To find the P-value for the hypothesis test regarding the proportion of children suffering from asthma, we need to follow several steps:
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Define the Hypotheses:
- Null hypothesis (\(H_0\)): \(p = 0.11\) (the proportion of children with asthma is equal to 11%)
- Alternative hypothesis (\(H_a\)): \(p \neq 0.11\) (the proportion of children with asthma is not equal to 11%)
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Collect the Data:
- Sample size (\(n\)) = 88
- Number of children with asthma (\(X\)) = 8
- Sample proportion (\(\hat{p}\)) = \(X/n = 8/88 = 0.0909\)
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Calculate the Test Statistic: The test statistic for a proportion is calculated using the formula:
\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]
where \(p_0 = 0.11\) (the hypothesized population proportion).
- \(\hat{p} = 0.0909\)
- \(p_0 = 0.11\)
- \(n = 88\)
First, calculate the standard error (SE):
\[ \text{SE} = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.11(1 - 0.11)}{88}} = \sqrt{\frac{0.11 \times 0.89}{88}} \approx \sqrt{\frac{0.0979}{88}} \approx \sqrt{0.0011136} \approx 0.03334 \]
Now, substituting into the z-formula:
\[ z = \frac{0.0909 - 0.11}{0.03334} \approx \frac{-0.0191}{0.03334} \approx -0.571 \]
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Calculate the P-value: Since this is a two-tailed test, we will find the P-value for both tails of the normal distribution. We can find the P-value using the standard normal distribution (z-table) or a statistical calculator.
The calculated \(z\) value is approximately \(-0.571\).
Using a standard normal distribution calculator or table, find the P-value associated with \(z = -0.571\).
- The area to the left of \(z = -0.571\) is approximately \(0.2843\).
- Because it's a two-tailed test, we multiply this area by 2:
\[ \text{P-value} = 2 \times 0.2843 \approx 0.5686 \]
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Conclusion: The P-value for the hypothesis test is 0.5686. Thus, the answer is:
B) 0.5686