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A researcher claims that the amounts of acetaminophen in a certain brand of cold tablets have a mean different from the 600 mg claimed by the manufacturer. Test this claim at the p < .05 level of significance. The mean acetaminophen content for a random sample of n = 41 tablets is 603.3 mg. Assume that the population standard deviation is 4.9 mg.Question 12Answera.Since the test statistic is greater than the critical z, there is sufficient evidence to reject thenull hypothesis and to support the claim that the mean content of acetaminophen is not600 mg.b.Since the test statistic is greater than the critical z, there is insufficient evidence to rejectthe null hypothesis and to support the claim that the mean content of acetaminophen isnot 600 mgc.Since the test statistic is greater than the critical z, there is sufficient evidence to acceptthe null hypothesis and to support the claim that the mean content of acetaminophen is600 mg.d.Since the test statistic is less than the critical z, there is sufficient evidence to reject thenull hypothesis and to support the claim that the mean content of acetaminophen is not600 mg.

Question

A researcher claims that the amounts of acetaminophen in a certain brand of cold tablets have a mean different from the 600 mg claimed by the manufacturer. Test this claim at the p < .05 level of significance. The mean acetaminophen content for a random sample of n = 41 tablets is 603.3 mg. Assume that the population standard deviation is 4.9 mg.Question 12Answera.Since the test statistic is greater than the critical z, there is sufficient evidence to reject thenull hypothesis and to support the claim that the mean content of acetaminophen is not600 mg.b.Since the test statistic is greater than the critical z, there is insufficient evidence to rejectthe null hypothesis and to support the claim that the mean content of acetaminophen isnot 600 mgc.Since the test statistic is greater than the critical z, there is sufficient evidence to acceptthe null hypothesis and to support the claim that the mean content of acetaminophen is600 mg.d.Since the test statistic is less than the critical z, there is sufficient evidence to reject thenull hypothesis and to support the claim that the mean content of acetaminophen is not600 mg.

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Solution

To answer this question, we first need to perform a hypothesis test. Here are the steps:

Step 1: State the null hypothesis and the alternative hypothesis. The null hypothesis (H0) is that the mean amount of acetaminophen is 600 mg. The alternative hypothesis (H1) is that the mean amount of acetaminophen is not 600 mg.

Step 2: Determine the level of significance. The level of significance is given as p < .05.

Step 3: Calculate the test statistic. We use the formula for the z-score: z = (X - μ) / (σ/√n), where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Plugging in the given values, we get z = (603.3 - 600) / (4.9/√41) = 1.48.

Step 4: Determine the critical z-value. For a two-tailed test with p < .05, the critical z-values are -1.96 and +1.96.

Step 5: Make a decision. Since the test statistic (1.48) is not greater than the critical z-value (+1.96), there is insufficient evidence to reject the null hypothesis.

So, the correct answer is:

b. Since the test statistic is not greater than the critical z, there is insufficient evidence to reject the null hypothesis and to support the claim that the mean content of acetaminophen is not 600 mg.

This problem has been solved

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