Answer to Question #213033 in Statistics and Probability for Daniel

Question #213033

A certain drug is claimed by its manufacturers to reduce overweight men by 4.75 kg per month,

with a standard deviation of 0.89 kg. Ten randomly chosen men reported losing an average of 4.25 kg

within a month. Does this data support the claim of the manufacturer at 0.05 level of significance?


1
Expert's answer
2021-07-05T13:34:31-0400

The following null and alternative hypotheses need to be tested:

H0:μ=4.75H_0:\mu=4.75

H1:μ4.75H_1:\mu\not=4.75

This corresponds to a two-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 two-tailed test is zc=1.96.z_c=1.96.

The rejection region for this two-tailed test is R={z:z>1.96}.R=\{z:|z|>1.96\}.

The z-statistic is computed as follows:


z=xˉμσ/n=4.254.750.89/101.77656z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{4.25-4.75}{0.89/\sqrt{10}}\approx -1.77656

Since it is observed that z=1.77656<1.96=zc,|z|=1.77656<1.96=z_c, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p=2P(Z<1.77656)=0.075641,p=2P(Z<-1.77656)=0.075641, and since p=0.078641<0.05=α,p=0.078641<0.05=\alpha, it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population μ\mu is different than 4.75,4.75, at the α=0.05\alpha=0.05 significance level.



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