Question

2.3 of 60 new entrants in a are given university found to have a mean height of 68.60 inches and 50 seniors, a mean height of 69.51 inches. Is the evidence conclusive that the mean height of seniors is greater then that of the new entrants? Assume the s.d. of the height to be 2.48 inches?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\leq\mu_2$

$H_1:\mu_1>\mu_2$

This corresponds to a right-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 $\alpha = 0.05,$ and the critical value for a right-tailed test is $z_c = 1.6449.$

The rejection region for this right-tailed test is $R = \{z: z > 1.645\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}}$$

$$=\dfrac{69.51-68.60}{\sqrt{\dfrac{2.48^2}{50}+\dfrac{2.48^2}{60}}}=1.9163$$

Since it is observed that $z = 1.9163 >1..6449= z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(z>1.9163)=0.027666,$ and since $p = 0.027666 < 0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is greater than $\mu_2,$ at the $\alpha = 0.05$ significance level.

Therefore, there is enough evidence to claim that the mean height of seniors is greater then that of the new entrants, at the $\alpha = 0.05$ significance level.