Answer to Question #349955 in Statistics and Probability for Sukesh

Question

Of all of the individuals who develop a certain rash, suppose the mean recovery time for individuals who do not use any form of treatment is 30 days with standard deviation equal to 8. A pharmaceutical company manufacturing a certain cream wishes to determine whether the cream shortens, extends, or has no effect on the recovery time. The company chooses a random sample of 100 individuals who have used the cream, and determines that the mean recovery time for these individuals was 28.5 days. Does the cream have any effect?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge30$

$H_1:\mu<30$

This corresponds to a left-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 left-tailed test is $z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z:z<-1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{28.5-30}{8/\sqrt{100}}=-1.875$$

Since it is observed that $z=-1.875<-1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z<-1.875)=0.030396,$ and since $p=0.030396<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 30, at the $\alpha = 0.05$ significance level.