a recent study indicated that 29% of the 100 women over age 55 in the study were widows. a. How large a sample must you take to be 90% confident that the estimate is within 0.05 of the true proportion of women over age 55 who are widows? b. If no estimate of the sample proportion is available, how large should the sample be?

To answer these questions, we need to use the concept of confidence intervals in statistics.

a. To determine how large of a sample we need to be 90% confident that the estimate is within 0.05 of the true proportion, we can use the formula:

n = (Z^2 * p * q) / E^2

Where:
n = sample size
Z = z-score corresponding to the desired confidence level (90% confidence level corresponds to a z-score of approximately 1.645)
p = estimated proportion of success (in this case, the proportion of widows which is 29% or 0.29)
q = estimated proportion of failure (1 - p, which is 1 - 0.29 = 0.71)
E = margin of error (0.05)

Plugging in the values, we get:

n = (1.645^2 * 0.29 * 0.71) / 0.05^2
n ≈ 201.377

Since you cannot have a fraction of a person, we round up to the nearest whole number. Therefore, you would need a sample size of at least 202 women over age 55 to be 90% confident that the estimate is within 0.05 of the true proportion of widows.

b. If no estimate of the sample proportion is available, we assume a conservative proportion of success of 0.5 (equal chance of success and failure). In this case, we can use the formula above with p = 0.5.

n = (1.645^2 * 0.5 * 0.5) / 0.05^2
n ≈ 169.236

Again, rounding up to the nearest whole number, you would need a sample size of at least 170 women over age 55 if no estimate of the sample proportion is available, in order to be 90% confident that the estimate is within 0.05 of the true proportion of widows.