Question

According to the U.S Census Bureau,

32%

of men who worked at home were college graduates. In a sample of

519

women who worked at home,

165

were college graduates.

(b) Construct a

99.8%

confidence interval for the proportion of women who work at home who are college graduates. Round the answer to at least three decimal places.

Answer 1

Answer 2

To construct a confidence interval for the proportion of women who work at home and are college graduates, we can use the sample proportion and the standard error.

Given: Sample size (n) = 519 Number of college graduates in the sample (x) = 165

First, calculate the sample proportion: Sample proportion (p̂) = x / n = 165 / 519 ≈ 0.317

Next, calculate the standard error: Standard error (SE) = √[(p̂ * (1 - p̂)) / n] = √[(0.317 * (1 - 0.317)) / 519] ≈ 0.016

To construct the confidence interval, we can use the formula: Confidence Interval = p̂ ± Z * SE

Where: p̂ is the sample proportion Z is the Z-score corresponding to the desired confidence level (99.8% confidence level corresponds to Z = 2.807) SE is the standard error

Plugging in the values, we have: Confidence Interval = 0.317 ± 2.807 * 0.016

Calculating the values: Confidence Interval = 0.317 ± 2.807 * 0.016 ≈ 0.317 ± 0.045

Rounding to at least three decimal places, the confidence interval is approximately: 0.272 < p < 0.362

Therefore, we can say with 99.8% confidence that the true proportion of women who work at home and are college graduates falls within the range of 0.272 to 0.362.