3.1 - Two-Sample Pooled t-Interval

If \(X_1,X_2,\ldots,X_n\sim N(\mu_X,\sigma^2)\) and \(Y_1,Y_2,\ldots,Y_m\sim N(\mu_Y,\sigma^2)\) are independent random samples, then a \((1-\alpha)100\%\) confidence interval for \(\mu_X-\mu_Y\), the difference in the population means is:

\((\bar{X}-\bar{Y})\pm (t_{\alpha/2,n+m-2}) S_p \sqrt{\dfrac{1}{n}+\dfrac{1}{m}}\)

where \(S_p^2\), the "pooled sample variance":

\(S_p^2=\dfrac{(n-1)S^2_X+(m-1)S^2_Y}{n+m-2}\)

is an unbiased estimator of the common variance \(\sigma^2\).

Proof

We'll start with the punch line first. If it is known that:

\(T=\dfrac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}}} \sim t_{n+m-2}\)

then the proof is a bit on the trivial side, because we then know that:

\(P\left[-t_{\alpha/2,n+m-2} \leq \dfrac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}}} \leq t_{\alpha/2,n+m-2}\right]=1-\alpha\)

And then, it is just a matter of manipulating the inequalities inside the parentheses. First, multiplying through the inequality by the quantity in the denominator, we get:

\(-t_{\alpha/2,n+m-2}\times S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}} \leq (\bar{X}-\bar{Y})-(\mu_X-\mu_Y)\leq t_{\alpha/2,n+m-2}\times S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}}\)

Then, subtracting through the inequality by the difference in the sample means, we get:

\(-(\bar{X}-\bar{Y})-t_{\alpha/2,n+m-2}\times S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}} \leq -(\mu_X-\mu_Y) \leq -(\bar{X}-\bar{Y})+t_{\alpha/2,n+m-2}\times S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}} \)

And, finally, dividing through the inequality by −1, and thereby changing the direction of the inequality signs, we get:

\((\bar{X}-\bar{Y})-t_{\alpha/2,n+m-2}\times S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}} \leq \mu_X-\mu_Y \leq (\bar{X}-\bar{Y})+t_{\alpha/2,n+m-2}\times S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}} \)

That is, we get the claimed \((1-\alpha)100\%\) confidence interval for the difference in the population means:

\((\bar{X}-\bar{Y})\pm (t_{\alpha/2,n+m-2}) S_p \sqrt{\dfrac{1}{n}+\dfrac{1}{m}}\)

Now, it's just a matter of going back and proving that first distributional result, namely that:

\(T=\dfrac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}}} \sim t_{n+m-2}\)

Well, by the assumed normality of the \(X_i\) and \(Y_i\) measurements, we know that the means of each of the samples are also normally distributed. That is:

\(\bar{X}\sim N \left(\mu_X,\dfrac{\sigma^2}{n}\right)\) and \(\bar{Y}\sim N \left(\mu_Y,\dfrac{\sigma^2}{m}\right)\)

Then, the independence of the two samples implies that the difference in the two sample means is normally distributed with the mean equaling the difference in the two population means and the variance equaling the sum of the two variances. That is:

\(\bar{X}-\bar{Y} \sim N\left(\mu_X-\mu_Y,\dfrac{\sigma^2}{n}+\dfrac{\sigma^2}{m}\right)\)

Now, we can standardize the difference in the two sample means to get:

\(Z=\dfrac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{\sqrt{\dfrac{\sigma^2}{n}+\dfrac{\sigma^2}{m}}} \sim N(0,1)\)

Now, the normality of the \(X_i\) and \(Y_i\) measurements also implies that:

\(\dfrac{(n-1)S^2_X}{\sigma^2}\sim \chi^2_{n-1}\) and \(\dfrac{(m-1)S^2_Y}{\sigma^2}\sim \chi^2_{m-1}\)

And, the independence of the two samples implies that when we add those two chi-square random variables, we get another chi-square random variable with the degrees of freedom (\(n-1\) and \(m-1\)) added. That is:

\(U=\dfrac{(n-1)S^2_X}{\sigma^2}+\dfrac{(m-1)S^2_Y}{\sigma^2}\sim \chi^2_{n+m-2}\)

Now, it's just a matter of using the definition of a \(T\)-random variable:

\(T=\dfrac{Z}{\sqrt{U/(n+m-2)}}\)

Substituting in the values we defined above for \(Z\) and \(U\), we get:

\(T=\dfrac{\dfrac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{\sqrt{\dfrac{\sigma^2}{n}+\dfrac{\sigma^2}{m}}}}{\sqrt{\left[\dfrac{(n-1)S^2_X}{\sigma^2}+\dfrac{(m-1)S^2_Y}{\sigma^2}\right]/(n+m-2)}}\)

Pulling out a factor of \(\frac{1}{\sigma}\) in both the numerator and denominator, we get:

\(T=\dfrac{\dfrac{1}{\sigma} \dfrac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{\sqrt{\dfrac{1}{n}+\dfrac{1}{m}}}}{\dfrac{1}{\sigma} \sqrt{\dfrac{(n-1)S^2_X+(m-1)S^2_Y}{(n+m-2)}}}\)

And, canceling out the \(\frac{1}{\sigma}\)'s and recognizing that the denominator is the pooled standard deviation, \(S_p\), we get:

\(T=\dfrac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}}}\)

That is, we have shown that:

\(T=\dfrac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{S_p\sqrt{\dfrac{1}{n}+\dfrac{1}{m}}}\sim t_{n+m-2}\)

And we are done.... our proof is complete!