Question

Tests Involving Means and Proportions

Problem:
A sample survey of 500 students, 300 from the first year level and 200 from the second year level, showed that 56% and 48%, respectively, were in favor of using Google Classroom for the delivery of online classes. At a level of significance of 0.05, test the hypothesis that
(a) there is a difference between the two year levels
(b) Google Classroom is preferred by the first year students
(c) Find the P values in tests (a) and (b)

Answer 1

(a) The null hypothesis $(H_0)$ for the test is that the proportions are the same.

The alternate hypothesis $(H_1)$ is that the proportions are not the same.

$H_0:p_1=p_2$

$H_1:p_1\not=p_2$

Since the null hypothesis states that $p_1=p_2,$ we use a pooled sample proportion $p$ to compute the standard error of the sampling distribution.

$$p={p_1n_1+p_2n_2\over n_1+n_2}$$

Given $n_1=300, n_2=200, p_1=0.56, p_2=0.48$

$$\hat{p}={0.56(300)+0.48(200)\over 300+200}=0.528$$

The test statistic is a z-score $(z)$ defined by the following equation.

$$z={(\hat{p_1}-\hat{p_2})-0\over \sqrt{\hat{p}(1-\hat{p})({1\over n_1}+{1\over n_2})}}$$

$$z={(0.56-0.48)-0\over \sqrt{0.528(1-0.528)({1\over 300}+{1\over 200})}}\approx1.755467$$

Use p-value

$$p=2P(Z>|z|)=2P(Z>1.755467)\approx0.079180$$

Since p-value $p=0.079180>0.05=\alpha$ we fail to reject the null hypothesis.

 The result is statistically nonsignificant.

(b)

$H_0:p_1\leq p_2$

$H_1:p_1>p_2$

$$\hat{p}={0.56(300)+0.48(200)\over 300+200}=0.528$$

$$z={(0.56-0.48)-0\over \sqrt{0.528(1-0.528)({1\over 300}+{1\over 200})}}\approx1.755467$$

Use p-value

$$p=P(Z>|z|)=P(Z>1.755467)\approx0.039590$$

Since p-value $p=0.039590<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that Google Classroom is preferred by the first year students​, at the 0.05 significance level.