Answer to Question #349596 in Statistics and Probability for ralphilagan

Question

2. The director of a secretarial school believes that its graduates can type more than75

words per minute. A random sample of 12 graduates has been found to have an

average of 77.2 words per minute with a standard deviation of 7.9 words per minute

in a typing test. Using the 0.05 level of significance, test the claim of the director.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le75$

$H_1:\mu>75$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=11$ and the critical value for aright-tailed test is $t_c =1.795885.$

The rejection region for thisright-tailed test is $R = \{t:t>1.795885\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{77.2-75}{7.9/\sqrt{12}}=0.9647$$

Since it is observed that $t=0.9647<1.795885=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, $df=11$ degrees of freedom, $t=0.9647$ is $p=0.177712,$ and since $p=0.177712>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is greater than 75, at the $\alpha = 0.05$ significance level.