Question

STAT2550 Winter 2002 13. A paired difference experiment produced the following data: n, = 16,21 = 143, x2 = 150,X) = -7,5) = 64 Calculate the value of t for which the null hypothesis, pp -0, would be rejected in favor of the alternative hypothesis Hp <0. Use a -0.10.(2) Conduct the paired difference test described in part a. Draw the appropriate conclusion. (3) Find a 90% confidence interval for the mean difference up (2)

Answer 1

Sol?: In the usual notations, we are given : no = 16, n = 143, n = 150, to = -7 s = 64 a> Null Hypothesis ; Ho: Mp=0 rei the difference is insignificant. Alternative hypothesis ; Hi & Moto (Left-Jailed) Test Statistic: Under Ho, the test statistic is : t = d = it= -7 -7 = -3.5 The tabulated value of t at 10% level for 15 d.f is 0.69 for one- tail test Conclusion: The critical region for the left to mil test is tc-0.69. Since talculated a w of t is less than -0.69, Ho is a ted at 10% level of significance of t, value at rejected

On 67 Paired t-test for Difference of Mears We setup the Hypothesis as - Null Hypothesis Ho: up = 0 ie; There is no significant difference between the sample means Alternative Hypothesis H, : Moro (Left- tailed) ici The sample means differ significantly. Test Statistic: under Ho, the test Statistic is : ta ł nt(n-1) ici t = * = -7 = -3.5 Sono Now, I abulated to. 1o for 15 dif for single-tail test is 0.09. Conclusion: The critical region for the Icht- tail test is tc- 0.09. Since calculated t is less than -0.69, Ho is rejected at 10.1 1.0.4 Mence, conclude that the sample means differ significantly teste critica carte tail we

difference c7 90% CI for mean UD : (d. f = 15) 90% fiducial limits given by : to.10 for up are [ TD= -7 S = 64 no = 16 to.no = 0.691 for is df - 7 1 0.6948 vīã = -7 + 0.69 x 2 = -7 + 1.38 = -5.62 and - 8.38 - - 8.38 < up < -5.62 in 90% c. I for meas difference Лp is (-5.62 , -8.38) DOMINICO