Question 1168837: Suppose 242 subjects are treated with a drug that is used to treat pain and 51 of them developed nausea. Use a 0.05 significance level to test the claim that more than 20% of users develop nausea.
Need to find hypotheis test
p-value
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this hypothesis test step-by-step.
**1. Define the Hypotheses**
* **Null Hypothesis (H₀):** The proportion of users who develop nausea is 20% or less (p ≤ 0.20).
* **Alternative Hypothesis (H₁):** The proportion of users who develop nausea is more than 20% (p > 0.20).
This is a right-tailed test.
**2. Set the Significance Level**
* α = 0.05
**3. Calculate the Sample Proportion**
* Sample size (n) = 242
* Number of subjects with nausea (x) = 51
* Sample proportion (p̂) = x/n = 51/242 ≈ 0.2107
**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 (0.2107)
* p₀ = hypothesized proportion (0.20)
* n = sample size (242)
$$ z = \frac{0.2107 - 0.20}{\sqrt{\frac{0.20(1-0.20)}{242}}} $$
$$ z = \frac{0.0107}{\sqrt{\frac{0.16}{242}}} $$
$$ z = \frac{0.0107}{\sqrt{0.000661157}} $$
$$ z = \frac{0.0107}{0.02571297} $$
$$ z \approx 0.4161 $$
**5. Calculate the P-value**
Since this is a right-tailed test, we need to find the area to the right of z = 0.4161 in the standard normal distribution.
Using a z-table or calculator, we find:
* P(Z > 0.4161) = 1 - P(Z < 0.4161)
* P(Z < 0.4161) ≈ 0.6612
* P(Z > 0.4161) ≈ 1 - 0.6612 ≈ 0.3388
Therefore, the p-value is approximately 0.3388.
**6. Make a Decision**
* Compare the p-value (0.3388) with the significance level (0.05).
* Since 0.3388 > 0.05, we fail to reject the null hypothesis.
**7. Conclusion**
There is not sufficient evidence at the 0.05 significance level to support the claim that more than 20% of users develop nausea.
**Summary**
* **Hypothesis test:** Right-tailed z-test for proportions.
* **p-value:** approximately 0.3388.
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