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

General
1

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 those 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?

2

What is the 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?

Answer:

To find the 99 percent confidence interval for the difference between the proportions of families that reported the wife as the primary wage earner in 1960 and 2011, we can follow these steps:

  1. Calculate the sample proportions for each year:

    • In 1960, 5.8% of 1,018 families reported the wife as the primary wage earner. This corresponds to 0.058.
    • In 2011, 22.3% of 1,013 families reported the wife as the primary wage earner. This corresponds to 0.223.

  2. Calculate the standard errors for each year:

    • The standard error for 1960 can be calculated as \sqrt{\frac{0.058 \times (1-0.058)}{1,018}}.
    • The standard error for 2011 can be calculated as \sqrt{\frac{0.223 \times (1-0.223)}{1,013}}.

  3. Calculate the standard error of the difference:

    • The standard error of the difference can be calculated as \sqrt{SE_{1960}^2 + SE_{2011}^2}.

  4. Find the margin of error:

    • With a 99 percent confidence level, the critical z-value is approximately 2.576.
    • The margin of error is calculated as 2.576 \times \text{standard error of the difference}.

  5. Calculate the confidence interval:

    • The confidence interval can be calculated by adding and subtracting the margin of error from the difference in sample proportions from 1960 to 2011.

By following these steps, you can determine the 99 percent confidence interval for the difference between the proportions of families where the wife was reported as the primary wage earner in 1960 and 2011.