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 will be using the formula for sample size calculation in estimating population proportions. The formula is:

n = (Z^2 * p * (1-p)) / (E^2)

Where:
- n is the required sample size
- Z is the z-value corresponding to the desired confidence level
- p is the estimated proportion or the sample proportion (0.29 in this case)
- 1-p is the complement of the estimated proportion
- E is the desired margin of error

a. To determine the required sample size to be 90% confident that the estimate is within 0.05 of the true proportion, we need to find the appropriate value of Z for the given confidence level.

To find the Z-value, we can use a standard normal distribution table or a calculator. For a 90% confidence level, the Z-value is approximately 1.645.

Substituting the values into the formula:

n = (1.645^2 * 0.29 * (1 - 0.29)) / (0.05^2)
n = (2.705 * 0.29 * 0.71) / 0.0025
n = 0.55 / 0.0025
n = 220

Therefore, you would need to take a sample size of at least 220 women over age 55 to be 90% confident that the estimate is within 0.05 of the true proportion.

b. If no estimate of the sample proportion is available, we can simply use a conservative estimate of 0.5 (maximum variability). This assumption ensures that the sample size will be large enough to provide a reasonable margin of error.

Substituting the values into the formula:

n = (1.645^2 * 0.5 * (1 - 0.5)) / (0.05^2)
n = (2.705 * 0.5 * 0.5) / 0.0025
n = 0.675 / 0.0025
n = 270

Therefore, without any estimate of the sample proportion, you would need to take a sample size of at least 270 women over age 55 to achieve a 90% confidence level with a margin of error of 0.05.