Question

Suppose you want to estimate the difference between two population means correct to within 1.8 with a 95% confidence interval. If prior information suggests that the population variances are approximately equal to ${{\sigma }_1}^2={{\sigma }_2}^2=14$ and you want to select independent random samples of equal size from the populations, how large should the sample sizes $n_1$, and $n_2$, be?

Short answer

The sample sizes andshould be at least 34.

Step-by-step solution

Given information

The sampling error is$SE=1.8$

The population variances are${{\sigma }_1}^2={{\sigma }_2}^2=14$.

Using the standard normal table, the critical value at a 95% confidence level is$Z_{\frac{\alpha }{2}}=1.96$

Sample size

The sample sizes are a crucial part of any statistical experiment. Considering a sufficient sample size per population is crucial to getting significant results. It can be calculated using the sampling error, population variances, and confidence level.

Calculating the sample size

Assume

$n_1=n_2=n$

The required sample size is

$\begin{array}{rcl}n& =& \frac{{\left(Z_{\frac{\alpha }{2}}\right)}^2\left[{{\sigma }_1}^2+{{\sigma }_2}^2\right]}{{\left(SE\right)}^2}\\ & =& \frac{{\left(1.96\right)}^2\left[14+14\right]}{{\left(1.8\right)}^2}\\ & =& \frac{{\left(1.96\right)}^2\left(28\right)}{{\left(1.8\right)}^2}\\ & =& 33.2\\ & \approx & 34\end{array}$

Hence the required sample size is $n_1=n_2=34$.