Question #349596

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.


1
Expert's answer
2022-06-10T14:14:25-0400

The following null and alternative hypotheses need to be tested:

H0:μ75H_0:\mu\le75

H1:μ>75H_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 α=0.05,\alpha = 0.05, df=n1=11df=n-1=11 and the critical value for aright-tailed test is tc=1.795885.t_c =1.795885.

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

The t-statistic is computed as follows:


t=xˉμs/n=77.2757.9/12=0.9647t=\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=tc,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=11df=11 degrees of freedom, t=0.9647t=0.9647 is p=0.177712,p=0.177712, and since p=0.177712>0.05=α,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 μ\mu

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


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