Given a sample of female students, n = 7, with SS = 550, and a sample of male st
ID: 3243171 • Letter: G
Question
Given a sample of female students, n = 7, with SS = 550, and a sample of male
students, n = 10. with SS = 720:
a. Find the pooled variance for the two samples. (3 pts)
b. Compute the estimated standard error for the sample mean difference. (3 pts)
c. If the sample mean difference is 5 points, is this enough to reject the Ho for an
independentmeasures hypothesis test using .01, onetailed test? (3 pts)
d. If the sample mean difference is 20 points, is this enough to reject the Ho for
an independentmeasures hypothesis test using .01, onetailed test? (3 pts)
e. Compute Cohen’s d to measure the effect size for the 5point difference and
for the 20point difference. Briefly evaluate the effect size of each difference.
(4 pts)
Explanation / Answer
Solution:-
Given,
Female --> n1 = 7, SS1 = 550
Male ---> n2 = 10. SS2 = 720
(a)
Pooled variance = {(n1 - 1) SS1 + (n2 - 1)SS2} / (n1 + n2 -2)
= {6(550) + 9(720)} / 15 = 652
Therefore, the pooled variance is 652.
(b)
Standard error = sqrt[S2 (1/n1 + 1/n2)]
= sqrt[652 (1/7 + 1/10)]
= sqrt[652 * (17/70)]
= sqrt (158.39)
= 12.58
(c)
Sample mean difference = 5
t = 5 / standard error
t = {xbar1 - xbar2} / standard error = 5 / 12.58
=> t = 0.3975
Critical value, t0.01,15 = 2.602
As, Calculated test statistic is less than 2.602, which is the critical value
DO NOT REJECT Ho.
(d)
t = {xbar1 - xbar2} / standard error
= 20 / 12.58
= 1.59
Since, t-statistics is less than the critical value.
DO NOT REJECT Ho.
(e)
d = {xbar1 - xbar2} / sqrt(pooled variance)
i) 5 / sqrt(652) = 0.1958
ii) 20 / sqrt(652) = 0.7833
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