In a random sample of males, it was found that 20 write with their left hands and 212 do not. In a random sample of fernales, it was found that 65 write with their left hands and 436 do not. Use a 0.01 significance level to test the claim that the rate of left-handedness among males is less than that among females. Complete parts (a) through (c) below. \[ \text { P-value }=.044 \] (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P-value is greater than the significance level of $\alpha=0.01$, so fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the rate of left-handedness among males is less than that among fernales. b. Test the claim by constructing an appropriate confidence interval. The $98 \%$ confidence interval is $\square<\left(p_{1}-p_{2}\right)<\square$
Solution
Step 1 ：Calculate the proportions of left-handedness for males and females. For males, \(p_1 = \frac{20}{20 + 212} = 0.086\). For females, \(p_2 = \frac{65}{65 + 436} = 0.130\).
Step 2 ：Calculate the sample sizes for males and females. For males, \(n_1 = 20 + 212 = 232\). For females, \(n_2 = 65 + 436 = 501\).
Step 3 ：Calculate the z-score for a 98% confidence level, which is \(z = 2.33\).
Step 4 ：Calculate the standard error using the formula \(\sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}}\), which gives \(se = 0.024\).
Step 5 ：Calculate the confidence interval using the formula \(p_1 - p_2 \pm z \times se\). The lower limit of the confidence interval is \(-0.099\) and the upper limit is \(0.012\).
Step 6 ：The final answer is: The 98% confidence interval for the difference in proportions of left-handedness between males and females is \(\boxed{-0.099 < (p_{1}-p_{2}) < 0.012}\).
