Answer to Question #335454 in Statistics and Probability for Jeve Quinones

Question

A new drug on the market is claimed by its manufacturers to reduce overweight women by 4.55 kg per month with a standard deviation of 0.91 kg. Forty women chosen at random have reported losing an average of 4.05 kg within a month.

Answer 1

1. The following null and alternative hypotheses need to be tested:

$H_0:\mu=4.55$

$H_a:\mu\not=4.55$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

2. 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 $R = \{z: |z| > 1.96\}.$

3. The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{4.05-4.55}{0.91/\sqrt{40}}\approx-3.475$$

4. Since it is observed that $|z| = 3.475 > 1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(Z<-3.475)=0.000511,$ and since $p=0.000511<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

5. Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 4.55, at the $\alpha = 0.05$ significance level.