chapter 1

Chapter 9: Inferences Based on Two Samples37.a. This exercise calls for paired analysis. First, compute the difference between indoor andoutdoor concentrations of hexavalent chromium for each of the 33 houses. These 33differences are summarized as follows: n = 33, d = −. 4239 , s = . 3868 , where d =(indoor value – outdoor value). Then t 2. 037 , and a 95% confidence interval. 025,32=for the population mean difference between indoor and outdoor concentration is⎛.3868⎞− . 4239 ±−⎝ 33 ⎠( 2.037) ⎜ ⎟ = −.4239± .13715 = ( −.5611,.2868)d. We can behighly confident, at the 95% confidence level, that the true average concentration ofhexavalent chromium outdoors exceeds the true average concentration indoors bybetween .2868 and .5611 nanograms/m 3 .b. A 95% prediction interval for the difference in concentration for the 34 th house is11( s d1+) = −.4239± ( 2.037)( .3868 1+33) = ( 1.224,.3758)d .± t. 025,32n−This prediction interval means that the indoor concentration may exceed the outdoorconcentration by as much as .3758 nanograms/m 3 and that the outdoor concentration mayexceed the indoor concentration by a much as 1.224 nanograms/m 3 , for the 34 th house.Clearly, this is a wide prediction interval, largely because of the amount of variation inthe differences.38.a. The median of the “Normal” data is 46.80 and the upper and lower quartiles are 45.55and 49.55, which yields an IQR of 49.55 – 45.55 = 4.00. The median of the “High” datais 90.1 and the upper and lower quartiles are 88.55 and 90.95, which yields an IQR of90.95 – 88.55 = 2.40. The most significant feature of these boxplots is the fact that theirlocations (medians) are far apart.Comparative Boxplotsfor Normal and High Strength Concrete Mix908070605040High:Normal:275