# Regarding a confidence interval for a population mean difference. All other things equal, the...

## Question:

Regarding a confidence interval for a population mean difference. All other things equal, the confidence interval will be narrower if the two sample sizes are smaller. Is this statement correct; or incorrect?

## Confidence interval

The confidence interval is the interval estimation where the confidence level provides the chances that the confidence interval contains the population parameter. The most common confidence level is 90%, 95% and 99%. As the confidence level increases the length of the confidence interval is also increases.

The formula of 90% confidence interval for the difference between two population when the population variance is known is given below.

{eq}P\left( {\left( {{{\bar X}_2} - {{\bar X}_1}} \right) - {Z_{\alpha /2}}\sqrt {\dfrac{{\sigma _1^2}}{{{n_1}}} + \dfrac{{\sigma _2^2}}{{{n_2}}}} < {\mu _1} - {\mu _2} < \left( {{{\bar X}_2} - {{\bar X}_1}} \right) + {Z_{\alpha /2}}\sqrt {\dfrac{{\sigma _1^2}}{{{n_1}}} + \dfrac{{\sigma _2^2}}{{{n_2}}}} } \right) = 0.90{/eq}

As the sample sizes are get smaller than the value of margin of error is increased, and it will make the confidence interval wider.

Therefore, the statement is incorrect.