Question

A soda manufacturer is interested in determining wheter its bottling machine tends to overfill. Each bottle is supposed to contain 12 ounces of fluid. A random sample of 25 bottles was taken and found that the mean amount of soda of the sample of bottles is 12.2 ounces with a standard deviation of 0.4 ounces. If the manufacturer decides on a significance level of 0.05 test, should the null hypothesis (u=12 ounces) be rejected?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=12$

$H_1:\mu\not=12$

This corresponds to a two-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 $\alpha = 0.05,$ $df=n-1=24$ and the critical value for a two-tailed test is $t_c = 2.063899.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.063899\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{12.2-12}{0.4/\sqrt{25}}=2.5$$

Since it is observed that $|t|=2.5> 2.063899=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=24$ degrees of freedom, $t=2.5$ is $p= 0.019654,$ and since $p= 0.019654<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 12, at the $\alpha = 0.05$ significance level.