Question
Find the P-value for the indicated hypothesis test.In a sample of 88 children selected randomly from one town, it is found that 8 of them suffer from asthma. Find the P-value for a test of the claim that the proportion of all children in the town who suffer from asthma is equal to 11%.(1 point)Responses0.28430.28430.56860.56860.21570.2157–0.2843
Answers
GPT-4o mini
To find the P-value for the given hypothesis test, we need to perform a hypothesis test for a population proportion. Here's the process step-by-step:
1. **Define the null and alternative hypotheses**:
- Null hypothesis (\(H_0\)): \(p = 0.11\) (the proportion of children suffering from asthma is 11%).
- Alternative hypothesis (\(H_a\)): \(p \neq 0.11\) (the proportion of children suffering from asthma is not 11%).
2. **Calculate the sample proportion**:
\[
\hat{p} = \frac{x}{n} = \frac{8}{88} = 0.0909
\]
3. **Calculate the standard error (SE)**:
\[
SE = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.11(1 - 0.11)}{88}} = \sqrt{\frac{0.11 \times 0.89}{88}} \approx \sqrt{\frac{0.0979}{88}} \approx \sqrt{0.00111364} \approx 0.03333
\]
4. **Calculate the Z-test statistic**:
\[
Z = \frac{\hat{p} - p_0}{SE} = \frac{0.0909 - 0.11}{0.03333} \approx \frac{-0.0191}{0.03333} \approx -0.573
\]
5. **Find the P-value**:
Since this is a two-tailed test, we need to find the probability of obtaining a Z-value as extreme as -0.573 (and its corresponding positive value +0.573) in a standard normal distribution.
Using a Z-table (or calculator), we find:
- The cumulative probability corresponding to \(Z = -0.573\) is approximately 0.2830.
- Since it’s a two-tailed test, we double this probability:
\[
P\text{-value} = 2 \times (1 - 0.2830) = 2 \times 0.7170 \approx 0.4340
\]
Thus, the P-value for the hypothesis test is approximately **0.4340**. However, it seems that the responses you've provided do not contain this exact value. The closest value listed is **0.2843** which might be reflecting calculations under slightly different conditions, but this should be verified based on the setup and calculations.
If you would like to adhere strictly to just the provided responses, please note that none of the available options correctly matches our calculated P-value of approximately **0.4340**.
1. **Define the null and alternative hypotheses**:
- Null hypothesis (\(H_0\)): \(p = 0.11\) (the proportion of children suffering from asthma is 11%).
- Alternative hypothesis (\(H_a\)): \(p \neq 0.11\) (the proportion of children suffering from asthma is not 11%).
2. **Calculate the sample proportion**:
\[
\hat{p} = \frac{x}{n} = \frac{8}{88} = 0.0909
\]
3. **Calculate the standard error (SE)**:
\[
SE = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.11(1 - 0.11)}{88}} = \sqrt{\frac{0.11 \times 0.89}{88}} \approx \sqrt{\frac{0.0979}{88}} \approx \sqrt{0.00111364} \approx 0.03333
\]
4. **Calculate the Z-test statistic**:
\[
Z = \frac{\hat{p} - p_0}{SE} = \frac{0.0909 - 0.11}{0.03333} \approx \frac{-0.0191}{0.03333} \approx -0.573
\]
5. **Find the P-value**:
Since this is a two-tailed test, we need to find the probability of obtaining a Z-value as extreme as -0.573 (and its corresponding positive value +0.573) in a standard normal distribution.
Using a Z-table (or calculator), we find:
- The cumulative probability corresponding to \(Z = -0.573\) is approximately 0.2830.
- Since it’s a two-tailed test, we double this probability:
\[
P\text{-value} = 2 \times (1 - 0.2830) = 2 \times 0.7170 \approx 0.4340
\]
Thus, the P-value for the hypothesis test is approximately **0.4340**. However, it seems that the responses you've provided do not contain this exact value. The closest value listed is **0.2843** which might be reflecting calculations under slightly different conditions, but this should be verified based on the setup and calculations.
If you would like to adhere strictly to just the provided responses, please note that none of the available options correctly matches our calculated P-value of approximately **0.4340**.