Question #345875

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


1
Expert's answer
2022-05-31T12:08:41-0400

The following null and alternative hypotheses need to be tested:

H0:μ24H_0:\mu\le24

H1:μ>24H_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 α=0.05,\alpha = 0.05, and the critical value for a right-tailed test is zc=1.6449.z_c = 1.6449.

The rejection region for this right-tailed test is R={z:z>1.6449}.R = \{z:z> 1.6449\}.

The z-statistic is computed as follows:


z=xˉμσ/n=16.3245/40=9.74z=\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=zc,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,p=P(Z>-9.74)=1, and since p=1>0.05=α,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 μ\mu

is greater than 24, at the α=0.05\alpha = 0.05 significance level.


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