Question

5. Conduct a test at the a=0,01 level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, and (c) the P-value. Assume the samples were obtained independently from a large population using simple random sampling. Test whether P, > P2. The sample data are xy = 116, n = 251, x2 = 132, and n2 = 301 (a) Choose the correct null and alternative hypotheses below. O A. Ho: P1 = P2 versus H: PHP2 OB. Ho: P = P2 versus Hy: P1 P2 OC. Ho: P1 = 0 versus H:P 0 OD. Ho: P = P2 versus H: P1 P2 (b) Determine the test statistic, Zo (Round to two decimal places as needed.) (c) Determine the P-value The P-value is (Round to three decimal places as needed.) What is the result of this hypothesis test? O A. Do not reject the null hypothesis because there is not sufficient evidence to conclude that p, P2 O B. Do not reject the null hypothesis because there is not sufficient evidence to conclude that P, #P2 O C. Do not reject the null hypothesis because there is not sufficient evidence to conclude that P1 P2- OD. Reject the null hypothesis because there is sufficient evidence to conclude that P, P2

Answer 1

Now , the estimate of the sample proportions are ,

$\hat{p_1}=\frac{X_1}{n_1}=\frac{116}{251}=0.4622$

X2 p2 132 = 0.4385 301 12

The pooled estimate is ,

$\hat{P}=\frac{X_1+X_2}{n_1+n_2}=\frac{116+132}{251+301}=0.4493$

(a) The null and alternative hypothesis is ,

$H_0:p_1=p_2$ Vs  $H_1:p_1>p_2$

The test is one-tailed test.

(b) The test statistic is ,

$Z_{stat}=\frac{p_1-p_2}{\sqrt{\hat{P}(1-\hat{P})(\frac{1}{n_1}+\frac{1}{n_2})}}=\frac{0.4622-0.4385}{\sqrt{0.4493(1-0.4493)(\frac{1}{251}+\frac{1}{301})}}=0.56$

(c) The p-value is ,

p-value=$P(Z>|Z_{stat}|)=P(Z>0.56)=1-P(Z\leq 0.56)=1-\Phi(0.56)$

$=1-0.7123=0.288$ ; From standard normal distribution table

Decision : Here , p-value=0.288>0.01

Therefore , Do not reject the null hypothesis

Conclsuion :

A. Do not reject the null hypothesis because there is not sufficient evidence to conclude that p1>p2