Suppose you have two populations: Population Along dashAll students at a state college (Nequals21,000) and Population Blong dashAll residents of a town in the state (Nequals21,000). You want to estimate the mean age of each population using two separate samples each of size nequals75. If you construct a 95% confidence interval for each population mean, will the margin of error for population A be larger, the same, or smaller than the margin of error for population B? Justify your reasoning.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

[tex]X \sim N(\mu, \sigma)[/tex]

We know that the margin of error for a confidence interval is given by:

[tex]Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

If we see the formula (1) at the same confidence level, for example 95% and with the same sample size the margin of error just depends of the deviation. If the population deviation for population A is higher than the population deviation for B then A will have more margin of error than B.

On the other case if the deviation for population A is lower than the deviation for population B, then we will have less margin of error for population A than population B.

And the other possible case if both population have the same deviation, then both have equal margin of error.