Chapter 18 Solutions

18.1: a) The population is college students. The parameter p is the proportion of college students who pray at least once in a while.; b) The estimate for p is 107/127.
18.2: a) The population is internet users. The parameter p is the proportion of internet users who would go on-line for information about health or medicine.; b) The estimate for p is 606/1318.
18.3: a) The mean of the sample proportion is .5, the standard deviation is .004; b) The z-score for .49 is (.49-.5)/.004 = -2.5. Using the symmetry of the Normal distribution and Table A, the probability is .9876 that p-hat will be between .49 and .51.
18.5 For sample size 1000 the probability is .47. For sample size 4000, the probability is .79. For sample size 16000, the probability is .99. This illustrates that as the sample size increases the varibility of the sample proportion decreases.
18.6 The population is not at least 10 times larger than the sample size.
18.7 The sample is not a random sample of all citizens in the city. It is limited by the fact that it was only those who watch the television news program and second by the fact that it was an optional call-in.
18.8 2673*.002= 5.346, so the number of 'successes' is to small for using the large-sample confidence interval.
18.9 A 99% confidence interval for the proportion of coaching among students who retake the SAT is given by 427/3160 ± 2.576 * .0061, or (.119, .151).
18.11: a) The sample was random and much smaller than the overall population. The number of successes and failures was greater than 15 in each case.; b) A 95% confidence interval for the true proportion of teenagers who have a television set in their room is .66 ± 1.96*.0146, or (.63, .69).; c) The estimated proportion is 66%, the 95% confidence interval has limits that are ± 3% from the estimate.
18.13: a) There are fewer than 15 people who are in the group of interest so computing a large-sample confidence interval would not be appropriate.; b) The sample size with the plus four method is 2677 and the count of successes is 7. The plus four estimate of p is 7/2677 = .0026.
18.15: a) A plus four 99% confidence interval for the proportion of all students who pray is (109/131) ± 2.576*.0327, or (.75, .92).; b) It seems reasonably likely that students who would tend to take psychology and communications classes may differ in their religious behavior from the population at large. Furthermore, if generalizations about the entire population of undergraduate students across the country wish to be made, the sample would need to reflect that (that is, it should not be from a single school, or region).
18.17 n = (1.645/.04)² * .25 * .75 = 317.11, so a sample size of 318 should be sufficient to get close to the desired margin of error.
18.18 H₀: p=.5, heads and tails are equally likely to occur when spinning a coin end up heads or tails.
H_a: p≠.5, heads and tails are not equally likely to occur when spinning a coin. The test statistic is -2.4. The p-value is between .01 and .02, so there is strong evidence that the spinning coin is not equally likely to end up heads.
18.20: a) 10 * .5 = 5, so the sample size is not large enough.; b) 200*(1-.99)= 2, so the sample size is not large enough.
18.22 The conditions are met for the plus four method for constructing confidence intervals. The plus four 95% confidence interval is given by (70/121) ± 1.96 * .045, or (.49,.67).
18.24 H₀: p=.5, half of the Hispanic female drivers in Boston wear seat belts.
H_a: p>.5, more than half of the Hispanic female drivers in Boston wear seat belts.
The test statistic z=1.76. The p-value for the one-sided test is .0392. There is strong statistical evidence that more than half of the Hispanic female drivers in Boston wear seat belts.
18.26: a) A plus four 95% confidence interval for the proportion of all drivers who ran a red light at one or more of the last ten lights they had drvien through is .19 ± 1.96 * .0133 or (.168, .221).; b) It seems likely that more than 171 of the 880 respondents really ran a red light. Running red lights is against the law, so people may be reluctant to admit having done so.
18.31 A plus four 99% confidence interval for the proportion of all adults who claim that they attended church or synagogue is .42 ± 2.576 * .0117 or (.39, .45).
18.32 0.5 is not included in the confidence interval so the results of the poll in the previous exercise do provide strong evidence that fewer than half of the population would claim to attend church or synagogue.
18.33 n = (2.576/.01)² * .5 * .5 = 16590. The confidence interval from Exercise 18.31 indicates that the true proportion is likely to be between .3 and .7, so it is reasonable to use .5 to get an approximate sample size needed for the desired margin of error.
18.35: a)H₀: p=.5, half of the people prefer the fresh-brewed coffee, half prefer instant.
H_a: p> .5, more than half of the people prefer the fresh-brewed coffee.
z=(.62-.5)/.07 = 1.71. The p-value for the one-sided test is .0436, so the test is significant at the 5% level. It seems likely that more people prefer fresh-brewed coffee than instant.; b) A 90% confidence interval for p is given by .62 ± 1.645 * .069, or (.507, .733).; c) The order of the coffee type given should be random.