Question

(1 point) Suppose you needed to test the claim that the two samples described below come from populations with the same mean. Assume that the samples are independent simple random samples. Sample 1: n = 18, x1 = 24.7, sı = 7.07 Sample 2: n2 = 5, X2 = 27, S2 = 7.99 Find: (a) The estimated degree of freedom is (b) The standardized test statistic is (use Sample 1 - Sample 2)

Answer 1

Hypotheses : To test null hypothesis Ho M1 - M2 = 0 against alternative hypothesis H1 :41-42 #0 This is a two tailed test as the alternative hypothesis contains '#' sign. Here, C1 = 24.7000, $i = 7.0700, n1 = 18, 22 = 27.0000, S2 = 7.9900, n2 = 5 The test statistic can be written as: += (1 – Ez) – 0 + which under Ho follows a t distribution with 6 df. where Game + entry ? (17.07* + 7.92 ) 2 - 6 7.072 2 7 .992 18 18-1 T 5-1 ni-1 ' no-1 and standard error for the difference between means 7.072 SE= se=11.12. - 792 – 3.242710 7.992 - = = 3.942710 V ni n2 18X

Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tstat) > to.025,df = to.025,6 = 2.446912 0.4 - 0.3 density 0.1- Nowo- WITT Value of the test statistic : The value of the test statistic is Istat =- =-0.583355 -0.583 (24.7000 – 27.0000) V 7.07002 + 7.9900" associated degrees of freedom = 6 The critical value = tcritical = + to.025.6 = 2.446912 2.447 P- value = P(|t6] > tobs) = 2 * P(to < -0.583355) = 2 * 0.290452 = 0.580904 ~ 0.5809 Conclusion: Since p-value > 0.05 and tobs * tcritical = 2.447, so We fail to reject H, at 0.05 level of significance Hence, we can conclude that there is no significant difference between two population means