1. What is the primary reason for applying a finite population correction coefficient?
A. When the sample is a very small portion of the population, the correction coefficient is required.
B. If you don’t apply the correction coefficient, your confidence intervals will be too broad, and thus less useful in decision making.
C. If you don’t apply the correction coefficient, your confidence intervals will be too narrow, and thus overconfident.
D. If you don’t apply the correction coefficient, you won’t have values to plug in for all the variables in the confidence interval formula.
2. A portfolio manager was analyzing the price-earnings ratio for this year’s performance. His boss said that the average price-earnings ratio was 20 for the many stocks that his firm had traded, but the portfolio manager felt that the figure was too high. He randomly selected a sample of 50 price-earnings ratios and
found a mean of 18.17 and a standard deviation of 4.60. Assume that the population is normally distributed, and test at the 0.01 level of significance. Which of the following is the correct decision rule for the 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 ratio for the stocks is less than 20.
B. Because –2.81 falls in the rejection region, reject H0. At the 0.01 level, the sample data suggest that the average priceearnings ratio for the stocks is less than 20.
C. If z > 2.33, reject H0. D. If t > 2.68 or if t < –2.68, reject H0. 3. When the confidence coefficient is large, which of the following is true? A. The confidence interval is narrow. B. Its value is 1.0 or larger. C. It's more likely that the test will lead you to reject the null hypothesis. D. Its value is close to 1.0, but not larger than 1.0. 4. If a teacher wants to test her belief that more than five students in college classes typically receive A as a grade, she'll perform A. one-tail testing of a proportion. B. one-tail testing of a mean. C. two-tail testing of a proportion. D. two-tail testing of a mean. 5. For 1996, the U.S. Department of Agriculture estimated that American consumers would have eaten, on average, 2.6 pounds of cottage cheese throughout the course of that year. Based on a longitudinal study of 98 randomly selected people conducted during 1996, the National Center for Cottage Cheese Studies found an average cottage cheese consumption of 2.75 pounds and a standard deviation of s = 14 ounces. Given this information, which of the following statements would be correct concerning a two-tail test at the 0.05 level of significance? A. We can conclude that we can't reject the claim that the average cottage cheese consumption in America is 2.6 pounds per person per year. B. We can conclude that the average cottage cheese consumption in America is at least 0.705 pound more or less than 2.75 pounds per person per year. C. We can conclude that the average cottage cheese consumption in America is actually 2.75 pounds per person per year. D. We can conclude that the average cottage cheese consumption in America isn't 2.6 pounds per person per year. 6. Consider a null hypothesis stating that the population mean is equal to 52, with the research hypothesis that the population mean is not equal to 52. Assume we have collected 38 sample data from which we computed a sample mean of 53.67 and a sample standard deviation of 3.84. Further assume the sample data appear approximately normal. What is the test statistic? A. –2.68 B. –2.64 C. 2.64 D. 2.68 7. Consider a null hypothesis stating that the population mean is equal to 52, with the research hypothesis that the population mean is not equal to 52. Assume we have collected 38 sample data from which we computed a sample mean of 53.67 and a sample standard deviation of 3.84. Further assume the sample data appear approximately normal. What is the p-value you would report for this test? A. 0.0037 B. 0.4963 C. 0.0041 D. 0.4959 8. Which of the following statements about hypothesis testing is false? A. In both the one-tailed and two-tailed tests, the rejection region is one contiguous interval on the number line. B. The rejection region is always given in units of standard deviations from the mean. C. A Type I error is the chance that the researcher rejects the null hypothesis when in fact the null hypothesis is true. D. The test will never confirm the null hypothesis, only fail to reject the null hypothesis 9 For one year, he has randomly selected 12 Sundays and found an average time of 26.42 minutes with a standard deviation of 6.69 minutes. Assuming that the population is normally distributed and using a 0.05 level of significance, he wishes to determine whether he is correct in thinking that the average length of sermons is more than 20 minutes. What is the test statistic? A.–3.32 B.6.69 C.3.32 D.0.95 10.Because of the popularity of movies as an entertainment medium for adolescents, an entrepreneur plans to do a national study of the average cost of a movie ticket. If you assume that s= $0.50, what sample size would the entrepreneur have to take to be 95% confident that the estimate was within $0.25 of the true mean ticket prices? A.16 B.4 C.15 D.8 11.Which of the following statements correctlycompares the t-statistic to the z-score when creating a confidence interval? A.You can use tall the time, but for n? 30 there is no need, because the results are almost identical if you use tor z. B.Using tis easier because you do not have to worry about the degrees of freedom, as you do with z. C.The value of zrelates to a normal distribution, while the value of trelates to a Poisson distribution. D.Use twhen the sample size is small, and the resulting confidence interval will be narrower. 12.What sample size is required from a very large population to estimate a population proportion within 0.05 with 95% confidence? Don't assume any particular value for p. A.767 B.271 C.385 D.38 13.In a simple random sample from a population of several hundred that's approximately normally distributed, the following data values were collected. 68, 79, 70, 98, 74, 79, 50, 102, 92, 96 Based on this information, the confidence level would be 90% that the population mean is somewhere between A.65.33 and 95.33. B.71.36 and 90.24. C.69.15 and 92.45. D.73.36 and 88.24. 14.In sampling without replacement from a population of 900, it's found that the standard error of the mean, ? is only two-thirds as large as it would have been if the population were x infinite in size. What is the approximate sample size? A.600 B.400 C.200 D.500 15.The commissioner of the state police is reported as saying that about 10% of reported auto thefts involve owners whose cars haven't really been stolen. What null and alternative hypotheses would be appropriate in evaluating this statement made by the commissioner? A.H0: p? 0.10 and H1: p> 0.10 B.H0: p> 0.10 and H1: p? 0.10
C.H0: p= 0.10 and H1: p? 0.10 D.H0: p? 0.10 and H1: p< 0.10 16.To schedule appointments better, the office manager for an ophthalmologist wants to estimate the average time that the doctor spends with each patient. A random sample of 49 is taken, and the sample mean is 20.3 minutes. Assume that the office manager knows from past experience that the standard deviation is 14 minutes. She finds that a 95% confidence interval is between 18.3 and 22.3 minutes. What is the point estimate of the population mean, and what is the confidence coefficient? A.18.3, 95% B.18.3, 0.95 C.20.3, 95% D.20.3, 0.95 17.Nondirectional assertions lead onlyto _______ tests. A.left-tail B.two-tail C.right-tail D.one-tail 18.A mortgage broker is offering home mortgages at a rate of 9.5%, but the broker is fearful that this value is higher than many others are charging. A sample of 40 mortgages filed in the county courthouse shows an average of 9.25% with a standard deviation of 8.61%. Does this sample indicate a smaller average? Use ? = 0.05 and assume a normally distributed population. A.Yes, because the sample mean of 9.25 is below 9.5. B.No, because the test statistic is –1.85 and falls in the rejection region. C.No, because the test statistic falls in the acceptance region. D.Yes, because the test statistic is greater than –1.645. 19.A human resources manager wants to determine a confidence interval estimate for the mean test score for the next office-skills test to be given to a group of job applicants. In the past, the test scores have been normally distributed with a mean of 74.2 and a standard deviation of 30.9. Determine a 95% confidence interval estimate if there are 30 applicants in the group. A.68.72 to 79.68 B.64.92 to 83.48 C.13.64 to 134.76 D.63.14 to 85.26 20.What is the purpose of sampling? A.To achieve a more accurate result than can be achieved by survey B.To estimate a target parameter of the population C.To create a point estimator of the population mean or proportion D.To verify that the population is approximately normally distributed