Answer: the correct answer is we are 95% confident that there is no statistically significant difference in the mean treatment methods for hostility.
Step-by-step explanation:
1) Test for the equality of variances in the two groups to choose the appropriate t-test.
H0: σ (1)^2 = σ (2)^2
Ha: σ (1)^2 ≠ σ (2)^2
Larger variance = 64
Smaller variance = 49
F = 1.30612
Degrees of freedom 15 and 9
Critical F from the table (with alpha=0.05) = 2.58
Calculated F is smaller than critical F, so we use the pooled variance t-test.
Sample 1 size 7
Sample 2 size 8
Sample 1 mean 79
Sample 2 mean 84
Sample 1 S.D. 7
Sample 2 S.D. 8
Pooled S.D = [(n1-1)s1^2+(n2-1)s2^2]/(n1+n2-2)
Pooled variance s = [(6)(49)+(7)(64))] / (13) = (294+448)/13=742/13=57.076923
Pooled variance s^2 = 57.076923
Standard error of difference in means = sqrt(1/n1+1/n2) times sqrt(s^2)
Standard error of difference in means = (0.517549)(7.554927) = 3.910046 (denominator of t)
Confidence interval = (mean1-mean2) +/- t SE
t is the critical t with 24 degrees of freedom = 2.056
(79 - 84) +/- (2.056) (3.910046)
= (-13.04, 3.04)
Interval encloses 0
We are 95% confident that there is no statistically significant difference in the mean treatment methods for hostility.
