Question

A maintenance supervisor is comparing the standard version of an instructional booklet with one
that has been claimed to be superior. An experiment is conducted in which 26 technicians are
divided into two groups, provided with one of the booklets, then given a test a week later. For the
13 using the standard version, the average exam score was 72.0, with a standard deviation of 9.3.
For the other 13 using the new version, the average score was 80.2, with the standard deviation of
10.1. Assume normal populations with equal standard deviations, and using the 5% level of
significance, does the standard version booklet appear to be better than the new version?

Answer 1

$\bar{x_1} = 72 \\ s_1 = 9.3 \\ n_1 = 13 \\ \bar{x_2} = 80.2 \\ s_2 = 10.1 \\ n_2 = 13$

The objective of the study is to test does the new booklet appear to be better than the standard version.

Under the claim the null and alternative hypotheses are

$H_0: \mu_1 = \mu_2 \\ H_1: \mu_1 < \mu_2 \\$

α=0.05

Test-statistic

$t = \frac{(\bar{x_1} - \bar{x_2}) -(\mu_1 -\mu_2)}{\sqrt{s^2_{p} \times (\frac{1}{n_1} + \frac{1}{n_2})}} \\ s^2_{p} = \frac{(n_1-1) \times s^2_1 + (n_2-1) \times s^2_2}{n_1+n_2-2} \\ = \frac{(13-1) \times 9.3^2 + (13-1) \times 10.1^2}{13+13-2} \\ = 94.25$

Substitute the values in test statistic as shown below:

$t = \frac{(72.0-80.2) -0}{\sqrt{94.25(\frac{1}{13} + \frac{1}{13})}} \\ = -2.15$

Degrees of freedom:

$df= n_1+n_2 -2 = 13+13 -2 = 24$

At the given level of significance 0.05 and the test is left-tailed, then from t-table corresponding to 24 degrees of freedom the critical value is equal to -1.711.

Here, the calculated t value -2.15 is less than the tabulated value -1.711, reject the null hypothesis.

Hence, it can be concluded that new booklet is better than the standard version.