would differ from the actual population proportion by more than 0.0429.
|
In response to budget cuts, county officials are interested in learning about the proportion of county residents who favor closure of a county park rather than closure of a county library. In a random sample of 500 county residents, 198 favored closure of a county park. For the statement below, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement: It is unlikely that the estimate p^ = 0.396 differs from the value of the actual population proportion by more than 0.0219.
|
Incorrect. About 68% of all possible sample proportions computed from random samples of size 500 will be within 0.0219 of the actual population proportion (this is the standard deviation of the sampling distribution). About 32% of the sample proportions would differ from the actual population proportion by more than 0.0219.
|
In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost ("What Supernatural Experiences We've Had," USA Today, February 8, 2010). Assume that this sample is representative of the population of adult Americans. Use the given information to estimate the proportion of adult Americans who would say they have seen a ghost.
|
p^ = 0.18
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In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost ("What Supernatural Experiences We've Had," USA Today, February 8, 2010). Assume that this sample is representative of the population of adult Americans. Verify that the conditions for use of the margin of error formula to be appropriate are met.
|
The two conditions are: (1) Large sample size: The number of successes and failures in the sample are at least 10. In this case, there are 722 successes, and 3,291 failures, which are both much greater than 10. (2) Random selection of the sample, or the sample is representative of the population: We are told that the sample is representative of the population.
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In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost ("What Supernatural Experiences We've Had," USA Today, February 8, 2010). Assume that this sample is representative of the population of adult Americans. Compute the margin of error.
|
margin of error = 0.0119
|
In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost ("What Supernatural Experiences We've Had," USA Today, February 8, 2010). Assume that this sample is representative of the population of adult Americans. Interpret the margin of error in context.
|
It is unlikely that the estimate p^ = 0.18 differs from the value of the actual proportion of adults Americans who say they have seen a ghost by more than 0.0119.
|
In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost ("What Supernatural Experiences We've Had," USA Today, February 8, 2010). Assume that this sample is representative of the population of adult Americans. Construct and interpret a 90% confidence interval for the proportion of all adult Americans who have seen a ghost.
|
(0.17, 0.19). You can be 90% confident that the actual proportion of all adult Americans who have seen a ghost is somewhere between 0.17 and 0.19.
|
In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost ("What Supernatural Experiences We've Had," USA Today, February 8, 2010). Assume that this sample is representative of the population of adult Americans. Would a 99% confidence interval be narrower or wider than a 90% confidence interval? Justify your answer.
|
A 99% confidence interval would be wider because the z critical value for 99% confidence (z = 2.58) is larger than the z critical value for 90% confidence (z = 1.645).
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