Answer to Question #342726 in Statistics and Probability for Van

Question

A researcher claims that a 20 year old women on a special diet will have an average weight of 110 pounds. A sample of 15 women has an average weight of 112.5 lbs and a standard deviation of 5 pounds. At 8 = 0.01 can the clean be rejected?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=110$

$H_a:\mu\not=110$

This corresponds to a two-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 $\alpha = 0.01,$ $df=n-1=14$ degrees of freedom, and the critical value for a two-tailed test is $t_c =2.976842.$The rejection region for this two-tailed test is $R = \{t:|t|> 2.976842\}.$

The t-statistic is computed as follows:

$$t=\dfrac{112.5-110}{5/\sqrt{15}}=1.9365$$

Since it is observed that $|t| = 1.9365<2.976842=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed $df=14$ degrees of freedom, $t=1.9365$ is $p=0.073257,$ and since $p=0.073257>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 110, at the $\alpha = 0.01$ significance level.