1. What is the purpose of sampling?A. To verify that the population is approximately normally distributedB. To estimate a target parameter of the populationC. To create a point estimator of the population mean or proportionD. To achieve a more accurate result than can be achieved by surveying the entire population2. The commissioner of the state police is reported as saying that about 10% of reported auto thefts involveowners whose cars haven’t really been stolen. What null and alternative hypotheses would be appropriate inevaluating this statement made by the commissioner?A. H0: p = 0.10 and H1: p ? 0.10B. H0: p = 0.10 and H1: pC. H0: p > 0.10 and H1: p = 0.10D. H0: p = 0.10 and H1: p > 0.103. In sampling without replacement from a population of 900 it’s found that the standard error of themean x is only two-thirds as large as it would have been if the population were infinite in size. What isthe approximate sample size?A. 200B. 500C. 400D. 6004. Consider a null hypothesis stating that the population mean is equal to 52 with the research hypothesisthat the population mean is not equal to 52. Assume we have collected 38 sample data from which wecomputed a sample mean of 53.67 and a sample standard deviation of 3.84. Further assume the sampledata appear approximately normal. What is the test statistic?A. 2.64B. –2.64C. 2.68D. –2.685. If the level of significance (a) is 0.005 in a two-tail test how large is the nonrejection region under thecurve of the t distribution?A. 0.995B. 0.005C. 0.050D. 0.99756. A portfolio manager was analyzing the price-earnings ratio for this year’s performance. His boss said thatthe average price-earnings ratio was 20 for the many stocks that his firm had traded but the portfoliomanager felt that the figure was too high. He randomly selected a sample of 50 price-earnings ratios andfound a mean of 18.17 and a standard deviation of 4.60. Assume that the population is normallydistributed and test at the 0.01 level of significance. Which of the following is the correct decision rule forthe manager to use in this situation?A. Because 2.81 is greater than 2.33 reject H0. At the 0.01 level the sample data suggest that the average price-earnings ratiofor the stocks is less than 20.B. If t > 2.68 or if tC. Because –2.81 falls in the rejection region reject H0. At the 0.01 level the sample data suggest that the average priceearningsratio for the stocks is less than 20.D. If z > 2.33 reject H0.7. Determine which of the following four population size and sample size combinations would not requirethe use of the finite population correction factor in calculating the standard error.A. N = 1500; n = 300B. N = 150; n = 25C. N = 15 000; n = 1 000D. N = 2500; n = 758. In the statement of a null hypothesis you would likely find which of the following terms?A. >B.C. =D. ?9. A human resources manager wants to determine a confidence interval estimate for the mean test scorefor the next office-skills test to be given to a group of job applicants. In the past the test scores have beennormally distributed with a mean of 74.2 and a standard deviation of 30.9. Determine a 95% confidenceinterval estimate if there are 30 applicants in the group.A. 63.14 to 85.26B. 64.92 to 83.48C. 13.64 to 134.76D. 68.72 to 79.6810. Consider a null hypothesis stating that the population mean is equal to 52 with the research hypothesisthat the population mean is not equal to 52. Assume we have collected 38 sample data from which wecomputed a sample mean of 53.67 and a sample standard deviation of 3.84. Further assume the sampledata appear approximately normal. What is the p-value you would report for this test?A. 0.0037B. 0.0041C. 0.0074D. 0.496311. Which of the following statements about hypothesis testing is false?A. The rejection region is always given in units of standard deviations from the mean.B. The test will never confirm the null hypothesis only fail to reject the null hypothesis.C. A Type I error is the chance that the researcher rejects the null hypothesis when in fact the null hypothesis is true.D. In both the one-tailed and two-tailed tests the rejection region is one contiguous interval on the number line.12. To schedule appointments better the office manager for an ophthalmologist wants to estimate theaverage time that the doctor spends with each patient. A random sample of 49 is taken and the samplemean is 20.3 minutes. Assume that the office manager knows from past experience that the standarddeviation is 14 minutes. She finds that a 95% confidence interval is between 18.3 and 22.3 minutes. Whatis the point estimate of the population mean and what is the confidence coefficient?A. 18.3 95%B. 20.3 0.95C. 20.3 95%D. 18.3 0.9513. What is the primary reason for applying a finite population correction coefficient?A. If you don’t apply the correction coefficient you won’t have values to plug in for all the variables in the confidence intervalformula.B. If you don’t apply the correction coefficient your confidence intervals will be too narrow and thus overconfident.C. When the sample is a very small portion of the population the correction coefficient is required.D. If you don’t apply the correction coefficient your confidence intervals will be too broad and thus less useful in decisionmaking.14. In a criminal trial a Type II error is made when a/anA. innocent person is convicted.B. innocent person is acquitted.C. guilty defendant is acquitted.D. guilty defendant is convicted.15. Which of the following statements correctly compares the t-statistic to the z-score when creating aconfidence interval?A. You can use t all the time but for n = 30 there is no need because the results are almost identical if you use t or z.B. Using t is easier because you do not have to worry about the degrees of freedom as you do with z.C. Use t when the sample size is small and the resulting confidence interval will be narrower.D. The value of z relates to a normal distribution while the value of t relates to a Poisson distribution.16. In a simple random sample from a population of several hundred that’s approximately normallyEnd of examdistributed the following data values were collected.68 79 70 98 74 79 50 102 92 96Based on this information the confidence level would be 90% that the population mean is some wherebetweenA. 71.36 and 90.24.B. 73.36 and 88.24.C. 69.15 and 92.45.D. 65.33 and 95.33.17. A federal auditor for nationally chartered banks from a random sample of 100 accounts found that theaverage demand deposit balance at the First National Bank of Arkansas was $549.82. If the auditor neededa point estimate for the population mean for all accounts at this bank what should he use?A. There’s no acceptable value available.B. The average of $54.98 for this sampleC. The average of $549.82 for this sampleD. The auditor should survey the total of all accounts and determine the mean.18. When the confidence coefficient is large which of the following is true?A. The confidence interval is narrow.B. It’s more likely that the test will lead you to reject the null hypothesis.C. Its value is close to 1.0 but not larger than 1.0.D. Its value is 1.0 or larger.19. Determine the power for the following test of hypothesis.H0 : µ = 950 vs. H1 : µ ? 950 given that µ = 1 000 a = 0.10 s = 200 and n = 25.A. 0.5062B. 0.6535C. 0.4938D. 0.346520. A mortgage broker is offering home mortgages at a rate of 9.5% but the broker is fearful that this valueis higher than many others are charging. A sample of 40 mortgages filed in the county courthouse shows anaverage of 9.25% with a standard deviation of 8.61%. Does this sample indicate a smaller average? Use a= 0.05 and assume a normally distributed population.A. No because the test statistic is –1.85 and falls in the rejection region.B. No because the test statistic falls in the acceptance region.C. Yes because the sample mean of 9.25 is below 9.5.D. Yes because the test statistic is greater than –1.645.
