Answer to Question #348570 in Statistics and Probability for max

Question

A random sample of 20 drinks from a soft-drink machine has an average content of 21.9 deciliters, with a standard deviation of 1.42 deciliters. At 0.05 level of significance, test the hypothesis that μ = 22. 2 against the alternative hypothesis μ < 22.2. Assume that the distribution is normal.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=22.2$

$H_1:\mu<22.2$

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 $\alpha = 0.05,$ $df=n-1=19$ and the critical value for a left-tailed test is $t_c =-1.729133.$

The rejection region for this left-tailed test is $R = \{t:t<-1.729133\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{21.9-22.2}{1.42/\sqrt{20}}=-0.9448$$

Since it is observed that $t=-0.9448>-1.729133=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed, $df=19$ degrees of freedom, $t=-0.9448$ is $p=0.178311,$ and since $p= 0.178311>0.05=\alpha,$ 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 22.2, at the $\alpha = 0.05$ significance level.