Answer to Question #279141 in Statistics and Probability for cess

Question

A researcher claims that 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 pounds and a standard deviation of 5 pounds. A α = .01, can the claim be rejected?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=110$

$H_1:\mu\not=110$

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 $\alpha = 0.01,$

$df=n-1=15-1=14$ 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:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}==\dfrac{112.5-110}{5/\sqrt{15}}\approx1.9365$$

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 $p = 0.073257 \ge 0.01=\alpha,$ 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.