What sample size is needed to find the percent of the customers at a store that are married, if 90% confidence is desired to be within a 5% margin of error?

To determine the sample size needed to find the percentage of married customers at a store with a desired confidence level and margin of error, you can use the formula:

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

Where:
n = sample size needed
Z = z-score for the desired confidence level
p = estimated proportion of married customers
E = desired margin of error as a decimal

First, let's calculate the z-score for a 90% confidence level. You can find this value from a standard normal distribution table or by using a calculator. For a 90% confidence level, the z-score is approximately 1.645.

Next, estimate the proportion of married customers. If there is no prior information, you can assume a proportion of 0.5, which represents an equal chance of being married or not married.

Now, substitute these values into the formula:

n = (1.645^2 * 0.5 * (1-0.5)) / 0.05^2

Simplifying:

n = (2.706025 * 0.25) / 0.0025

n = 10.7651 / 0.0025

n ≈ 4,306.04

Therefore, you would need a sample size of at least 4,307 customers to estimate the percentage of married customers at the store with a 90% confidence level and within a 5% margin of error. Note that usually sample sizes are rounded up to the nearest whole number, so in this case, you would need a sample size of at least 4,307 customers.