chapter 1

Chapter 12: Simple Linear Regression and Correlation35.ˆa. We want a 95% CI for β 1: β ± t ⋅ s 1 .025,15 β ˆ . First, we need our point estimate,11Using the given summary statistics,( 222.1)Sxx= 3056.69 − = 155.019 ,17( 222.1)( 193)S 238.112Sxy= 2759 .6 −= 238.112 , and ˆ1 xy= = 1. 53617S 115.019=xx193 − ( 1.536)( 222.1)βˆ0== −8.715to calculate the SSE:17SSE = 2975 − −8.715193 − 1.536 2759.6 = 418. . ThenWe need( )( ) ( )( ) 24942ˆβ .β .418.24945.28s == 5.28 and sˆ= = . 424 . Withβ. 025, 15= 2.131,151155.0191.536± 2.131⋅.424 = ( .632, 2.440). With 95% confidence, we estimate thatCI is ( )t ourthe change in reported nausea percentage for every one-unit change in motion sicknessdose is between .632 and 2.440.b. We test the hypotheses : β = 101.536= = 3.6226.424H vs H : β ≠ 10oa, and the test statistic ist . With df=15, the two-tailed p-value = 2P( t > 3.6226) = 2( .001)= .002. With a p-value of .002, we would reject the null hypothesis at most reasonablesignificance levels. This suggests that there is a useful linear relationship betweenmotion sickness dose and reported nausea.c. No. A regression model is only useful for estimating values of nausea % when usingdosages between 6.0 and 17.6 – the range of values sampled.d. Removing the point (6.0, 2.50), the new summary stats are: n = 16, , x = 216. 1,Σ iy = 191.5 , x 2 = 3020. 69 , y 2 = 2968. 75 , Σ x = 2744. 6Σ iΣiΣiy i iβˆ1= 1.561, βˆ0= −9. 118 , SSE = 430.5264, s = 5. 55 , s . 551is 1.5612.145⋅(.551)ˆ =β1± , or ( .379, 2.743). The interval is a little wider. But, and then, and the new CIremoving the one observation did not change it that much. The observation does notseem to be exerting undue influence.368