Answer to Question #253485 in Statistics and Probability for mave

Question

performance of 40 teachers in the Mathematics department showed a mean of 95 with a standard deviation of 7.3.

Answer 1

The performance of 50 teachers from the faculty of engineering showed a mean of 98 with a standard deviation of 9.2, while, a sample of 40 teachers from the college of science showed an average performance of 95 with a standard deviation of 7.3. Is there a difference in performance between the two samples? Use α = 0.05

The following null and alternative hypotheses need to be tested:

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

This corresponds to a two-tailed test.

Level of significance =0.05

Sincesample sizes are greater than 30 so z-test for two means, with unknown population standard deviations will be used.

Under null hypothesis the test statistic is obtained as:

$Z = \frac{\bar{x_1} - \bar{x_2}}{\sqrt{s^2_1/n_1 + s^2_2/n_2}} \\ = \frac{98-95}{\sqrt{(9.2)^2/50 + (7.3)^2/40}} \\ = 1.725$

Based on the information provided, the significance level is 0.05, and the critical value for a two-tailed test is $Z_c =1.96$

The rejection region for this two-tailed test is

$R = {Z:|Z|>1.96}$

Since it is observed that

$|Z| = 1.725 < Z_c = 1.96$

It is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2$ at 0.05 significance level.