Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

a company that makes cola drinks states that the mean caffeine content …

Question

a company that makes cola drinks states that the mean caffeine content per 12 - ounce bottle of cola is 41 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 is 7.7 milligrams. at \\( \alpha=0.03 \\), can you reject the companys claim? complete parts (a) through (d).
(a) find the standardized test statistic.
\\( z=0.36 \\) (round to two decimal places as needed.)
(b) 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.
(c) interpret the decision in the context of the original claim.
at the \\( 3 \\% \\) significance level, there \\( \square \\) enough evidence to \\( \square \\) the companys claim that the mean caffeine content per 12 - ounce bottle of cola \\( \square \\) milligrams.

Explanation:

Step1: State the null and alternative hypotheses

The company claims that the mean caffeine content per 12 - ounce bottle of cola is \( \mu = 44\) milligrams. So, \(H_0:\mu = 44\) and \(H_1:\mu
eq44\) (two - tailed test). The significance level \( \alpha=0.03\), so the critical values are \(z_{\alpha/2}=\pm z_{0.015}\). From the standard normal table, \(z_{0.015}\approx\pm 2.17\).

Step2: Calculate the test statistic

The formula for the \(z\) - test statistic for a sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Given \(\bar{x} = 44.5\), \(\mu = 44\), \(\sigma=1.7\), and \(n = 30\).

$$ LATEXBLOCK0 $$

Step3: Make a decision

Since \(z = 1.61\) and the critical values are \(z=\pm2.17\), and \(|z|=1.61<2.17\), the test statistic \(z\) is not in the rejection region.

Answer:

For part (b), the correct option is: Since \(z\) is not in the rejection region, fail to reject the null hypothesis.
For part (c), at the \(3\%\) significance level, there is enough evidence to support the company's claim that the mean caffeine content per 12 - ounce bottle of cola is equal to \(44\) milligrams.