Question

a company that makes cola drinks states that the mean caffeine content per 12 - ounce bottle of cola is 35 milligrams. you want to test this claim. during your tests, you find that a random sample of thirty 12 - ounce bottles of cola has a mean caffeine content of 33.9 milligrams. assume the population is normally distributed and the population standard deviation is 6.4 milligrams. at \\( \alpha=0.02 \\), can you reject the companys claim? complete parts (a) through (e). at the 2% significance level, there is not enough evidence to support the companys claim that the mean caffeine content per 12 - ounce bottle of cola is equal to 35 milligrams.

Explanation:

Step1: State the hypotheses

The null hypothesis \(H_0:\mu = 35\) (the company's claim), and the alternative hypothesis \(H_1:\mu
eq35\) (two - tailed test).

Step2: Calculate the test statistic

The formula for the \(z\) - test statistic is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
Given \(\bar{x} = 33.9\), \(\mu = 35\), \(\sigma=6.4\), \(n = 30\).

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Step3: Find the critical values

For a two - tailed test with \(\alpha=0.02\), the critical values are \(z_{\alpha/2}=\pm z_{0.01}\).
From the standard normal table, \(z_{0.01}\approx\pm 2.33\).

Step4: Make a decision

Since \(- 2.33< - 0.94<2.33\) (the test statistic \(z=-0.94\) does not fall in the rejection region), we fail to reject the null hypothesis.

Answer:

At the \(2\%\) significance level, there is not enough evidence to reject the company's claim that the mean caffeine content per \(12 -\) ounce bottle of cola is equal to \(35\) milligrams.