Answer to Question #213033 in Statistics and Probability for Daniel

Question

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?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=4.75$

$H_1:\mu\not=4.75$

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 $\alpha=0.05,$ and the critical value for a two-tailed test is $z_c=1.96.$

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

The z-statistic is computed as follows:

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

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

Using the P-value approach:

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

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