Hypothesis Test for the Difference in Two Proportions Test the claim that the proportion of men who own cats is significantly different tha women who own cats at the 0.05 significance level. The null and alternative hypothesis would be: The test is: right-tailed two-tailed left-tailed \( \sigma^{\varangle} \checkmark \) Based on a sample of 20 men, \( 45 \% \) owned cats Based on a sample of 80 women, \( 70 \% \) owned cats The test statistic is: (to 2 decimals) The \( \mathrm{p} \)-value is: (to 2 decimals) Racad an thic wa. Type here to search

Step 1 ：Step 1 Null Hypothesis (H0): \( p_m = p_w \) Alternative Hypothesis (H1): \( p_m \neq p_w \) This is a two-tailed test.

Step 2 ：Step 2 Calculate pooled proportion:\[p_p = \frac{x_m + x_w}{n_m + n_w} = \frac{0.45 * 20 + 0.70 * 80}{20 + 80} = \frac{9 + 56}{100} = 0.65\] Calculate Standard error:\[SE = \sqrt{\frac{p_p * (1-p_p)}{n_m} \times \frac{p_p * (1-p_p)}{n_w}} = 0.0791\]

Step 3 ：Step 3 Calculate the test statistic:\[z = \frac{(p_m - p_w) - 0}{SE} = \frac{0.45 - 0.70}{0.0791} = -3.16\] Find the p-value from z-table:\[p = 2 * (1-P(-3.16)) = 0.0016\] Since the p-value is less than 0.05, we reject the null hypothesis. Thus, the proportion of men who own cats is significantly different than women who own cats at the 0.05 significance level.