In a prior sample of U.S. adults, the Center for Disease Control (CDC), found that 10% of the people in this sample had pinworm but the margin of error for the population estimate was too large. They want an estimate that is in error by no more than 1.5 percentage points at the 95% confidence level. Enter your answers as whole numbers.

Question

In a prior sample of U.S. adults, the Center for Disease Control (CDC), found that 10% of the people in this sample had pinworm but the margin of error for the population estimate was too large. They want an estimate that is in error by no more than 1.5 percentage points at the 95% confidence level. Enter your answers as whole numbers.

(a) What is the minimum sample size required to obtain this type of accuracy? Use the prior sample proportion in your calculation.
The minimum sample size is U.S. adults.

(b) What is the minimum sample size required to obtain this type of accuracy when you assume no prior knowledge of the sample proportion?
The minimum sample size is U.S. adults.

Answer 1

(a) To calculate the minimum sample size required to obtain a margin of error of 1.5 percentage points at the 95% confidence level, we need to use the formula:

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

Where:
n = sample size
Z = Z-score corresponding to the desired confidence level (95% confidence level corresponds to a Z-score of 1.96)
p = estimated proportion from the prior sample (10% or 0.10)
E = margin of error (1.5% or 0.015)

Using these values, we can calculate the minimum sample size:

n = (1.96^2 * 0.10 * (1-0.10)) / (0.015^2)
n = (3.8416 * 0.09) / 0.000225
n = 0.345744 / 0.000225
n = 1537.32

Rounded up to the nearest whole number, the minimum sample size required to obtain a margin of error of 1.5 percentage points at the 95% confidence level is 1538 U.S. adults.

(b) When assuming no prior knowledge of the sample proportion, we use a more conservative estimation and assume a 50% proportion (0.50) to calculate the minimum sample size. The formula remains the same:

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

Using the same values for Z and E as before:

n = (1.96^2 * 0.50 * (1-0.50)) / (0.015^2)
n = (3.8416 * 0.25) / 0.000225
n = 0.9604 / 0.000225
n = 4268.444

Rounded up to the nearest whole number, the minimum sample size required to obtain a margin of error of 1.5 percentage points at the 95% confidence level, assuming no prior knowledge of the sample proportion, is 4269 U.S. adults.