SOLUTION: Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 6083 patients treated with this drug, 159 developed the adverse react
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Question 1168836: Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 6083 patients treated with this drug, 159 developed the adverse reaction of nausea. Use a 0.10 significance level to test the claim that 3% of users develop nausea. Does nausea appear to be a problematic adverse reaction?
Need to find hypothesis test
p-value
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Absolutely, let's analyze this hypothesis test.
**1. Define the Hypotheses**
* **Null Hypothesis (H₀):** The proportion of users who develop nausea is 3% (p = 0.03).
* **Alternative Hypothesis (H₁):** The proportion of users who develop nausea is not 3% (p ≠ 0.03).
This is a two-tailed test.
**2. Set the Significance Level**
* α = 0.10
**3. Calculate the Sample Proportion**
* Sample size (n) = 6083
* Number of patients with nausea (x) = 159
* Sample proportion (p̂) = x/n = 159/6083 ≈ 0.02614
**4. Calculate the Test Statistic**
We'll use the z-test for proportions:
$$ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} $$
Where:
* p̂ = sample proportion (0.02614)
* p₀ = hypothesized proportion (0.03)
* n = sample size (6083)
$$ z = \frac{0.02614 - 0.03}{\sqrt{\frac{0.03(1-0.03)}{6083}}} $$
$$ z = \frac{-0.00386}{\sqrt{\frac{0.0291}{6083}}} $$
$$ z = \frac{-0.00386}{\sqrt{0.00000478382}} $$
$$ z = \frac{-0.00386}{0.002187195} $$
$$ z \approx -1.7648 $$
**5. Calculate the P-value**
Since this is a two-tailed test, we need to find the area in both tails beyond z = -1.7648 and z = 1.7648.
* P(Z < -1.7648) ≈ 0.0388
* P(Z > 1.7648) ≈ 0.0388
* P-value = 2 * 0.0388 ≈ 0.0776
**6. Make a Decision**
* Compare the p-value (0.0776) with the significance level (0.10).
* Since 0.0776 < 0.10, we reject the null hypothesis.
**7. Conclusion**
There is sufficient evidence at the 0.10 significance level to reject the claim that 3% of users develop nausea.
**Does nausea appear to be a problematic adverse reaction?**
The sample proportion of nausea (approximately 2.61%) is statistically different from the claimed 3%. Although the difference is small in magnitude, it is statistically significant at the 0.10 level. Whether it is "problematic" depends on the severity of the nausea and the benefits of the drug. A 0.39% difference in nausea may not be clinically significant. However, it is important to report the results of the clinical trials accurately.
**Summary**
* **Hypothesis test:** Two-tailed z-test for proportions.
* **p-value:** approximately 0.0776.
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