Consider the hypothesis test
Null Hypothesis: H0: p1 – p2 ≤ 0
Alternative Hypothesis: Ha: p1 – p2 > 0
The following results are for independent samples taken from the two populations:
Sample 1: n1 = 200, p1 = 0.22
Sample 2: n2 = 300, p2 = 0.16
(a) Calculate the test statistic. Round your answer to two decimal places.
(b) What is the p-value? Round your answer to four decimal places.
(c) With a significance level of alpha = 0.05, what is your hypothesis testing conclusion?
- Reject the null hypothesis: There is sufficient evidence to conclude that p1 – p2 > 0.
- Reject the null hypothesis: There is insufficient evidence to conclude that p1 – p2 > 0.
- Do not reject the null hypothesis: There is insufficient evidence to conclude that p1 – p2 > 0.
- Do not reject the null hypothesis: There is sufficient evidence to conclude that p1 – p2 > 0.
Answer:
Given Data:
For Sample 1:
Sample proportion $(\bar{p_1}) = 0.22$
Sample size $(n_1) = 200$
For Sample 2:
Sample proportion $(\bar{p_2}) = 0.16$
Sample size $(n_2) = 300$
Level of Significance $(\alpha) = 0.05 $
Solution:
The null and alternative hypothesis:
$H_0: p_1 – p_2 \leq 0$
$H_a: p_1 – p_2 > 0$
(A) The test statistic $(z):$
$z = \frac{\bar{p_1} – \bar{p_2}}{\sqrt{\frac{\bar{p_1}(1 – \bar{p_1})}{n_1} + \frac{\bar{p_2}(1 – \bar{p_2})}{n_2}}}$
$z = \frac{0.22 – 0.16}{\sqrt{\frac{0.22(1 – 0.22)}{200} + \frac{0.16(1 – 0.16)}{300}}}$
$z = 1.66027$
$\approx 1.66$
The p-value:
$p$-value $= P(z > 1.66)$
$ = 0.0485 $ [ use adbhutah p-value calculator ]
(b) Decision Rule:
Reject the null hypothesis if p-value is less than the significance level.
Therefore, here the p-value is less than the significance level therefore we reject the null hypothesis.
Conclusion:
Reject $H_0$. There is sufficient evidence to conclude that $p_1-p_2 \leq 0 $
