Answer to Question #347188 in Statistics and Probability for Carol

Question

The average zone of inhibition (in mm) for mouthwash L as tested by the medical technology students has been known to be 9mm. A random sample of 10 mouthwash L was tested and the test yielded an average zone of inhibition of 7.5mm with a variance of 25 mm. Is there enough reason to believe that the anti-bacterial 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.

A. State the hypotheses.

B. Determine the test statistic to use.

C. Determine the level of significance, critical value, and the decision rule.

D. Compute the value of the test statistic.

E. Make a decision.

F. Draw a conclusion.

Answer 1

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

$H_0:\mu\ge9$

$H_1:\mu<9$

B.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.

C. 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 $t_c =-1.8331.$

The rejection region for this left-tailed test is $R = \{t:t<-1.8331\}.$

D. The t-statistic is computed as follows:

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

E. Since it is observed that $t=-0.9487>-1.8331=t_c,$ 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.

F. Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 9, at the $\alpha = 0.05$ significance level.