How to Solve Sampling and Limit Theorems — Sampling — variance of sample means and their difference Questions on Exam P
Sample Practice Problem
ID: #9UBJ2
Independent samples of sizes \(18\) and \(24\) are drawn from populations with respective means \(30.4\) and \(32.1\) and standard deviations \(12\) and \(14\). Let \(\overline X\) and \(\overline Y\) be their sample means. Calculate \(\Pr(\overline X>\overline Y)\) using the normal approximation.
📖 Worked Solution & Strategy
Independent sample means are approximately normal with their population means and variances divided by sample size. Their difference therefore has the difference of means and the sum of sampling variances. The difference \(\overline X-\overline Y\) has expected value \(\operatorname{E}[\overline X-\overline Y]=30.4-32.1=-1.7\). By sample independence, variances add: \(\operatorname{Var}(\overline X-\overline Y)=\frac{12^2}{18}+\frac{14^2}{24}\). The standard error is \(SE=\sqrt{\frac{144}{18}+\frac{196}{24}}=4.0208\). The event \(\overline X>\overline Y\) is equivalent to \(\overline X-\overline Y>0\). Standardizing zero gives \(z=\frac{0-(-1.7)}{SE}=0.4228\). Using the normal CDF, \(\Pr(\overline X-\overline Y>0)=1-\Phi(0.4228)=0.3362\approx0.3362\).