The federal government would like to estimate the proportion of families with no savings. A pilot sample of 60 families found that 18 had no savings. The total sample size needed to construct a 95% confidence interval for the proportion of families with no savings with a margin of error equal to 6% is_________ .

Question

Quizplus · Accepted Answer

The sample size needed to construct a 95% confidence interval for the proportion of families with no savings with a margin of error equal to 6% is 397. This can be calculated using the formula: n = (Z^2 * p * q) / e^2, where Z is the z-score for the desired confidence level, p is the estimated proportion of families with no savings, q is the complement of p, and e is the margin of error. In this case, Z = 1.96, p = 18/60 = 0.3, q = 1 - p = 0.7, and e = 0.06. Plugging these values into the formula, we get n = (1.96^2 * 0.3 * 0.7) / 0.06^2 = 396.64. Rounding up to the nearest whole number, we get a sample size of 397.

The Federal Government Would Like to Estimate the Proportion of Families