Question

Independent random samples were selected from two quantitative populations, with sample sizes, means, and variances given below. Sample Size Sample Mean Sample Variance Population 1 2 34 45 9.8 7.5 10.83 16.49 State the null and alternative hypotheses used to test for a difference in the two population means. O Ho: (41 - H2) = 0 versus Ha: (41 - M2) > 0 Ho: (41 – 12) # O versus Ha: (H1 - H2) = 0 HO: (41 – My) < 0 versus Hy: (My – My) > 0 HO: (41 - H2) = 0 versus Hy: (My – My) < 0 Ho: (My - H2) = 0 versus Ha: (41 - H2) = 0 Calculate the necessary test statistic. (Round your answer to two decimal places.) z = Calculate the rejection region with a = 0.01. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.) 2 > z < Draw the appropriate conclusion. OH, is rejected. There is insufficient evidence to indicate a difference in mean. Ho is not rejected. There is insufficient evidence to indicate a difference in mean. He is not rejected. There is sufficient evidence to indicate a difference in mean. He is rejected. There is sufficient evidence to indicate a difference in mean.

Answer 1

o X-X2 1 = 34 h = 45. * = 9.2 ng = 7.5 1s = 10.83 8 2 = 16:49 6 Null keypothesis, Ho: M, H2) = 0 Vs Alternative hypothesis,, '64,-M2)70 1134 10.33 16.49 45 11 0.82763 1121 = 13.29 1:12l at g9 % Confidence internal oral 17 level of significance is 2.58, aztery Czeut Hence null hypothesis is rejected at 17 level of significaure. if the 2=0.01 or 11. lovel of significance 22 = 2.33 (Right tailed) 22 = -2.33 (left tailed Hence P[272.33J = 0.01. P[2. (-2.33]=0.01 (for Right tailed) (for left tailed) fence, Ho is rejected. There is sufficient evidence to indicate a difference in mean.