Question

According to a recent report, 44% of college student internships are unpaid. A recent survey of 100 college interns at a local university found that 55 had unpaid internships. a. Use the five-step p-value approach to hypothesis testing and a 0.01 level of significance to determine whether the proportion of college interns that had unpaid internships is different from 0.44. b. Assume that the study found that 61 of the 100 college interns had unpaid internships and repeat (a). Are the conclusions the same? a. Let it be the population proportion. Determine the null hypothesis, Ho, and the alternative hypothesis, H- Hy: Vo (Type integers or decimals. Do not round.)

Answer 1

a.

The null and alternative hypothesis are:

$H_0: \pi = 0.44$

$H_1: \pi \ne 0.44$

Calculations of the test:

The following information is provided: The sample size is N=100, the number of favorable cases is X=55, and the sample proportion is

$\bar p = \frac{X}{N} = \frac{ 55}{ 100} = 0.55$, and the significance level is =0.01

(1) Null and Alternative Hypotheses

The null and alternative hypothesis are:

$H_0: \pi = 0.44$

$H_1: \pi \ne 0.44$

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

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the critical value for a two-tailed test is zc​=2.58.

The rejection region for this two-tailed test is R={z:∣z∣>2.58}

(3) Test Statistics

The z-statistic is computed as follows:

$z = \frac{\bar p - \pi}{\sqrt{\pi(1-\pi)/n}} = \frac{ 0.55 - 0.44 }{\sqrt{ 0.44(1- 0.44)/100}} = 2.216$

(4) The decision about the null hypothesis

Since it is observed that ∣z∣=2.216≤zc​=2.58, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0267, and since p=0.0267≥0.01, 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 is different than $\pi$ ​, at the α=0.01 significance level.

Graphically

$normaldistributiongrapher.php?mean=0&sig$

b.

When X=61

Test Statistics

The z-statistic is computed as follows:

$z = \frac{\bar p - p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{ 0.61 - 0.44 }{\sqrt{ 0.44(1- 0.44)/100}} = 3.425$

The p-value is p = 0.0006 and we reject the null hypothesis. So the conclusions are different.

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