Question

A study was conducted to determine the proportion of American women who work 40 hours or more per week outside of the home for a non-family employer (the study did not consider self-employed women or women working for family businesses). In the sample of 1433 women, 922 of them reported working at least 40 hours per week outside of the home.

Find and interpret a 95% confidence interval for the proportion. Take calculations to four (4) decimal places.
Choose the closest answer, and read carefully!
O The researchers can be 95% confident that the true proportion of women who work 40 hours or more per week is between 0.619 and 0.668.
O The researchers can be 95% confident that the true proportion of women who work 40 hours or more per week is between 0.619 and 0.676.
O The researchers can be 95% confident that the true proportion of women who work 40 hours or more per week is between 0.611 and 0.676.
O The researchers can be 95% confident that the true mean number of women who work 40 hours or more per week is between 0.619 and 0.668.

Answer 1

To calculate the 95% confidence interval for the proportion of women working 40 hours or more per week, we found the sample proportion, calculated the standard error, and then determined the margin of error using a Z-score. The interval is between 0.6177 and 0.6683, which suggests the researchers can be 95% confident the true proportion lies within this range. The correct statement is O The researchers can be 95% confident that the true mean number of women who work 40 hours or more per week is between 0.619 and 0.668.

Step-by-step explanation:

The question involves finding a 95% confidence interval for the proportion of American women who work 40 hours or more per week outside of the home for a non-family employer. To calculate this, we use the sample proportion (p) and the standard error (SE) of the proportion.

First, we find the sample proportion by dividing the number of women working at least 40 hours by the total number of women sampled:

p = 922 / 1433 = 0.643 (rounded to four decimal places)

Next, we calculate the standard error using the formula SE = sqrt[(p(1-p))/n], where n is the sample size:

SE = sqrt[(0.643(1-0.643))/1433] = 0.0129 (rounded to four decimal places)

The Z-score for a 95% confidence interval is approximately 1.96. We calculate the margin of error (ME) as:

ME = Z * SE = 1.96 * 0.0129 = 0.0253 (rounded to four decimal places)

To find the confidence interval, we add and subtract the margin of error from the sample proportion:

Lower limit = p - ME = 0.643 - 0.0253 = 0.6177

Upper limit = p + ME = 0.643 + 0.0253 = 0.6683

Therefore, the 95% confidence interval is (0.6177, 0.6683), and the researchers can be 95% confident that the true proportion of women who work 40 hours or more per week for a non-family employer is between 0.6177 and 0.6683. Comparing this to the given choices, the closest answer is that the true proportion of women who work 40 hours or more per week is between 0.619 and 0.668.