Solve for Sample Size - Prob/stats

Question

You work for a marketing firm that has a large client in the automobile industry. You have been asked to estimate the proportion of households in Chicago that have two or more vehicles. You have been assigned to gather a random sample that could be used to estimate this proportion to within a 0.02 margin of error at a 99% level of confidence.



a) With no prior research, what sample size should you gather in order to obtain a 0.02 margin of error? Round your answer up to the nearest whole number.

n = ____________ households

b) Your firm has decided that your plan is too expensive, and they wish to reduce the sample size required. You conduct a small preliminary sample, and you obtain a sample proportion of p^=0.24 . Using this new information. what sample size should you gather in order to obtain a 0.02 margin of error? Round your answer up to the nearest whole number.

n = ___________ households

Answer 1

To determine the sample size needed to estimate the proportion of households in Chicago with two or more vehicles, we can use the formula:

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

Where:

n = sample size

Z = Z-score corresponding to the desired confidence level (for a 99% confidence level, Z ≈ 2.576)

p = estimated proportion (unknown in this case, but we can assume p = 0.5 for maximum sample size)

E = margin of error (0.02)

Given:

Z ≈ 2.576

p = 0.5 (maximum sample size assumption)

E = 0.02

Using the formula, we can calculate the sample size:

n = (2.576^2 * 0.5 * (1-0.5)) / 0.02^2

n = 4147.36

Round up to 4148 households

Part 2:

Use the same formula:

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

Where:

n = sample size

Z = Z-score corresponding to the desired confidence level (for a 99% confidence level, Z ≈ 2.576)

p = estimated proportion

E = margin of error (0.02)

Given:

Z ≈ 2.576

p = 0.24

E = 0.02

Using the formula, we can calculate the sample size:

n = (2.576^2 * 0.24 * (1-0.24)) / 0.02^2

n = 3025.913856

Round up to 3026 households