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: