A cold medicine lists

500500

milligrams of acetaminophen per fluid ounce as an active ingredient. A test of

5555

​one-ounce samples of the medicine finds that the mean amount of acetaminophen for the sample is

495495

milligrams with a standard deviation of

2020

milligrams. Test the claim that the medicine does not contain the required amount of acetaminophen. Use a 0.05 significance level.

To test the claim that the medicine does not contain the required amount of acetaminophen, we'll perform a hypothesis test. Here's the step-by-step process:

### 1. **State the Hypotheses**

- **Null Hypothesis (\(H_0\))**: The mean amount of acetaminophen in the medicine is 500500 milligrams per ounce.
  \[
  H_0: \mu = 500500 \text{ mg}
  \]
- **Alternative Hypothesis (\(H_1\))**: The mean amount of acetaminophen in the medicine is not 500500 milligrams per ounce.
  \[
  H_1: \mu \neq 500500 \text{ mg}
  \]

This is a two-tailed test because we are testing for a difference in either direction.

### 2. **Determine the Significance Level**
The significance level (\(\alpha\)) is given as 0.05.

### 3. **Calculate the Test Statistic**

The test statistic for a sample mean is calculated using the formula:

\[
t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
\]

where:
- \(\bar{x}\) is the sample mean = 495495 mg
- \(\mu_0\) is the population mean under the null hypothesis = 500500 mg
- \(s\) is the sample standard deviation = 2020 mg
- \(n\) is the sample size = 5555

Let's calculate the t-statistic:

\[
t = \frac{495495 - 500500}{\frac{2020}{\sqrt{5555}}}
\]

Now, let's compute it:

\[
t \approx \frac{-5005}{\frac{2020}{\sqrt{5555}}}
\]
\[
t \approx \frac{-5005}{26.95} \approx -185.66
\]

### 4. **Determine the Critical Value(s)**

Since this is a two-tailed test at a significance level of 0.05, the critical value of \(t\) for 5554 degrees of freedom (df = n - 1) can be approximated using the standard t-distribution table. For such a large sample size, the critical values approximate to \(\pm 1.96\).

### 5. **Make the Decision**

Compare the calculated t-statistic to the critical t-values:
- If \(t \leq -1.96\) or \(t \geq 1.96\), reject the null hypothesis.
- Otherwise, fail to reject the null hypothesis.

Here, \(t \approx -185.66\), which is much less than -1.96, so we **reject the null hypothesis**.

### 6. **Conclusion**

At the 0.05 significance level, there is sufficient evidence to support the claim that the medicine does not contain the required amount of acetaminophen.

Would you like any details or have any questions?

### Further Questions:
1. What are Type I and Type II errors in this context?
2. How would the conclusion change if the significance level were 0.01?
3. What assumptions are made in this t-test?
4. How can we calculate the confidence interval for the mean amount of acetaminophen?
5. How would a different sample size affect the t-statistic?

**Tip:** Always check the sample size to decide whether to use a t-test or a z-test.

Explore a hypothesis test to determine if a medicine contains the required amount of acetaminophen. Includes step-by-step calculation of t-statistic and critical values with real-world sample data.

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