Problems and Solutions 1. Consider a normal population distribution with the value of σ known. (from 8th edition, #1, p. 275) a) What is the confidence level for the interval ̅ ± 2.81 ? 99.5% √
b) What is the confidence level for the interval ̅ ± 1.44 ? 85% √
c) What value of zα/2 in the CI formula: ̅ ± zα/2 results in a confidence level of 99.7%? 2.96 √
d) Answer the question posed in part (c) for a confidence level of 75%. 1.15 2. Each of the following is a confidence interval for µ = true average (i.e., population mean) resonance frequency (Hz) for all tennis rackets of a certain type: (114.4, 115.6) (114.1, 115.9) (from 8th edition, #2, p.275) a) What is the value of the sample mean resonance frequency? 115, found by taking the average of the upper and lower bounds. b) Both intervals were calculated from the same sample data. The confidence level for one of these intervals is 90% and for the other is 99%. Which of the intervals has the 90% confidence level, and why? The 90% confidence interval is (114.4, 115.6) because it has the smaller width. Since the interval is smaller, we are less confident that the interval will contain the true population mean. c) Which interval is more precise? Explain your reasoning. (114.4, 115.6) is more precise because it is pinpointing a smaller range of values. A less precise interval has endpoints that are farther apart. 3. Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let µ denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence interval is (7.8, 9.4). (from 8th edition, #3, p.275) a) Would a 90% confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning. The interval would be narrower than the given interval because we gain precision when we have a lower confidence level. b) Consider the following statement: There is a 95% chance that µ is between 7.8 and 9.4. Is this statement correct? Why or why not? This statement is not correct because the population mean either is or isn’t between those two values. Therefore it has a probability of 0 or 1 of being between 7.8 and 9.4. c) Consider the following statement: We can be highly confident that 95% of all bottles of this type of cough syrup have an alcohol content that is between 7.8 and 9.4. Is this statement correct? Why or why not? This statement is not correct because confidence intervals do not tell us about individuals in the population, they are about population means. We are 95% confident that the true population mean is between 7.8 and 9.4, but we cannot make a statement about all bottles in the population using a confidence interval. 4. A CI is desired for the true average stray‐load loss µ (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of 1500 rpm. Assume that stray‐load loss is normally distributed with σ =3.0. (from 8th edition,#4, p.276) a) Compute a 95% CI for µ when n = 25 and ̅ = 58.3. (57.124, 59.476) We are 95% confident that the true average stray‐load loss for this type of induction motor , when the line current is held at 10 amps for a speed of 1500 rpm, falls between 57.124 and 59.476 watts. b) Compute a 95% CI for µ when n = 100 and ̅ = 58.3. (57.712, 58.888) c) Compute a 99% CI for µ when n = 25 and ̅ = 58.3. (56.755, 59.845) d) Compute an 82% CI for µ when n = 25 and ̅ = 58.3. (57.496, 59.104) e) How large must n be if the width of the 99% interval for µ is to be 1.0? 238.7, round up to 239. 5. By how much must the sample size n be increased if the width of the CI is to be halved? If the sample size is increased by a factor of 25, what effect will this have on the width of the interval? Justify your assertions. (from 8th edition, #7, p.276) If the width is to be halved, the sample size must be multiplied by 4. If the sample size is multiplied by 25, the width decreases by a factor of 5. 6. Consider the next 1000 95% CIs for µ that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these 1000 intervals do you expect to capture the corresponding value of µ? We expect about .95(1000) = 950 intervals to capture mu. 7. Determine the confidence level for each of the following large‐sample one‐sided confidence bounds: (from 8th edition, #15, p.284) a) Upper bound: + 0.84 =0.84 and Φ(0.84) ≈ .7995≈.80, so the confidence level is 80%. b) Lower bound: – 2.05 =2.05 and Φ(2.05) ≈ .9798≈.98, so the confidence level is 98%. c) Upper bound: + 0.67 =0.67 and Φ(0.67) ≈ .7486≈.75, so the confidence level is 75%. √
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8. This is taken from Chapter 1, problem 36 of your text. A sample of 26 offshore workers took part in a simulated escape exercise, for which data was gathered on time (sec) to complete the escape (from the article “Oxygen Consumption and Ventilation During Escape from an Offshore Platform,” Ergonomics, 1997: 281‐292). The sample mean escape time for 26 observations is 370.69, and σ = 24.36. (from 8th edition, #36, p. 292) a) Calculate an upper confidence bound for population mean escape time using a confidence level of 95%. 370.69+
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*
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= 370.79 + 1.64*4.78=378.63 We are 95% confident that the mean escape time for an oil rig worker will fall below 378.63 units of time. b) Calculate an upper prediction bound for the escape time of a single additional worker using a prediction level of 95%. How does this bound compare with the confidence bound of part (a)? 370.69+
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*24.36* 1
= 370.79 + 1.64*24.82=411.49. We are 95% confident that the escape time for an individual oil rig worker will fall below 411.49 units of time. The upper prediction bound for the escape time of a single additional worker is larger than the upper confidence bound for population mean escape time because there are two sources of uncertainty: the fact that we are predicting the value of a random variable, and the fact that we have estimated the mean.