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Let $X_1, X_2, \ldots, X_{n_1}$ and $Y_1, Y_2, \ldots, Y_{n_2}$ be two different independent i.i.d. samples from $N(\mu_1, \sigma^2)$ and $N(\mu_2, \sigma^2)$, respectively. Note the variances are the same. How can I find a $100(1 - \alpha)\%$ confidence interval for $4\mu_1 - 3\mu_2$?


I know the following derivation for a confidence interval for difference of means (however, the variables not necessarily have to be normal).

If we have variables $X$ and $Y$, and we define $\overline{X} = n_1^{-1} \sum_{i = 1}^{n_1} X_i$ and $\overline{Y} = n_2^{-1} \sum_{i = 1}^{n_2} Y_i$ to be rthe sample means, then $\hat{\Delta} = \overline{X} - \overline{Y}$ is an unbiased estimator for $\Delta = \mu_1 - \mu_2$. Furthermore, by the independence of the samples, we have $\text{Var}(\hat{\Delta}) = \frac{\sigma^2}{n_1} + \frac{\sigma^2}{n_2}$ assuming the variances are the same.

Now if we define $S_1^2 = (n_1 - 1)^{-1} \sum_{i = 1}^{n_1} (X_i - \overline{X})^2$ and $S_2^2 = (n_2 - 1)^{-1} \sum_{i = 1}^{n_2} (Y_i - \overline{Y})^{2}$ to be the sample variances, we can estimate the variances by the sample variances and consider the random variable $Z = (\hat{\Delta} - \Delta)/\left(\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}\right)$, which follows a $N(0, 1)$ distribution due to the Central Limit Theorem. This leads to a $(1 - \alpha)100\%$ confidence interval for $\Delta = \mu_1 - \mu_2$ given by

$$\left((\overline{x} - \overline{y} - z_{\alpha/2}\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}, (\overline{x} - \overline{y}) \right) + z_{\alpha/2}\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}},$$

where the quantity $\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$ represents the standard error of $\overline{X} - \overline{Y}$.

However, I'm struggling to extend this to find a confidence interval for $4\mu_1 - 3\mu_2$ specifically when the samples come from normal distributions with a common sample variance. Can someone please help me approach this problem? Note that I would like to actually derive the interval myself and not just use R or some other software.

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  • $\begingroup$ Do you know the Delta method? $\endgroup$ – doubled Mar 28 at 20:09
  • $\begingroup$ Yes, I am aware of the Delta method, but I'm not sure how it can be applied here? $\endgroup$ – user295786 Mar 28 at 20:20
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    $\begingroup$ You don't need the delta method for this: you can directly (and easily) compute the distribution of the estimator $4\bar X - 3\bar Y.$ $\endgroup$ – whuber Mar 28 at 20:34
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You could estimate $4\mu_1 - 3\mu_2$ by $\hat \Delta =4\bar X_1 - 3\bar X_2).$ Then $$E(\hat\Delta) = E(4\bar X_1 - 3\bar X_2) = 4\mu_1 - 3\mu_2,$$ $$Var(\hat\Delta) = Var(4\bar X_1 - 3\bar X_2) = 16\frac{\sigma_1^2}{n_1} + 9\frac{\sigma_2^2}{n_2},$$ and $$SD(\hat\Delta) = \sqrt{16\frac{\sigma_1^2}{n_1} + 9\frac{\sigma_2^2}{n_2}} = \sigma\sqrt{\frac{16}{n_1}+\frac{9}{n_2}},$$ because the two variances are the same. In a balanced design with $n = n_1 = n_2,$
$SD(\hat\Delta) = 5\sigma/\sqrt{n}.$

You might estimate $\sigma$ by $\hat\sigma_p = \sqrt{\frac{(n_1-1)S_1^2+(n_2 - 1)S_1^2}{n_1+n_2-2}},$ or in the balanced case, $\hat\sigma_p =\sqrt{(S_1^2 + S_2^2)/2}.$

Thus a 95% confidence interval for $\Delta$ would be of the form $\hat\Delta \pm t^*\left[\widehat{SD}(\hat\Delta)\right],$ where $t^*$ cuts probability $0.025$ from the upper tail of $\mathsf{T}(\nu =n_1+n_2-2),$ Student's t distribution with degrees of freedom $\nu = n_1+n_2-2,$ and $\widehat{SD}(\hat\Delta)$ is the estimated standard error of $\hat\Delta.$

Simulation in R with $n_1 = n_2 = 16, \sigma=3, \mu_1 = 50, \mu_2 = 53,$ illustrates normal distribution of $\Delta.$

n = 16; sg=3
set.seed(1234)
Dlt.est = replicate(10^6, 4*mean(rnorm(n,50,sg))
                          -3*mean(rnorm(n,53,sg)))
mean(Dlt.est)
[1] 40.99765    # aprx 41
4*50-3*53
[1] 41          # exact
var(Dlt.est)
[1] 14.08624
sd(Dlt.est)
[1] 3.753164   # aprx 3.75
5*sg/sqrt(16)
[1] 3.75       # exact

hist(Dlt.est, prob=T, br=30, col="skyblue2")
curve(dnorm(x, 41, 3.75), add=T, col="orange", lwd=2)

enter image description here

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  • $\begingroup$ How did you get the expression for $\hat{\sigma_p}?$ And would $\overline{SD}(\hat{\Delta})$ just be equal to $\hat{\sigma_p}$ here? $\endgroup$ – user295786 Mar 28 at 22:29
  • $\begingroup$ I got $\hat \sigma_p$ as the square root of of the "pooled variance" $\hat \sigma_p^2.$ In the balanced case $\sigma_p^2$ is just the average of the two sample variances $S_1^2$ and $S_2^2.$ If sample sizes are unequal, the slightly messier formula gives greater weight to the larger sample in estimating the common variance $\sigma^2.$ // The answer to your second question is NO. The estimate of $SD(\hat \Delta)$ needs to take into account how $\Delta$ is defined in terms of the two samples (with coefficients 4 and 3, instead of just the usual difference). $\endgroup$ – BruceET Mar 28 at 22:55
  • $\begingroup$ Thank you. How can I compute $\overline{SD}(\hat{\Delta})$? $\endgroup$ – user295786 Mar 28 at 23:01
  • $\begingroup$ System went down to be maintained as I was answering your comment and fixing an (unrelated) typo in my answer. I think I have restored everything, but ask again if unclear. $\endgroup$ – BruceET Mar 28 at 23:03
  • $\begingroup$ I gave the formula for the standard error $SD(\hat\Delta)$ in terms of $\sigma.$ Then you get the estimated standard error $\widehat{SD}(\hat\Delta)$ by plugging in $\hat\sigma_p$ for $\sigma.$ $\endgroup$ – BruceET Mar 28 at 23:07

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