Answer to Question #346290 in Statistics and Probability for pogi

Question

A company produces mosquito repellants that have a period of effectiveness that is approximately normally distributed with a mean of 300 hours and a standard deviation of 30 hours. A sample of 40 repellants was taken and the average period of effectiveness is computed as 290 hours. Is this sufficient evidence to conclude that the company’s claim is true at α=0.01?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=300$

$H_1:\mu\not=300$

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 $\alpha = 0.01,$ and the critical value for a two-tailed test is $z_c =2.5758.$

The rejection region for this two-tailed test is $R = \{z:|z|> 2.5758\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{290-300}{30/\sqrt{40}}=-2.1082$$

6. Since it is observed that $|z|=2.1082<2.5758=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed is $p=2P(Z<-2.1082)=0.035014,$ and since $p=0.035014>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 300, at the $\alpha = 0.01$ significance level.