Question

a manufacturer claims that a certain brand of cold tablets contains an average 600 mg of acetaminophen. to test this claim, a researcher creates a random sample of 36 tablets. the mean of this sample is 576.4 mg, and the standard deviation is 76.7 mg. test the claim using a level of significance of 10%.
a. what type of test will be used in this problem? select an answer
b. identify the null and alternative hypotheses?
$h_0$: select an answer?
$h_a$: select an answer?
c. is the original claim located in the null or alternative hypothesis? select an answer
d. calculate your test statistic. write the result below, and be sure to round your final answer to two decimal places.

Explanation:

Step1: Determine test type

Since the population standard - deviation is unknown and we are testing a claim about the population mean with a sample, we use a one - sample t - test.

Step2: State hypotheses

The manufacturer claims that the average is 600 mg. The null hypothesis \(H_0\) is the claim we are testing, so \(H_0:\mu = 600\). The alternative hypothesis \(H_a\) can be two - tailed (since we are just testing if the claim is wrong without specifying a direction), so \(H_a:\mu
eq600\).

Step3: Locate claim

The original claim is \(\mu = 600\), which is in the null hypothesis.

Step4: Calculate test statistic

The formula for the one - sample t - test statistic is \(t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\), where \(\bar{x} = 576.4\), \(\mu = 600\), \(s = 76.7\), and \(n = 36\).
First, calculate \(\frac{s}{\sqrt{n}}=\frac{76.7}{\sqrt{36}}=\frac{76.7}{6}\approx12.78\).
Then, \(t=\frac{576.4 - 600}{12.78}=\frac{- 23.6}{12.78}\approx - 1.85\).

Answer:

a. One - sample t - test
b. \(H_0:\mu = 600\), \(H_a:\mu
eq600\)
c. Null hypothesis
d. \(-1.85\)