CHAPTER 10 Section 10-2 10-1. a) 1) 2) 3) 4) The parameter of interest is the difference in means H 0 : H 1 : 1 1 - 2 . Note that 0 = 0. - 2 = 0 or 1 = 2 2 - 1 0 or 2 1 The test statistic is z 0 = ( x 1 - x 2 ) - 0 2 1 2 2 n 1 5) 6) x 1 = 4.7 x 2 = 7.8 1 + n 2 Reject H 0 if z 0 < -z /2 = -1.96 or z 0 > z /2 = 1.96 for = 0.05 . = 10 = 5 ( 4.7 -7 .8 ) 2 n 1 = 10 n 2 = 15 z 0 = 7) = -0.9 (10) 2 (5) 2 + 10 15 Conclusion: Because ­1.96 < ­0.9 < 1.96, do not reject the null hypothesis. There is not sufficient evidence to conclude that the two means differ at = 0.05. P-value = 2(1-(0.9) ) = 2(1-0.8 159 50) = 0.36 8 2 1 /2 2 2 2 1 /2 2 2 b) ( x 1 - x 2 ) -z n 1 + n 2 1 - 2 x 1 - x 2 +z - ( ) n 1 + n 2 (10) 2 (5) 2 + 10 15 ( 4.7 -7 .8 ) -1.9 6 -9 .7 9 1 (10) 2 (5) 2 + 10 15 1 2 4.7 -7 .8 +1.9 6 ( ) - 2 3.59 c) With 95% confidence, the true difference in the means is between -9.79 and 3.59. Because zero is contained in this interval, we conclude there is no significant difference between the means. We fail to reject the null hypothesis. - 0 - 0 - -z /2 - = z /2 - 2 2 2 2 1 + 2 1 + 2 n 1 n 2 n 1 n 2 3 = 1.9 6 - 2 (10) (5) 2 + 10 15 3 - -1.9 6 - 2 (10) (5) 2 + 10 15 = 1.08 - -2.8 3 = 0.8599 - 0.0023 = 0.86 Power = 1 - 0.86 = 0.14 ( ) ( ) 10-1