Question

a company that makes cola drinks states that the mean caffeine content per 12 - ounce bottle of cola is 45 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 44.5 milligrams. assume the population is normally distributed and the population standard deviation 7.7 milligrams. at \\( \alpha=0.03 \\), can you reject the companys claim? complete parts (a) through (e)
(c) find the standardized test statistic
\\( z=-0.36 \\) (round to two decimal places as needed)
(d) decide whether to reject or fail to reject the null hypothesis.
a. since \\( z \\) is in the rejection region, fail to reject the null hypothesis.
b. since \\( z \\) is in the rejection region, reject the null hypothesis.
c. since \\( z \\) is not in the rejection region, fail to reject the null hypothesis.
d. since \\( z \\) is not in the rejection region, reject the null hypothesis.
(e) interpret the decision in the context of the original claim.
at the \\( 3 \\% \\) significance level, there enough evidence to the companys claim that the mean caffeine content per 12 - ounce bottle of cola milligrams

Explanation:

Step1: Determine the type of test

Since the company claims the mean is 45 (a specific value) and we are testing against this claim, it's a two - tailed test. For a two - tailed test with \(\alpha = 0.03\), the critical \(z\) - values are \(z=\pm z_{\alpha/2}\). Calculate \(z_{\alpha/2}\), where \(\alpha/2=0.015\). Looking up in the standard normal table, \(z_{0.015}\approx\pm 2.17\).

Step2: Analyze the position of \(z=- 0.36\)

The value of the test statistic \(z = - 0.36\). Since \(-2.17<-0.36 < 2.17\), the \(z\) - value is not in the rejection region.

Answer:

C. Since \(z\) is not in the rejection region, fail to reject the null hypothesis.