Answer to Question #350889 in Statistics and Probability for vrix

Question

A soda manufacturer is interested in determining whether it's bottling machine tends to overfill. Each bottle is supposed to contain 15 ounces of fluid. A random sample of 30 bottles was taken and found that the mean amount of soda of the sample of bottles is 13.4 ounces with a standard deviation of 2.98 ounces. if the manufacturer decides on a significance level od 0.05 test, should the null hypothesis be rejected?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le 15$

$H_1:\mu>15$

This corresponds to a right-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 $\alpha = 0.05,$ $df=n-1=29$ and the critical value for a right-tailed test is $t_c =1.699127.$

The rejection region for this right-tailed test is $R = \{t:t>1.699127\}.$

The t-statistic is computed as follows:

$$t=\dfrac{11.9-10}{1.8/\sqrt{25}}=\dfrac{13.4-15}{2.98/\sqrt{30}}=-2.9408$$

Since it is observed that $t=-2.9408<1.699127=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, $df=29$ degrees of freedom, $t=-2.9408$ is $p=0.996814,$ and since $p=0.996814>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is greater than 15, at the $\alpha = 0.05$ significance level.