chapter 1

Chapter 9: Inferences Based on Two Samplesb. This data is paired because the two measurements are taken for each of 15 test conditions.Therefore, we have to work with the differences of the two samples. A quantile of the 15differences shows that the data follows (approximately) a straight line, indicating that it isreasonable to assume that the differences follow a normal distribution. Takingdifferences in the order “Normal” – “High” , we find d = −42. 23 , and s = 4. 34 .With t 2. 145 , a 95% confidence interval for the difference between the. 025,14=population means is⎛ 4.34 ⎞− 42.23±−⎝ 15 ⎠( 2.145) ⎜ ⎟ = −42.23±2.404 = ( − 44.63, 39.83)d. Because 0 isnot contained in this interval, we can conclude that the difference between the populationmeans is not 0; i.e., we conclude that the two population means are not equal.39.a. A normal probability plot shows that the data could easily follow a normal distribution.b. We test : 0t =H µ vs. H : µ ≠ 0 , with test statistic0 d=d − 0 167.2 − 0= = 2.74 ≈ 2.7s / n 228 / 14Dad. The two-tailed p-value is 2[ P( t > 2.7)] =2[.009] = .018. Since .018 < .05, we can reject H o . There is strong evidence to supportthe claim that the true average difference between intake values measured by the twomethods is not 0. There is a difference between them.40.t or t ≤ −2. 947a. H o will be rejected in favor of H a if either ≥ t .= 2. 947summary quantities are = −. 544005,15− .544d , and sd= . 714 , so = = −3.05.17853.05≤ −2.Because − 947 , H o is rejected in favor of H a .. Thet .− .544sp, sp= 2. 70 , and t = = −.57.96b. 2 = 7. 31incorrect analysis yields an inappropriate conclusion., which is clearly insignificant; the41. We test : 0H0µd= vs. Ha: µd> 0 . With d = 7. 600 , andd= 4. 1787.600 − 5 2.6t = = = 1.87 ≈1.9. With degrees of freedom n – 1 = 8, the4.178/ 9 1.39s ,corresponding p-value is P( t > 1.9 ) = .047. We would reject H o at any alpha level greaterthan .047. So, at the typical significance level of .05, we would (barely) reject H o , andconclude that the data indicates that the higher level of illumination yields a decrease of morethan 5 seconds in true average task completion time.276