chapter 1

25. We calculate the degrees of freedomChapter 9: Inferences Based on Two Samplesν=2 2 25.5 7.8(28+31)2 225.5 7.8( ) ( )2827+31302= 53.95, or about 54 (normallywe would round down to 53, but this number is very close to 54 – of course for this largenumber of df, using either 53 or 54 won’t make much difference in the critical t value) so the25.5 7.891 .5−88.3 ± 1.68 +desired confidence interval is ( )28 313 .2 ± 2.931 = (.269,6.131)= . Because 0 does not lie inside this interval, we can bereasonably certain that the true difference2population means are not equal. For a 95% interval, the t value increases to about 2.01 or so,which results in the interval 3 .2 ± 3. 506 . Since this interval does contain 0, we can nolonger conclude that the means are different if we use a 95% confidence interval.1µ2µ − is not 0 and, therefore, that the two26. Let µ1= the true average potential drop for alloy connections and let µ2= the true averagepotential drop for EC connections. Since we are interested in whether the potential drop ishigher for alloy connections, an upper tailed test is appropriate. We test H µ − µ 00:1 2=vs. H µ − µ 0 . Using the SAS output provided, the test statistic, when assuminga:1 2>unequal variances, is t = 3.6362, the corresponding df is 37.5, and the p-value for our uppertailed test would be ½ (two-tailed p-value) = (.0008) . 00041= . Our p-value of .0004 is2less than the significance level of .01, so we reject H o . We have sufficient evidence to claimthat the true average potential drop for alloy connections is higher than that for ECconnections.27. The approximate degrees of freedom for this estimate are2 2 211.3 8.3(6+8)2 2211.3 8.3( ) ( )893.59ν == = 8.83 , which we round down to 8, so t2. 025,8= 2. 306101.17568+5 7and the desired interval is ( 40.321.4) 2.30611.3 2 28.3− ±6+8= 18.9 ± 2.306( 5.4674)= 18 .9 ± 12.607 = ( 6.3,31.5). Because 0 is not contained in this interval, there is strongevidence that2µ1− µ is not 0; i.e., we can conclude that the population means are not equal.µ − would change only the order of subtraction ofCalculating a confidence interval for1the sample means, but the standard error calculation would give the same result as before.2µµ − would be ( -31.5, -6.3), just the negatives ofTherefore, the 95% interval estimate of1the endpoints of the original interval. Since 0 is not in this interval, we reach exactly the sameconclusion as before; the population means are not equal.2µ270