Step 1 :We are given two samples, one of males and one of females. In the male sample, 22 are left-handed and 222 are not. In the female sample, 64 are left-handed and 464 are not. We are asked to test the claim that the rate of left-handedness among males is less than that among females at a 0.01 significance level.
Step 2 :To test this claim, we will perform a hypothesis test for two proportions. The null hypothesis \(H_{0}\) is that the proportions of left-handedness among males and females are equal, i.e., \(p_{1}=p_{2}\). The alternative hypothesis \(H_{1}\) is that the proportion of left-handedness among males is less than that among females, i.e., \(p_{1} Step 3 :The test statistic for this problem is a z-score, which can be calculated using the formula: \[z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}}\] where \(\hat{p}_1\) and \(\hat{p}_2\) are the sample proportions of left-handedness among males and females respectively, \(\hat{p}\) is the pooled sample proportion, and \(n_1\) and \(n_2\) are the sample sizes for males and females respectively. Step 4 :Substituting the given values into the formula, we get: \[n1 = 244, n2 = 528, \hat{p}_1 = 0.09016393442622951, \hat{p}_2 = 0.12121212121212122, \hat{p} = 0.11139896373056994\] Step 5 :Calculating the z-score, we get: \[z = -1.2748100646672456\] Step 6 :Rounding to two decimal places, the test statistic is \(\boxed{-1.27}\).