18. Twenty students randomly assigned to an experimental group receive an instru
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18. Twenty students randomly assigned to an experimental group receive an instructional program; 30 in a control group do not. After 6 months, both groups are tested on their knowledge. The experimental group has a mean of 38 on the test (with an estimated population standard deviation of 3); the control group has a mean of 35 (with an estimated population standard deviation of 5). Using the .05 level, what should the experimenter conclude? (a) Use the steps of hypothesis testing, (b) sketch the distributions involved, and (c) explain your answer to someone who is familiar with the t test for a single sample but not with the t test for independent means.Explanation / Answer
H0: No difference in means of experimental and control groups H1: There is a difference. n1=20 (number of cases in experimental group) n2=30 (number of cases in control group) mean 1 = 38 s1 = 3 mean 2 = 35 s2 = 5 level of significance = 0.05 First, test for the equality of population variances between the two groups. F= s2^2/s1^2 = 5^2/3^2 = 25/9 = 2.78 Compare with F(19,29) table value = 1.9446 Since F-computed exceeds F-critical, the population variances are unequal. Use the two sample t-test (assuming unequal variances) Sample 1 size 20 Sample 2 size 30 Sample 1 mean 38 Sample 2 mean 35 Sample 1 S.D. 3 Sample 2 S.D. 5 t = (x-bar1 - x-bar2) / sqrt [ s1^2/n1+ s2^2/n2] sqrt [ s1^2/n1+ s2^2/n2] = 1.132843 t = (38 - 35) = 3 / 1.132843 t =2.6482 Degree of freedom = [s1^2/n1+s2^2/n2]^2 / [(s1^2/n1)^2/(n1-1)+(s2^2/n2)^2/(n2-1)] = [9/20 + 25/30]^2 / [(9/20)^2 /19+ (25/30)^2/29] Degree of freedom 47.5937 Compare the computed t 2.6482 with critical t (48 degrees of freedom and 0.05 level of significance) t-critical = 2.011 Computed t > critical t Decision: Reject the null hypothesis H0. The experimenter must conclude that there is a difference between the experimental group and the control group. The instructional program has an effect.
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