Question

Independent random samples selected from two normal populations produced the sample means and standard deviations shown to the right. a. Assuming equal variances, conduct the test Ho ??-?2):0 against Ha : (??-?2)#0 using ?:010. b. Find and interpret the 90% confidence interval for ( 1- 2)- Sample 1 Sample 2 n1 18 n2 13 x1-5.2 x27.7 s1 3.7 s2 4.3 a. Find the test statistic. The test statistic is (Round to two decimal places as needed.)

Answer 1

A Test: μ1 * μ2 using equal variance assumption Hypothesis test (Null Hypothesis) (Alternative Hypothesis, also called Hı) This is two tailed test sample1sample2 5.22 7.7 13.7824.3 n118 n-13 Significance Level, a = 0.1 Degrees of freedom: = df = n1 + n2-2-29 The pooled estimate of Variance is: n+(15.676207 Test statistic, tobserved - 5.2 7.7 15.676207/18+ 15.676207/13 -1.735 Using excel we find p-value-Plt! > 1.735) = 2*t.dist(-1735. 29.1) = 0.0933966004 Critical Value, teriticalta/2dfto.o5,d-29 1.699 (from t-table, two-tails, d.f.- 29 Rejection criteria: Rejection region is in direction of Alternative hypothesis, since Ha : μ1-12 0, rejection rule is: Reject Ho if Ito> te or equivalently if p-value < α Decision : Since Ito> tc or equivalently since p-value < α, we reject the null hypothesis. Hence at 10% significance level, we have sufficient evidence to conclude that μι is different from 090% CI for μί-μ2 using equal variance as- sumption

We want to construct confidence interval of mean difference for independent samples from data sample1sample2 x = 5.2 | = 7.7 s = 3.7 | 82-4.3 Significance level-α-1-0.9-0.1 Since Variances are assumed equal, but knowm, we use pooled variance Degrees of freedomdf -ni +n2 -2-29 The pooled estimate of Variance is: = 82 = (m-1)si + (n2-1) 15.676207 (ni -1)+ The standard error is sil-h- 그 + 2 1.441098 Critical Value = ta/2df-t0.05df-29-1.699 (from t-table, two-tails. d.f. = 29 、ㅣ 그 +-p 2 Margin of Error = E-ta/2,df × - 1.699 x 1.441098 E 2.448 Margin of Error, E 2.448 Point Estimate of difference: 2.5 Limits of 90% confidence interval are given by: Lower limit 2-E--2.5 _ 2.448425 =-4.948425 Upper limit-Ti-7, + E =-2.5 + 2.448425 =ー0.051575 90% confidence interval is:--22士E--2.5士2.448425 (-4.948425,-0.051575) 90% CI using t-dist: 4.948 < -μ2 <-0.0516 Note that exact answer using technology is: -4.9486079740, -0.0513920260)