The following results are for independent samples taken from the two populations.
$\text{Sample 1} \quad n_1 = 100 \quad \overline{p}_1 = 0.28$
$\text{Sample 2} \quad n_2 = 300 \quad \overline{p}_2 = 0.16$
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,
$$H_0 : p_1 – p_2 \leq 0$$ $$\quad H_a : p_1 – p_2 > 0$$
The singnificance level, $𝛼=0.05$
Sample 1:
The sample proportion, $\overline{p}_1 = 0.28$
The sample size, $n_1 = 100$
Sample 2:
The sample proportion, $\overline{p}_2 = 0.16$
The sample size, $n_2 = 300$
The pooled estimator of the population proportion can be calculated as,
$$\therefore \overline{p} = \frac{\overline{p}_1 n_1 + \overline{p}_2 n_2}{n_1 + n_2}$$ $$= \frac{(0.28 \times 100) + (0.16 \times 300)}{100 + 300}$$ $$= \frac{28 + 48}{400}$$ $$\therefore \overline{p} = 0.19$$
(a) The test statistic can be calcualted as,
$$\therefore z = \frac{\overline{p}_1 – \overline{p}_2}{\sqrt{\overline{p}(1 – \overline{p}) \times \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$$ $$= \frac{0.28 – 0.16}{\sqrt{0.19(1 – 0.19) \left(\frac{1}{100} + \frac{1}{300}\right)}}$$ $$= \frac{0.12}{0.0453}$$ $$\therefore z \approx 2.649$$
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
