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Suppose 245 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.10 significance level to test the claim that more than 20% of users develop nausea.

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Answer:

To test the claim that more than 20% of users develop nausea when treated with the drug, you can set up a hypothesis test. Here are the null and alternative hypotheses:

Null Hypothesis (H0): The proportion of users who develop nausea (p) is less than or equal to 20%, i.e., p ≤ 0.20.

Alternative Hypothesis (Ha): The proportion of users who develop nausea (p) is greater than 20%, i.e., p > 0.20.

Now, you can perform a hypothesis test using the given information and a significance level (α) of 0.10. The test statistic for a proportion can be calculated using the z-test formula:

\[z = \frac{(\hat{p} - p)}{\sqrt{\frac{p(1-p)}{n}}}\]

Where:

- \(\hat{p}\) is the sample proportion of users who developed nausea.

- \(p\) is the hypothesized population proportion (0.20 in this case).

- \(n\) is the sample size (245 in this case).

First, calculate \(\hat{p}\), the sample proportion of users who developed nausea:

\(\hat{p} = \frac{\text{Number of users who developed nausea}}{\text{Total sample size}} = \frac{54}{245} \approx 0.2204\)

Now, calculate the test statistic \(z\):

\[z = \frac{(0.2204 - 0.20)}{\sqrt{\frac{0.20(1-0.20)}{245}}} \approx \frac{0.0204}{\sqrt{\frac{0.20(0.80)}{245}}}\]

Now, calculate the value of the square root part:

\(\sqrt{\frac{0.20(0.80)}{245}} \approx 0.0654\)

Now, calculate \(z\):

\[z \approx \frac{0.0204}{0.0654} \approx 0.3118\]

Next, you'll want to find the critical value for a one-tailed test at a significance level of 0.10. You can use a standard normal distribution table or calculator to find this value. For a significance level of 0.10, the critical z-value is approximately 1.28.

Since your calculated test statistic \(z\) (0.3118) is less than the critical value (1.28), you fail to reject the null hypothesis.

Conclusion:

At a significance level of 0.10, there is not enough evidence to support the claim that more than 20% of users develop nausea when treated with the drug.

Explanation:

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