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.

Answer 1

P-value is P(|z| > 1.094) = 0.1369775 0.137

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SOLUTION

The sample size is N = 130, the sample proportion is 0.32, and the significance level is α = 0.05

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used.

(2) Rejection Region

The significance level is α=0.05, and the critical value for a right-tailed test is zc​ = 1.64.

The rejection region for this right-tailed test is R = { z : z > 1.64}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that z = 1.094 ≤ zc ​= 1.64, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is P(|z| > 1.094) = 0.1369775 0.137, and since p = 0.137 ≥ 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is greater than p0​, at the α=0.05 significance level.

Therefore, there is not enough evidence to claim that more than 28% of its students plan to go into general practice.