Question

The null and alternate hypotheses are: He: Th = 12 Hg: Th Th A sample of 200 observations from the first population indicated that X1 is 170. A sample of 150 observations from the second population revealed x2 to be 110. Use the 0.05 significance level to test the hypothesis. a. State the decision rule. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.) The decision rule is to reject HO if z b. Compute the pooled proportion. (Do not round the intermediate values. Round your answer to 2 decimal places.) The pooled proportion

Answer 1

Since , the null and alternatine hypothesis is ,

$H_0:\pi_1=\pi_2$

$H_1:\pi_1\neq \pi_2$

The test is two-tailed test.

Now , the estimate of the sample proportion is ,

$\hat{p_1}=\frac{X_1}{n_1}=\frac{170}{200}=0.85$

$\hat{p_2}=\frac{X_2}{n_2}=\frac{110}{150}=0.7333$

a. The critical values are , $Z_{\alpha/2}=Z_{0.05/2}=\pm 1.96$ ; From Z-table

The decision rule is ,

Reject Ho , if Z>1.96 or Z<-1.96

i.e. The decision rule is to be reject Ho if Z does not contain the (-1.96,1.96)

b. The pooled proportion is ,

$\hat{P}=\frac{X_1+X_2}{n_1+n_2}=\frac{170+110}{200+150}=0.80$

c. The value of the test statistic is ,

$Z=\frac{\hat{p_1}-\hat{p_2}}{\sqrt{\hat{P}(1-\hat{P})(\frac{1}{n_1}+\frac{1}{n_2})}}=\frac{0.85-0.7333}{\sqrt{0.80(1-0.80)(\frac{1}{200}+\frac{1}{150})}}=2.70$

d. Here , Z=2.70>1.96

Therefore , reject the null hypothesis.