Answer to Question #348566 in Statistics and Probability for xxx

Question

You are given the following information of a company that produces cola fizzy drinks. The

company states that the mean caffeine content per bottle of a fizzy drink is 40 mg with a

standard deviation of 7.5 mg. The quality controller is convinced that it is lower. A sample of

30 randomly drawn bottles has a mean caffeine content of 39.2 mg. Can the quality

controller reject the claim? Conduct a hypothesis test at 0.05 level of significance.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge40$

$H_1:\mu<40$

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a left-tailed test is $z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z:z<-1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{39.2-40}{7.5/\sqrt{30}}=-0.5842$$

Since it is observed that $z=-0.5842>-1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(z<-0.5842)= 0.279543,$ and since $p= 0.279543>0.05=\alpha,$ 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 40, at the $\alpha = 0.05$ significance level.