F. To compare the strengths of two pressure plywood samples of ten of each were selected and tested for breaking strength. The following data were obtained: r 1 r 2 Mean 35 3t] 21:2 12443 $1 6? (mg: It 12250 soon a} What is the 99% condence interval estimate of the true difference between means of the two plywood? {2 Points} b) Can you conclude that the difference between the true means is significant at the 1% level without conducting a test of hypothesis? (2 Points] c} What is the basis for your conclusion in lo} '? (1 Points) null hypothesis, Ho: H1- H2= 0 Alternative hypothesis, Ha: M1- 12 0 a) The 99% confidence interval estimate for the difference between means, H1- H2 = (X1-X2)I t(a/2,n1+n2-2) (n1-1)sit(n2-1)sz + nitn2-2 n2 2(23)2 Exi- -11 12443-12250 =21.44 nj-1 10-1 nz 9167-9000 S =18.56 n2-1 10-1 The critical value of t for n_1+n_2-2 = 10+10-2 = 18 df for two tailed test at 99% confidence level is 2.88H1 + /2 = (35 - 30) + 2.88 (10-1) (21.44)+(10-1)(18.56) + 10+10-2 10- M1 - My" =(35-30) +2.88X2 =(-0.76,10.76) b) Since the 99% confidence interval estimate for the difference in means include 0, the result is not statistically significant, and we reject the null hypothesis and conclude that there is no significant difference between the true means. c) To make the conclusion in (b), the confidence interval approach is used. First, we create the confidence interval for a specified significance level and then according to the null hypothesis (in this case difference is 0) we check if the statement mentioned can be true in confidence interval or not. If it is not possible to satisfy the null hypothesis then the result is significant, and we reject Ho otherwise result is not significant and Ho is rejected