Chapter 16 Solutions

16.1 The standard error is 9.3/5.2 = 1.788.
16.2 The standard deviation is 1.732*.01 = .01732
16.3: a) 2.015; b) 2.518
16.5: a) 2.262; b) 2.861; c) 1.440
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A 90% t confidence interval for the mean percent change in blood polyphenols is (3.94, 7.06).
16.8: a) There are 14 degrees of freedom.; b) Two critical values that bracket t=1.82 are 1.761 and 2.145 with right-tail probabilities .05 and .025 respectively.; c) The p-value for this test is between .05 and .025.
16.9: a) There are 24 degrees of freedom.; b) 1.12 falls between 1.059 (p=0.15) and 1.318 (p=.10).; c) Since the test is two sided, the p-value is between 0.3 and 0.2.; d) The value t=1.12 is not significant at the 10% level or the 5% level.
16.11: a) The students should be randomly assigned which instrument to use first.; b) μ is the mean paired difference in time it takes a subject to move the indicator a fixed distance via left-hand or right-hand thread.
H₀: μ=0. There is no difference in time between the left-hand and right-hand thread instruments.
H_a: μ> 0. The it takes longer to use the left-hand thread instrument than the right-hand thread instruemnt.; c) The test statistic t=2.9037 with 24 degrees of freedom has a p-value of .0039. We will reject H₀: μ=0 and conclude that it does take longer to use the left-hand thread instrument.
16.13 A 90% confidence interval for the mean time advantage of right-hand over left-hand threads is (5.47, 21.17) seconds. The mean time for right-hand thread is 104.1, which is 88.7% of the mean time for left-hand thread. This seems to a practically significant difference.
16.14: a) H₀: μ_D=0. On average there is no difference between the growth of trees in the control plots and the growth of the trees in the matched treated plots.
H_a: μ_D>0. On average growth of the trees in the control plots is less than the growth of the trees in the matched treated plots.
The alternative is one-sided because the question of interest is whether or not the increase in CO₂ in the atmosphere is likely to result in an increase of plant growth.; b) t = 1.92/(1.05/1.732) = 3.17. The p-value for t=3.17 with two degrees of freedom is between .05 and .025. This test is significant at the .05 level. There is a less than 5% chance that the observed differences were due to random variation with no true difference.; c) The sample size is very small, so it is not possible to check the important assumption of Normality.
16.16: a) H₀: μ=0; there is no difference in the amount charged when no credit card fee is charged.
H_a: μ>0; customers charge more when no credit card fee is charged.
(332-0)/(108/14.14) = 43.47, so the p-value is less than .0005 This is significant at the 1% level.; b) 332 ± 2.626 * 108/14.14 = (311.94 ,352.06) is a conservative 99% confidence interval for the mean amount of increase.; c) The sample size is larger than 40, so the t procedures should be okay to use.; d) Cardholders could be randomly assigned to receive the no-fee offer or not. In this case, the design would not be a matched pair design, but it would remove the problem of confounding economic changes between years.
16.18 The critical value for 99% confidence given a t-distribution with 667 degrees of freedom is 2.583, so A 99% confidence interval for the mean number of unexcused absences for all workers is given by 9.88 ± 2.583 * 17.847/25.83 or (8.10, 11.66).
16.26 A 95% confidence interval for the mean HAV angle is given by 25.42 ± 2.021 * 7.47/6.16 or (22.97, 27.87).
16.27: a) 24.76 ± 2.021 * 6.34/6.08 or (22.7, 26.9).; b) In this case the most important effect of removing the outlier seems to be a reduction in width of the interval.
16.31: b) A 95% confidence interval for the mean length of great white sharks is (14.81, 16.36). This implies that we would reject the claim that great white sharks have an average length of 20 feet at the 5% level.; c) It is necessary to know the population from which the sample was drawn.