Question

Question 11 --/2 View Policies Current Attempt in Progress Use the t-distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distributions are relatively normal. Test Ho : µr = Hc vs H. : Hr < µc using the fact that the treatment group (T) has a sample mean of 8.6 with a standard deviation of 4.1 while the control group (C) has a sample mean of 11.2 with a standard deviation of 3.4. Both groups have 25 cases. (a) Give the test statistic and the p-value. Round your answer for the test statistic to two decimal places and your answer for the p-value to three decimal places. test statistic = p-value = i Hint eTextbook and Media (b) What is the conclusion of the test at a 5% significance level? O Reject Ho. O Do not reject Ho.

Answer 1

Here, 21 = 8.60, $i = 4.10, ni = 25, C2 = 11.20, S2 = 3.40, n2 = 25 The test statistic can be written as : += (61 – 12) – 0 which under Ho follows at distribution with 46 df. where Decision rule : We reject Ho at 0.05 level of significance if P-value < 0.05 or if tobs < -10.05,df Now, The value of the test statistic = -2.440696 ~ -2.441 associated degrees of freedom = 46 The critical value =-1.678660 -1.679 P-value = P(+46 < tobs) = P(+46 < -2.440696) = 0.009284 Since p-value < 0.05 and tobs < tcritical = -1.678660, so we reject Ho at 0.05 level of significance Hence, we can conclude that mean for first population is significantly less than mean for second population.