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

1 answer

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.