In 1960 sociologists studied a random sample of 1,018 families that consisted of a husband, a wife, and at least one child. Of those families, 5.8 percent reported that the wife was the primary wage earner of the family. In 2011 the study was replicated with a random sample of 1,013 families that consisted of a husband, a wife, and at least one child. Of these families, 22.3 percent reported that the wife was the primary wage earner of the family. Which of the following represents a 99 percent confidence interval for the difference between the proportions of families that consisted of a husband, a wife, and at least one child from 1960 to 2011 that would have reported the wife as the primary wage earner?

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We are looking for the confidence interval corresponding to the difference between two population proportions.

$$ \text{Confidence interval} : \text{statistic} \pm (\text{critical value})(\text{standard error of statistic}) $$

The statistic is the population difference:

$$ \text{statistic} = \hat{p}_{2011}-\hat{p}_{1960} = 0.223-0.058$$

The critical value can be found in the table. A 99 percent confidence corresponds to a 0.005 tail probability or 2.576 critical value.

The standard error of statistic for the difference between two population proportions:

$$ s_{\hat{p}_{2011}-\hat{p}_{1960}} = \sqrt{ \frac{\hat{p}_{2011}(1-\hat{p}_{2011}) }{n_{2011}} + \frac{\hat{p}_{1960}(1-\hat{p}_{1960})}{n_{1960}}} $$

Putting it all together:

$$ \text{Confidence interval} : \text{statistic} \pm (\text{critical value})(\text{standard error of statistic}) $$ $$ (0.223-0.058) \pm 2.576\sqrt{\frac{(0.223)(0.777)}{1{,}013} + \frac{(0.058)(0.942)}{1{,}018}} $$