Consider the hypothesis test below. $H_0:p_1–p_2\le 0$ $H_a:p_1–p_2>0$

The following results are for independent samples taken from the two populations.

$\text{Sample 1}{n}_1=100{\bar{p}}_1=0.28$

$\text{Sample 2}{n}_2=300{\bar{p}}_2=0.16$

Use pooled estimator of p.

a) what is the p-value?

b) with $\alpha =0.05$ what is your hypothesis testing conclusion?

Answer:

The null and alternative hypothesis,

$$H_0:p_1–p_2\le 0$$ $$H_a:p_1–p_2>0$$

The singnificance level, $𝛼=0.05$

Sample 1:

The sample proportion, ${\bar{p}}_1=0.28$

The sample size, $n_1=100$

Sample 2:

The sample proportion, ${\bar{p}}_2=0.16$

The sample size, $n_2=300$

The pooled estimator of the population proportion can be calculated as,

$$\therefore \bar{p}=\frac{{\bar{p}}_1{n}_1+{\bar{p}}_2{n}_2}{n_1+n_2}$$ $$=\frac{(0.28\times 100)+(0.16\times 300)}{100+300}$$ $$=\frac{28+48}{400}$$ $$\therefore \bar{p}=0.19$$

(a) The test statistic can be calcualted as,

$$\therefore z=\frac{{\bar{p}}_1–{\bar{p}}_2}{\sqrt{\bar{p}(1–\bar{p})\times (\frac{1}{n_1}+\frac{1}{n_2})}}$$ $$=\frac{0.28–0.16}{\sqrt{0.19(1–0.19)(\frac{1}{100}+\frac{1}{300})}}$$ $$=\frac{0.12}{0.0453}$$ $$\therefore z\approx 2.649$$

The p-value can be obtained as,

$$p\text{-value}=P(z>2.649)=1–P(z<2.649)=1–0.99596348\approx 0.0040$$

(b)

Decision rule: Reject the null hypothesis, if $p\text{-value}<\alpha$

As the p-value is less than the significance level of 0.05, we reject the null hypothesis.

Decision: Reject the null hypothesis

Conclusion: With $\alpha =0.05$, the hypothesis test conclude the difference between the population proportions is greater than 0