ECO 045 - Practice Problems for Final Exam - Part2_Spring 2023

1 Lehigh University ECO 045 Practice Problems for Final Exam – Part2 Yuval Erez Spring 2023 1. What type of error occurs if you fail to reject H 0 when, in fact, it is not true? a. Type II b. Type I c. either Type I or Type II, depending on the level of significance d. either Type I or Type II, depending on whether the test is one-tailed or two-tailed ANSWER: a 2. For a given sample size in hypothesis testing, a. the smaller the Type I error, the smaller the Type II error will be. b. the smaller the Type I error, the larger the Type II error will be. c. Type II error will not be effected by Type I error. d. the sum of Type I and Type II errors must equal to 1. ANSWER: b 3. In hypothesis testing, the tentative assumption about the population parameter is a. the alternative hypothesis. b. the null hypothesis. c. either the null or the alternative. d. neither the null nor the alternative. ANSWER: b 4. The p -value is a. the same as the z statistic. b. a sample statistic. c. a distance. d. a probability. ANSWER: d

2 5. For a lower (one) tail test, the p -value is the probability of obtaining a value for the test statistic a. at least as small as that provided by the sample. b. at least as large as that provided by the sample. c. at least as small as that provided by the population. d. at least as large as that provided by the population. ANSWER: a 6. The level of significance is the a. maximum allowable probability of Type II error. b. maximum allowable probability of Type I error. c. same as the confidence coefficient. d. same as the p -value. ANSWER: b 7. Which of the following does not need to be known in order to compute the p -value? a. knowledge of whether the test is one-tailed or two-tailed b. the value of the test statistic c. the level of significance d. the probability distribution of the test statistic ANSWER: c 8. As the test statistic becomes larger, the p -value a. gets smaller. b. becomes larger. c. goes beyond 1. d. becomes negative. ANSWER: a 9. For a lower tail test, the test statistic z is determined to be zero. The p -value for this test is a. zero. b. -.5. c. +.5. d. 1. ANSWER: c

4 13. A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a standard deviation of 2 years. We want to determine if the average age of all the students at the university is significantly more than 24. Assume the distribution of the population of ages is normal. The test statistic is a. 1.96. b. 2.00. c. 1.65. d. .05. ANSWER: b 14. A random sample of 16 students selected from the student body of a large university had an average age of 25. The population standard deviation is estimated to be 2 years. We want to determine if the average age of all the students at the university is significantly more than 24. Assume the distribution of the population of ages is normal. Using α = .05, and the critical value approach, it can be concluded that the population mean age is a. not significantly different from 24. b. significantly different from 24. c. significantly less than 24. d. significantly more than 24. ANSWER: d 15. A grocery store has an average sales of $8000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8300 per day. From past information, it is known that the standard deviation of the population is $1200. The p- value is a. 2.000. b. .9772. c. .0228. d. .5475. ANSWER: c

5 16. The average price of homes sold in the U.S. in 2012 was $240,000. A sample of 144 homes sold in Chattanooga in 2012 showed an average price of $246,000. It is known that the standard deviation of the population ( σ ) is $36,000. We are interested in determining whether or not the average price of homes sold in Chattanooga is significantly more than the national average. a. State the null and alternative hypotheses to be tested. b. Compute the test statistic. c. The null hypothesis is to be tested at the 5% level of significance. Determine the critical value(s) for this test. d. What do you conclude? ANSWER: a. H 0 : μ ≤ $240,000 H a : μ > $240,000 b. Test statistic z = 2 c. z .05 = 1.645 d. Reject H 0 and conclude that the average price in Chattanooga is higher than the national average. 17. The average U.S. daily internet use at home is two hours and twenty minutes . A sample of 64 homes in Soddy-Daisy showed an average usage of two hours and 50 minutes. The population standard deviation is estimated to be 80 minutes . We are interested in determining whether or not the average usage in Soddy-Daisy is significantly different from the U.S. average. a. State the null and alternative hypotheses to be tested. b. Compute the test statistic. c. The null hypothesis is to be tested using α = .05. Determine the critical value(s) for this test. d. What do you conclude? ANSWER: a. H 0 : μ = 140 minutes H a : μ ≠ 140 minutes b. Test statistic z = 3 c. z .025 = 1.96 and - z .025 = -1.96 d. Reject H 0 and conclude that the average usage in Soddy-Daisy is significantly different from the national average of 140 minutes.

7 20. A producer of various kinds of batteries has been producing "D" size batteries with a life expectancy of 87 hours. Due to an improved production process, management believes that there has been an increase in the life expectancy of their "D" size batteries. A sample of 36 batteries showed an average life of 88.5 hours. It is known that the standard deviation of the population is 9 hours. a. Give the null and the alternative hypotheses. b. Compute the test statistic. c. At the 1% level of significance using the critical value approach, test management's belief. d. What is the p -value associated with the sample results? What is your conclusion based on the p -value using α = .01? e. What is your conclusion based on the p -value using α = .05? ANSWER: a. H 0 : μ < 87 hours H a : μ > 87 hours b. 1.00 c. Since z = 1 < 2.33, do not reject H 0 d. p -value = .1587 > .01; therefore, do not reject H 0 and conclude that there is insufficient evidence to support the management's belief. e. p -value = .1587 > .05; therefore, do not reject H 0 ; no evidence to support the management's belief using the 5% level of significance. 21. Identify the null and alternative hypotheses for the following problems. a. The manager of a restaurant believes that it takes a customer less than or equal to 25 minutes to eat lunch. b. Economists have stated that the marginal propensity to consume is at least 90¢ out of every dollar. c. It has been stated that 75 out of every 100 people who go to the movies on Saturday night buy popcorn. ANSWER: a. H 0 : μ < 25 H a : μ > 25 b. H 0 : p > .9 H a : p < .9 c. H 0 : p = .75 H a : p ≠ .75

8 22. A sample of 64 account balances from a credit company showed an average daily balance of $1040. The standard deviation of the population is known to be $200. We are interested in determining if the mean of all account balances (i.e., population mean) is significantly different from $1000. a. Develop the appropriate hypotheses for this problem. b. Compute the test statistic. c. Compute the p -value. d. Using the p -value approach and α = .05, test the above hypotheses. What is your conclusion? e. Using the critical value approach and α = .05, test the hypotheses. What is your conclusion? ANSWER: a. H 0 : μ = $1000 H a : μ ≠ $1000 b. 1.60 c. .1096 d. p -value = .1096 > α = .05. The null hypothesis is not rejected; no evidence to show that the mean is significantly different from $1000. e. z = 1.60 is between -1.96 and 1.96; the null hypothesis cannot be rejected; there is no evidence that the mean is significantly different from $1000. 23. UHON Research Group has tested the hypotheses regarding the IQ of university honor students. They provided the following information. H 0 : µ ≤ 144 H a : µ > 144 (Genius) Sample size n = 121 Sample mean = 145 p -value = .0054 a. Using α = .015, would you reject or not reject the null hypothesis? Explain how you arrived at your answer and what you can conclude about the IQ scores of UHON students? b. The researchers failed to report the standard deviation. Determine the standard deviation. Use the normal distribution table (since df > 100) to answer this question and show your complete work. ANSWER: a. Reject H 0 because p -value = .0054 < α = .015. It can be concluded that the population mean IQ score of UHON students is more than 144. b. p -value = .0054 results in test statistic value = 2.55 Solving 2.55 = gives s = 4.314 (rounded)