Answer to Question #347803 in Statistics and Probability for Gracia

Question

The amount of coffee dispensed by a vending machine is assumed to be normally distributed, A vendor claims that the mean amount of coffer dispensed by the machine is 6 ounces per cup. The office manager doubted this claim. He randomly selected 25 cups of coffee from this machine and recorded the amount of coffee dispensed in each. The sample mean is 5.8 ounces with a standard deviation of 0.8 ounces, do the data provide sufficient evidence to indicate that the mean amount of coffee dispensed is not equal to 6 ounces per cup use 0.05 level of significance.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=6$

$H_1:\mu\not=6$

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:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{5.8-6}{0.8/\sqrt{25}}=-1.25$$

Since it is observed that $|t|=1.25<2.063899=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, $df=24$ degrees of freedom, $t=-1.25$ is $p=0.223351,$ and since $p=0.223351>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 6, at the $\alpha = 0.05$ significance level.