Answer to Question #348501 in Statistics and Probability for Maybel

Question

The average zone of inhibition (in mm) for mouthwash H as tested by medical technology

students have been known to be 9mm. A random sample of 10 mouthwash H was tested

and the test yielded an average zone of inhibition of 7.5 mm with a variance of 25mm. Is

there enough reason to believe that the antibacterial property of the mouthwash has

decreased? Test the hypothesis that the average zone of inhibition of the mouthwash is

no less than 9mm using 0.05 level of significance.

Answer 1

The parameter is average content of fruit concentrate per bottle.

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge 9$

$H_1:\mu<9$

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

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=9$ and the critical value for a left-tailed test is $t_c =-1.833113.$

The rejection region for this left-tailed test is $R = \{t:t<-1.833113\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{7.5-9}{5/\sqrt{10}}=-0.9487$$

Since it is observed that $t=-0.9487>-1.833113=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed, $df=9$ degrees of freedom, $t=-0.9487$ is $p=0.183775,$ and since $p=0.183775>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

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

is less than 9, at the $\alpha = 0.05$ significance level.