Answer to Question #156629 in Statistics and Probability for Neha

Question

A pharmaccetuical firm maintains that the mean time for a drug to take effect is 24 minutes.In a sample of 400 trails ,the mean time is 26 mintues with a standard deviation of 4minutes.test the hypothesis that the mean time is 24 minute against the alternative that it is not equal to 24 min.use a level of signifixance

Answer 1

The provided sample mean is $\bar{x}=26$ and the sample standard deviation is $s=4.$

The sample size is $n=400.$ The number of degrees of freedom are $df=n-1=399,$

and the significance level is $\alpha=0.01,$

The following null and alternative hypotheses need to be tested:

$H_0: \mu=24$

$H_1: \mu\not =24$

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

Based on the information provided, the critical value for a two-tailed test is $t_c=2.5882.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.5882\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{26-24}{4/\sqrt{400}}=10$$

Since it is observed that $|t|=10>2.5882=t_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$  is different than 24, at the 0.01 significance level.

Using the P-value approach: The p-value is $p=0,$ and since $p=0<0.01,$ it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$  is different than 24, at the 0.01 significance level.