Question
Find the P-value for the indicated hypothesis test.
A medical school claims that more than 28% of its students plan to go into general practice. It is found that among a random sample of 130 of the school's students, 32% of them plan to go into general practice. Find the P-value for a test of the school's claim.
A) 0.3461
B) 0.3078
C) 0.1635
D) 0.1539
A medical school claims that more than 28% of its students plan to go into general practice. It is found that among a random sample of 130 of the school's students, 32% of them plan to go into general practice. Find the P-value for a test of the school's claim.
A) 0.3461
B) 0.3078
C) 0.1635
D) 0.1539
Answers
GPT-4o mini
To find the P-value for the hypothesis test regarding the medical school's claim that more than 28% of its students plan to go into general practice, we can follow these steps:
1. **Set up the hypotheses**:
- Null hypothesis (\(H_0\)): \(p \leq 0.28\)
- Alternative hypothesis (\(H_1\)): \(p > 0.28\)
2. **Calculate the sample proportion**:
The sample proportion \(\hat{p} = 0.32\) (which means 32% of the students plan to go into general practice).
3. **Sample size**:
The sample size \(n = 130\).
4. **Calculate the test statistic**:
The test statistic for a proportion can be calculated using the formula:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]
where:
- \(\hat{p} = 0.32\) (the sample proportion)
- \(p_0 = 0.28\) (the hypothesized proportion)
- \(n = 130\)
Plugging in the values:
\[
z = \frac{0.32 - 0.28}{\sqrt{\frac{0.28 \times (1 - 0.28)}{130}}}
\]
Calculate \( \sqrt{\frac{0.28 \times 0.72}{130}} \):
\[
= \sqrt{\frac{0.2016}{130}} = \sqrt{0.0015507692} \approx 0.0393
\]
Now, substituting this back into the z formula:
\[
z = \frac{0.04}{0.0393} \approx 1.018
\]
5. **Find the P-value**:
Since this is a one-tailed test (we are testing if the proportion is greater than 0.28), we look for the area to the right of our z value in the standard normal distribution.
Using a standard normal distribution table or calculator, we look up the z-score of 1.018:
\[
P(Z > 1.018) = 1 - P(Z < 1.018) \approx 1 - 0.8452 = 0.1548
\]
6. **Final Result**:
The P-value is approximately \(0.1539\).
Given the answer choices:
A) 0.3461
B) 0.3078
C) 0.1635
D) 0.1539
The correct answer is **D) 0.1539**.
1. **Set up the hypotheses**:
- Null hypothesis (\(H_0\)): \(p \leq 0.28\)
- Alternative hypothesis (\(H_1\)): \(p > 0.28\)
2. **Calculate the sample proportion**:
The sample proportion \(\hat{p} = 0.32\) (which means 32% of the students plan to go into general practice).
3. **Sample size**:
The sample size \(n = 130\).
4. **Calculate the test statistic**:
The test statistic for a proportion can be calculated using the formula:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]
where:
- \(\hat{p} = 0.32\) (the sample proportion)
- \(p_0 = 0.28\) (the hypothesized proportion)
- \(n = 130\)
Plugging in the values:
\[
z = \frac{0.32 - 0.28}{\sqrt{\frac{0.28 \times (1 - 0.28)}{130}}}
\]
Calculate \( \sqrt{\frac{0.28 \times 0.72}{130}} \):
\[
= \sqrt{\frac{0.2016}{130}} = \sqrt{0.0015507692} \approx 0.0393
\]
Now, substituting this back into the z formula:
\[
z = \frac{0.04}{0.0393} \approx 1.018
\]
5. **Find the P-value**:
Since this is a one-tailed test (we are testing if the proportion is greater than 0.28), we look for the area to the right of our z value in the standard normal distribution.
Using a standard normal distribution table or calculator, we look up the z-score of 1.018:
\[
P(Z > 1.018) = 1 - P(Z < 1.018) \approx 1 - 0.8452 = 0.1548
\]
6. **Final Result**:
The P-value is approximately \(0.1539\).
Given the answer choices:
A) 0.3461
B) 0.3078
C) 0.1635
D) 0.1539
The correct answer is **D) 0.1539**.