Answer to Question #345875 in Statistics and Probability for John

Question

3. A diet clinic states that there is an average loss of 24 pounds for those who stay on the program for 20 weeks. The standard deviation is 5 pounds. The clinic tries a new diet, reducing salt intake to see whether that strategy will produce a greater weight loss. A group of 40 volunteers loses as as average of 16.3 pounds each over 20 weeks. Should the clinic change the new diet? Use a 0.05

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le24$

$H_1:\mu>24$

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation 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 $R = \{z:z> 1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{16.3-24}{5/\sqrt{40}}=-9.74$$

6. Since it is observed that $z=-9.74<1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed is $p=P(Z>-9.74)=1,$ and since $p=1>0.05=\alpha,$ 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 24, at the $\alpha = 0.05$ significance level.