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CI for normal variance. The confidence interval for the variance $\sigma^2$ of a normal population is based on the fact that $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu=n-1).$ The CI is of the form $\left(\frac{(n-1)S^2}{U},\frac{(n-1)S^2}{L}\right),$ where $L$ and $U$ cut probability $0.025$ from the lower and upper tails, respectively, of $\mathsf{Chisq}(n-1).$ The 95% CI fpr $\sigma^2$ is $(119.7,\,266.5).)$ Notice that the point estimate $\widehat{\sigma^2} = 171.6.$ does not lie at the center of this CI (because the chi-squared distribution is not symmetrical). Take square roots of the endpoint of the CI for $\sigma^2$ to get a 95% CI $(10.84,\,16.32)$ for the population standard deviation $\sigma.$

There ere theoretical CIs for $\mu$ based on estimates of parameters the shape and rate of such a distribution, but the formulas are not as simple as the ones we have seen above.

Then a 95% nonparametric bootstrap CI can give useful information. Here is how one style of bootstrap CI can be computed. By repeated re-sampling with replacement from the sample, we can get an idea of the variability of $\bar X$ as a point estimate of $\mu$ and use that information to make a 95% CI for $\mu:$ $(4.28,\, 15.74).$

# gamma data
set.seed(519)
y = rgamma(500, 3, 1/5)
summary(y);  sd(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.3141  8.4593 13.1476 14.9594 19.4401 52.2321 
[1] 8.913885

# 95% nonparametric bootstrap CI
set.seed(1234)
a.obs = mean(y)
d = replicate(200, mean(sample(y,500,rep=T))-a.obs)
UL = quantile(d, c(.975,.025))
a.obs - UL
   97.5%     2.5% 
14.27896 15.74119 

CI for normal variance. The confidence interval for the variance $\sigma^2$ of a normal population is based on the fact that $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu=n-1).$ The CI is of the form $\left(\frac{(n-1)S^2}{U},\frac{(n-1)S^2}{L}\right),$ where $L$ and $U$ cut probability $0.025$ from the lower and upper tails, respectively, of $\mathsf{Chisq}(n-1).$ The 95% CI for $\sigma^2$ is $(119.7,\,266.5).)$ Notice that the point estimate $\widehat{\sigma^2} = 171.6.$ does not lie at the center of this CI (because the chi-squared distribution is not symmetrical). Take square roots of the endpoint of the CI for $\sigma^2$ to get a 95% CI $(10.84,\,16.32)$ for the population standard deviation $\sigma.$

There ere theoretical CIs for $\mu$ based on estimates of parameters (the shape and rate) of such a distribution, but the formulas are not as simple as the ones we have seen above.

Then a 95% nonparametric bootstrap CI can give useful information. Here is how one style of bootstrap CI can be computed. By repeated re-sampling with replacement from the sample, we can get an idea of the variability of $\bar X$ as a point estimate of $\mu$ and use that information to make a 95% CI $(4.28,\, 15.74)$ for $\mu.$

# simulate fictitious gamma data
set.seed(519)
y = rgamma(500, 3, 1/5)
summary(y);  sd(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.3141  8.4593 13.1476 14.9594 19.4401 52.2321 
[1] 8.913885

# 95% nonparametric bootstrap CI
set.seed(1234)
a.obs = mean(y)
d = replicate(200, mean(sample(y,500,rep=T))-a.obs)
UL = quantile(d, c(.975,.025))
a.obs - UL
   97.5%     2.5% 
14.27896 15.74119 

However, for various populations other than normal and population parameters other than $\mu,$ confidence intervals for various population parameters may be of different styles. Here are examples of yet other styles of CIs that do not explicitly use endpoints based on a margin of error (based on standarberror) above and below the sample mean

However, for various populations other than normal and population parameters other than $\mu,$ confidence intervals for various population parameters may be of different styles. Here are examples of yet other styles of CIs that do not explicitly use endpoints based on a margin of error (based on standard error) above and below the sample mean

CI for normal varance. The confidence interval for the variance $\sigma^2$ of a normal population is based on the fact that $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu=n-1).$ The CI is of the form $\left(\frac{(n-1)S^2}{U},\frac{(n-1)S^2}{L},\right),$ where $L$ and $U$ cut probability $0.025$ from the lower and upper tails, respectively, of $\mathsf{Chisq}(n-1).$ The 95% CI fpr $\sigma^2$ is $(119.7,\,266.5).)$ Notice that the point estimate $\widehat{\sigma^2} = 171.6.$ does not lie at the center of this CI (because the chi-squared distribution is not symmetrical). Take square roots of the endpoint of the CI for $\sigma^2$ to get a 95% CI $(10.84,\,16.32)$ for the population standard deviation $\sigma.$

CI for normal variance. The confidence interval for the variance $\sigma^2$ of a normal population is based on the fact that $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu=n-1).$ The CI is of the form $\left(\frac{(n-1)S^2}{U},\frac{(n-1)S^2}{L}\right),$ where $L$ and $U$ cut probability $0.025$ from the lower and upper tails, respectively, of $\mathsf{Chisq}(n-1).$ The 95% CI fpr $\sigma^2$ is $(119.7,\,266.5).)$ Notice that the point estimate $\widehat{\sigma^2} = 171.6.$ does not lie at the center of this CI (because the chi-squared distribution is not symmetrical). Take square roots of the endpoint of the CI for $\sigma^2$ to get a 95% CI $(10.84,\,16.32)$ for the population standard deviation $\sigma.$