Answer to Question #346275 in Statistics and Probability for John Lloyd

Question

A coffee vending machine is designed to dispense 180 ml of coffee but its owner 

suspects that it is dispensing more than what is designed for. He took a random 

sample of 40 and found out that the mean is 192 ml with a standard deviation of 

4 ml. do you think the owner is right about his suspicion? Test at 0.05 level of 

significance.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le180$

$H_1:\mu>180$

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 $\alpha = 0.05,$ $df=n-1=39$ and the critical value for a right-tailed test is $t_c =1.684875.$

The rejection region for this right-tailed test is $R = \{t:t>1.684875\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{192-180}{4/\sqrt{40}}=2.1082$$

Since it is observed that $t=2.1082>1.684875=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=39$ degrees of freedom, $t=2.1082$ is $p=0.041491,$ and since $p= 0.041491<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is greater than 180, at the $\alpha = 0.05$ significance level.