Question

Two methods of teaching statistics are being tried by a professor. A class of 40 students is taught by method A and a class of 36 is taught by method B. the two classes are given the same final examination. The scores are: Using the .01 significance level, can we conclude that the average final examination scores produced by the two methods are different if the population standard deviation is 5?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\not=\mu_2$

This corresponds to a two-tailed test, and a z-test for two means, with known population standard deviations will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a two-tailed test is $z_c = 2.5758.$

The rejection region for this two-tailed test is $R = \{z: |z| > 2.5758\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\sigma^2_1/n_1+\sigma^2_2/n_2}}$$

$$=\dfrac{78-74}{\sqrt{(5)^2/40+(5)^2/36}}=3.4823$$

Since it is observed that $|z| = 3.4823 >2.5758= z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z>3.4823)=0.000497,$  and since 

$p = 0.000497 < 0.01=\alpha,$ ​it is concluded that the null hypothesis is rejected

Therefore, there is enough evidence to claim that the population mean $\mu_1$

is different than $\mu_2,$ at the $\alpha = 0.01$ significance level.