Calculating a 95% confidence interval for the difference of two random variables

Question

Let $ x_1, ..., x_9 $ and $ y_1, ..., y_8 $ be two random samples of two populations. $ \bar x = 7 $ is the mean of the first and $ \bar y = 11 $ the mean of the second sample. The sample standard deviations are $ s_x = 2 $ and $ s_y = 3.5 $. Now I want to calculate a 95% confidence interval for the difference of the mean of the two populations. I know how to calculate a 95% confidence interval for the mean for the populations: $$ \bar x -2*\left(\frac{s_x}{\sqrt 9}\right) $$ and $$ \bar y -2*\left(\frac{s_y}{\sqrt 8}\right) $$ But don't know how to proceed from here.

Answer 1

HINT: You have shown that you know how to find a confidence interval for the mean of a population, but not for the difference of the means of two populations. However, if $X$ and $Y$ are two random variables and a large number of samples is taken, then the random variable $\bar{X}-\bar{Y}$ is approximately normally distributed, with a sample mean of $\bar{x}-\bar{y}=7-11=-4$ and a standard deviation approximately equal to $$s_{\bar{x}-\bar{y}}\approx \sqrt{\frac{s_x^2}{n_x}+\frac{s_y^2}{n_y}}$$ Can you use this to construct a confidence interval?

My only concern is that your sample sizes (of 8 and 9) are relatively small, so you may need to use the student’s t-distribution rather than the normal distribution.