95% limits of a normally distributed parameter

Question

How do I find the 95% limits of the population distribution of a normally distributed parameter? I've taken the mean and SD from 10 different readings of the parameter. Will the 95% limits be mean +/-1.96SD as per the normal distribution; or mean +/-2.262SD as per the t distribution as it is a small sample?

Answer 1

Suppose $X_1, X_2, \dots, X_{10}$ is a random sample from a normal distribution with unknown mean $\mu$ and unknown SD $\sigma.$

CI for normal $\sigma$ with $\mu$ unknown. Then the 'pivotal quantity' $Q = \frac{9\,S^2}{\sigma^2} \sim \mathsf{Chisq}(df = 9),$ where $S^2$ is the sample variance. Thus $$P(2.700 < Q < 19.023) = P\left(\frac{9\,S^2}{19.023} < \sigma^2 < \frac{9\,S^2}{2.700}\right) = 0.95,$$ where percentage points .025 and .975 of the chi-squared distribution can be found from printed tables or using software (such as R below).

Thus a 95% CI for $\sigma^2$ is of the form $\left(\frac{9\,S^2}{19.023},\; \frac{9\,S^2}{2.700}\right)$ and a 95% CI for $\sigma$ can be found by taking square roots of the endpoints of the CI for $\sigma^2.$

qchisq(c(.025, .975), 9)
[1]  2.700389 19.022768

Numerical example. For example, the R code below generates $n = 10$ observations from $\mathsf{Norm}(\mu=100,\, \sigma=20),$ for which the sample SD is $S = 24.51.$ A 95% CI for $\sigma$ is $(16.86,\, 44.75).$ In this example where $\sigma$ is known because we simulated the data, we can see that the CI includes the true value $\sigma = 20.$

set.seed(1122);  n = 10;  mu = 100;  sig = 20
x = rnorm(n, mu, sig);  s = sd(x);  s
[1] 24.51219
CI = sqrt((n-1)*s^2/qchisq(c(.975,.025), n-1));  CI[1] 
[1] 16.86035 44.74971

Distribution of $Q.$ The relationship $Q = \frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\text{df} = n-1)$ can be proved using moment generating functions or by a multi-dimensional orthogonal transformation. The simulation below, illustrates this relationship for the parameter values used in this Answer.

set.seed(1118);  m=10^5;  n = 10;  mu = 100; sig = 20
q = replicate(m, (n-1)*sd(rnorm(n, mu, sig))^2/sig^2)
mean(q); var(q)
[1] 9.000164  # aprx E(Q) = 9
[1] 17.9578   # aprx Var(Q) = 18

HDR = "Simulated Dist'n of Q with Density of CHISQ(9)"
hist(q, prob=T, br=30, col="skyblue2", xlab="Q", main=HDR)
curve(dchisq(x, n-1), 0, max(q), add=T, lwd=2, col="red")