Question

10-66. Consider the following set of samples obtained from two normally distributed populations whose variances are equal: Sample 1: 11.2 11.2 7.4 8.7 8.5 13.5 4.5 11.9 Sample 2: 11.7 9.5 15.6 16.5 11.3 17.6 17.0 8.5 a. Suppose that the samples were independent. Perform a test of hypothesis to determine if there is a difference in the two population means. Use a significance level of 0.05.

Answer 1

a.

Two-Sample T-Test and CI: Sample 1, Sample 2

Two-sample T for Sample 1 vs Sample 2

N Mean StDev SE Mean
Sample 1 8 9.61 2.89 1.0
Sample 2 8 13.46 3.62 1.3

Difference = mu (Sample 1) - mu (Sample 2)
Estimate for difference: -3.85
95% CI for difference: (-7.36, -0.34)
T-Test of difference = 0 (vs not =): T-Value = -2.35 P-Value = 0.034 DF = 14
Both use Pooled StDev = 3.2736

b.

Paired T-Test and CI: Sample 1, Sample 2

Paired T for Sample 1 - Sample 2

N Mean StDev SE Mean
Sample 1 8 9.61 2.89 1.02
Sample 2 8 13.46 3.62 1.28
Difference 8 -3.85 5.41 1.91

95% CI for mean difference: (-8.38, 0.68)
T-Test of mean difference = 0 (vs not = 0): T-Value = -2.01 P-Value = 0.084

c.

For part a., since p-value=0.034<0.05 so we reject null hypothesis at 5% level of significance and conclude that there is a significant difference between the two population mean.

Whereas for part b. since p-value=0.084>0.05 so we fail to reject null hypothesis at 5% level of significance and conclude that there is a insignificant difference between the two population mean.

These two are different conclusion because Part a is considered for two independent samples whereas for part b is considered as bivariate sample i.e. two populations are collected from same indivual i.e. there exist some significant relationship.