Homework SourseThe difference for the scores of Group A minus Group B has a
The difference for the scores of Group A minus Group B has a mean of 3.2 which a standard deviation of 4.96.
Does it appear that early exposure to exercise increases artistic ability? Us appropriate statistical evidence to support your answer
Group A (Exercise Program) 86 49 Twin Set Group B 2 3 4 5 6 7 8 9 10 (No Exercise Program) 85 53 70 60 53 84 79 85 65 70 72 70 63 59 83 67 73 78 74 86 73 69 12 13 14 15 71 60 The mean score for those having participated in the exercise program is 74.2 with a standard deviation of 12.17. The mean score for those that did not participate in the exercise program is 71 with a standard deviation of 11.48.Solution
Formulating the null and alternative hypotheses,
Ho: u1 - u2 <= 0
Ha: u1 - u2 > 0
At level of significance = 0.05
As we can see, this is a right tailed test.
Calculating the means of each group,
X1 = 74.2
X2 = 71
Calculating the standard deviations of each group,
s1 = 12.17
s2 = 11.48
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 15
n2 = sample size of group 2 = 15
Thus, df = n1 + n2 - 2 = 28
Also, sD = 4.319716812
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = 0.740789302
where uD = hypothesized difference = 0
Now, the critical value for t is
tcrit = 1.701130934
Thus, comparing t and tcrit, we decide to WE FAIL TO REJECT THE NULL HYPOTHESIS.
Also, using p values,
p = 0.232494353
Comparing this to the significance level, WE FAIL TO REJECT THE NULL HYPOTHESIS.
Thus, there is no signficant evidence at 0.05 level that early exposure to exercise increases artistic ability. [CONCLUSION]
