Introduction to Inference Tutorial PDF

Chapter 6: Introduction to Inference © 2017 W.H. Freeman and Company 6.1-1 A researcher wants to know if the average time in jail for robbery has increased from what it was when the average sentence was 7 years. He obtains data on 400 more-recent robberies and finds an average time...

Chapter 6: Introduction to Inference © 2017 W.H. Freeman and Company 6.1-1 A researcher wants to know if the average time in jail for robbery has increased from what it was when the average sentence was 7 years. He obtains data on 400 more-recent robberies and finds an average time served of 7.5 years. A 95% confidence interval for the average time served is (assume jail time follows a Normal distribution with standard deviation 3 years) a. (7.25, 7.75). b. (7.21, 7.79). c. (7.11, 7.89). 6.1 Estimating with Confidence 6.1-1 answer A researcher wants to know if the average time in jail for robbery has increased from what it was when the average sentence was 7 years. He obtains data on 400 more-recent robberies and finds an average time served of 7.5 years. A 95% confidence interval for the average time served is (assume jail time follows a Normal distribution with standard deviation 3 years) a. (7.25, 7.75). 3 7.5 1.96  b. (7.21, 7.79). (correct) 400 c. (7.11, 7.89). 6.1 Estimating with Confidence 6.1-2 The manager at a movie theater would like to estimate the true mean amount of money spent by customers on popcorn only. He selects a simple random sample of 26 receipts and calculates a 92% confidence interval for true mean to be ($12.45, $23.32). The confidence interval can be interpreted to mean that, in the long run, a. 92% of similarly constructed intervals would contain the population mean. b. 92% of similarly constructed intervals would contain the sample mean. c. 92% of all customers who buy popcorn spend between $12.45 and $23.22. 6.1 Estimating with Confidence 6.1-2 answer The manager at a movie theater would like to estimate the true mean amount of money spent by customers on popcorn only. He selects a simple random sample of 26 receipts and calculates a 92% confidence interval for true mean to be ($12.45, $23.32). The confidence interval can be interpreted to mean that, in the long run, a. 92% of similarly constructed intervals would contain the population mean. (correct) b. 92% of similarly constructed intervals would contain the sample mean. c. 92% of all customers who buy popcorn spend between $12.45 and $23.22. 6.1 Estimating with Confidence 6.1-3 Lumber intended for building houses and other structures must be monitored for strength. A random sample of 25 specimens of Southern Pine is selected, and the mean strength is calculated to be 3700 pounds per square inch. Strengths are known to follow a Normal distribution with standard deviation 500 pounds per square inch. A 90% confidence interval for the true mean strength of Southern Pine is a. (3200, 3647). b. (3402, 3794). c. (3536, 3865). 6.1 Estimating with Confidence 6.1-3 answer Lumber intended for building houses and other structures must be monitored for strength. A random sample of 25 specimens of Southern Pine is selected, and the mean strength is calculated to be 3700 pounds per square inch. Strengths are known to follow a Normal distribution with standard deviation 500 pounds per square inch. A 90% confidence interval for the true mean strength of Southern Pine is a. (3200, 3647). 3700 (1.645)(500 ) b. (3402, 3794). 25 c. (3536, 3865). (correct) 6.1 Estimating with Confidence 6.1-4 You want to rent an unfurnished one-bedroom apartment in Boston next year. A 95% confidence interval for the true mean monthly rent of all one-bedroom apartments in Boston is (1263.64, 1536.36). What is the margin of error for this interval? a. 126.43 b. 136.36 c. 143.46 6.1 Estimating with Confidence 6.1-4 answer You want to rent an unfurnished one-bedroom apartment in Boston next year. A 95% confidence interval for the true mean monthly rent of all one-bedroom apartments in Boston is (1263.64, 1536.36). What is the margin of error for this interval? a. 126.43 b. 136.36 (correct) MOE = ½(length of the interval) = ½(1536.36 – 1263.64) c. 143.46 6.1 Estimating with Confidence 6.1-5 A student curious about the average number of chocolate chips in a commercial brand of cookie estimated the standard deviation to be 8. If he wants to be 99% confident in his results, how many chocolate chip cookies will he need to sample to estimate the mean to within 2? a. 11 b. 107 c. 62 6.1 Estimating with Confidence 6.1-5 answer A student curious about the average number of chocolate chips in a commercial brand of cookie estimated the standard deviation to be 8. If he wants to be 99% confident in his results, how many chocolate chip cookies will he need to sample to estimate the mean to within 2? a. 11 2 b. 107 (correct)  z   n   106.17 c. 62  m  6.1 Estimating with Confidence 6.1-6 Lumber intended for building houses and other structures must be monitored for strength. Good lumber is known to have a standard deviation of 500 lb per square inch. At 80% confidence, what is the required sample size for the margin of error to be +/-250 lb? a. 6 b. 7 c. 8 6.1 Estimating with Confidence 6.1-6 answer Lumber intended for building houses and other structures must be monitored for strength. Good lumber is known to have a standard deviation of 500 lb per square inch. At 80% confidence, what is the required sample size for the margin of error to be +/-250 lb? a. 6 * 2 z b. 7 (correct) n   c. 8  m  6.1 Estimating with Confidence 6.1-9 To construct a 92% confidence interval, the correct z* to use is a. 1.75. b. 1.41. c. 1.645. 6.1 Estimating with Confidence 6.1-9 answer To construct a 92% confidence interval, the correct z* to use is a. 1.75. (correct) b. 1.41. c. 1.645. 6.1 Estimating with Confidence 6.1-14 We would like to construct a confidence interval to estimate the mean μ of some variable X. Which of the following combinations of confidence level and sample size will produce the shortest interval? a. 90% confidence, n = 10 b. 95% confidence, n = 10 c. 99% confidence, n = 10 6.1 Estimating with Confidence 6.1-14 answer We would like to construct a confidence interval to estimate the mean μ of some variable X. Which of the following combinations of confidence level and sample size will produce the shortest interval? a. 90% confidence, n = 10 (correct) b. 95% confidence, n = 10 c. 99% confidence, n = 10 6.1 Estimating with Confidence 6.2-1 A researcher reports that a test is “significant at 5%.” This test will be a. significant at 1%. b. not significant at 1%. c. significant at 10%. 6.2 Tests of Significance 6.2-1 answer A researcher reports that a test is “significant at 5%.” This test will be a. significant at 1%. b. not significant at 1%. c. significant at 10%. (correct) 6.2 Tests of Significance 6.2-2 The response times of technicians of a large heating company follow a Normal distribution with a standard deviation of 10 minutes. A supervisor suspects that the mean response time has increased from the target of 30 minutes. He takes a random sample of 25 response times and calculates the sample mean response time to be 33.8 minutes. What is the value of the test statistic for the appropriate hypothesis test? a. Z = 1.65 b. Z = 2.09 c. Z = 0.77 d. Z = 1.90 e. Z = 1.83 6.2 Tests of Significance 6.2-2 answer The response times of technicians of a large heating company follow a Normal distribution with a standard deviation of 10 minutes. A supervisor suspects that the mean response time has increased from the target of 30 minutes. He takes a random sample of 25 response times and calculates the sample mean response time to be 33.8 minutes. What is the value of the test statistic for the appropriate hypothesis test? a. Z = 1.65 b. Z = 2.09 33.8  30 z 1.90 c. Z = 0.77 10 d. Z = 1.90 (correct) 25 e. Z = 1.83 6.2 Tests of Significance 6.2-3 A researcher wants to know if the average time in jail for robbery has increased from what it was several years ago when the average sentence was 7 years. He obtains data on 400 more-recent robberies and finds an average time served of 7.5 years. If we assume the standard deviation is 3 years, what is the P-value of the test? a. 0.0004 b. 0.0008 c. 0.9996 6.2 Tests of Significance 6.2-3 answer A researcher wants to know if the average time in jail for robbery has increased from what it was several years ago when the average sentence was 7 years. He obtains data on 400 more-recent robberies and finds an average time served of 7.5 years. If we assume the standard deviation is 3 years, what is the P-value of the test? a. 0.0004 (correct) 7.5  7 z 3.33 P ( z  3.33) 3 b. 0.0008 400 c. 0.9996 6.2 Tests of Significance 6.2-5 What is the P-value for a test of the hypotheses H0: μ = 10 against Ha: μ ≠ 10 if the calculated statistic is z = 2.56? a. 0.0052 b. 0.0104 c. 0.9948 6.2 Tests of Significance 6.2-5 answer What is the P-value for a test of the hypotheses H0: μ = 10 against Ha: μ ≠ 10 if the calculated statistic is z = 2.56? a. 0.0052 b. 0.0104 (correct) 2 P( z   2.56) c. 0.9948 6.2 Tests of Significance 6.2-6 The times for untrained rats to run a standard maze has an N (65, 15) distribution where the times are measured in seconds. The researchers hope to show that training improves the times. The alternative hypothesis is a. Ha: µ > 65. b. Ha: > 65. c. Ha: µ < 65. d. Ha: < 65. 6.2 Tests of Significance 6.2-6 answer The times for untrained rats to run a standard maze has an N (65, 15) distribution where the times are measured in seconds. The researchers hope to show that training improves the times. The alternative hypothesis is a. Ha: µ > 65. (correct) b. Ha: > 65. c. Ha: µ < 65. d. Ha: < 65. 6.2 Tests of Significance 6.2-7 Suppose we are testing the null hypotheses H0: µ = 50 Ha: µ ≠ 50 for a Normal population with s = 6. The 95% confidence is (51.3, 54.7). Then the p-value for the test a. is greater than 0.05. b. is less than 0.05. c. could be greater or less than 0.05. It cannot be determined without knowing the sample size. 6.2 Tests of Significance 6.2-7 answer Suppose we are testing the null hypotheses H0: µ = 50 Ha: µ ≠ 50 for a Normal population with s = 6. The 95% confidence is (51.3, 54.7). Then the p-value for the test a. is greater than 0.05. b. is less than 0.05. (correct) c. could be greater or less than 0.05. It cannot be determined without knowing the sample size. 50  (51.3, 54.7) Therefore, p  value  0.05 6.2 Tests of Significance 6.2-8 Suppose we are testing that the true mean of some variable is equal to 40 against a two-tailed alternative. It is known that the standard deviation of the population is 10. A random sample of 16 observations is drawn from the population and a sample mean is calculated to be 43. What is the P-value of the test? a. 0.7642 b. 0.3821 c. 0.2302 d. 0.1179 e. 0.1151 6.2 Tests of Significance 6.2-8 answer Suppose we are testing that the true mean of some variable is equal to 40 against a two-tailed alternative. It is known that the standard deviation of the population is 10. A random sample of 16 observations is drawn from the population and a sample mean is calculated to be 43. What is the P-value of the test? a. 0.7642 43  40 b. 0.3821 z 1.20 10 c. 0.2302 (correct) 16 d. 0.1179 e. 0.1151 2P(Z | z |) 2(.1151) 0.2302 6.2 Tests of Significance