Question

In a test given to two groups of students to marks were as follows:

First group: 9, 11, 13, 11, 15, 9, 12, 14

Second group: 10, 12, 10, 14, 9, 8, 10

Examine the significance of difference μ1-μ2 under the assumption on the samples are drawn from normal population with σ12=σ22.

Answer 1

First group:

Sample size = = 8

Mean= = (9+11+13+11+15+9+12+14)/8 = 11.75

Standard deviation= =

=2.188

Second group:

Sample size = = 7

Mean= = (10+12+10+14+9+8+10)/7 = 10.429

Standard deviation= =

=1.988

The null hypothesis and alternate hypothesis are as below:

Assuming a significance level is = 0.05,

and the degrees of freedom are df = 8 + 7 -2 = 13.

The critical t value at = 0.05, df = 13, is

The rejection region for this two-tailed test is .

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

=1.217

Decision about the null hypothesis

Since it is observed that , it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.2454, and since p = 0.2454 0.05, it is concluded that the null hypothesis is not rejected.

Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean of first group is different than ​, of the second group at the 0.05 significance level.