# Confidence Interval for the Variance

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We want to derive a confidence interval for the unknown variance ${\displaystyle \sigma ^{2}}$ of a population under the following assumption:

1. The expectation ${\displaystyle E(X)=\mu }$ is unknown.

As we have seen above, an unbiased estimator of the unknown variance ${\displaystyle \sigma ^{2}}$ is given by: ${\displaystyle s^{2}={\frac {1}{n-1}}\sum \limits _{i=1}^{n}(X_{i}-{\bar {x}})^{2}\,.}$ It has been shown (see distribution of the sample variance) that${\displaystyle {\frac {(n-1)s^{2}}{\sigma ^{2}}}={\frac {1}{\sigma ^{2}}}\sum \limits _{i=1}^{n}(X_{i}-{\bar {x}})^{2}=\sum \limits _{i=1}^{n}\left({\frac {X_{i}-{\bar {x}}}{\sigma }}\right)^{2}}$ has a chi-square distribution with ${\displaystyle \ n-1}$ degrees of freedom. We can now make probability statements of the following form:${\displaystyle P\left({\frac {\sigma ^{2}\chi _{{\frac {\alpha }{2}};n-1}^{2}}{n-1}}\leq s^{2}\leq {\frac {\sigma ^{2}\chi _{1-{\frac {\alpha }{2}};n-1}^{2}}{n-1}}\right)=1-\alpha }$ Here, ${\displaystyle \chi _{{\frac {\alpha }{2}};n-1}^{2}}$ is the ${\displaystyle \ \alpha /2}$ - quantile and ${\displaystyle \chi _{1-{\frac {\alpha }{2}};n-1}^{2}}$ the ${\displaystyle \ (1-\alpha /2)}$ - quantile of the chi-square distribution with ${\displaystyle n-1}$ degrees of freedom. By algebraic manipulation we may isolate ${\displaystyle \sigma ^{2}}$ in the middle of the probability statement: ${\displaystyle P\left({\frac {(n-1)s^{2}}{\chi _{1-{\frac {\alpha }{2}};n-1}^{2}}}\leq \sigma ^{2}\leq {\frac {(n-1)s^{2}}{\chi _{{\frac {\alpha }{2}};n-1}^{2}}}\right)=1-\alpha \,.}$ The corresponding confidence interval is ${\displaystyle \left[{\frac {(n-1)s^{2}}{\chi _{1-{\frac {\alpha }{2}};n-1}^{2}}}\,,\,\,\ \,{\frac {(n-1)s^{2}}{\chi _{{\frac {\alpha }{2}};n-1}^{2}}}\right]}$ The interpretation is the same as before: a proportion ${\displaystyle 1-\alpha }$ of confidence intervals constructed in this fashion will contain the true parameter value ${\displaystyle \sigma ^{2}}$.

• By construction, these confidence intervals assign equal probability mass to the tails: ${\displaystyle P\left(\sigma ^{2}<{\frac {(n-1)s^{2}}{\chi _{1-{\frac {\alpha }{2}};n-1}^{2}}}\right)={\frac {\alpha }{2}}\ ,\,\,\,\ P\left({\frac {(n-1)s^{2}}{\chi _{{\frac {\alpha }{2}};n-1}^{2}}}<\sigma ^{2}\right)={\frac {\alpha }{2}}\,.}$
• The confidence interval is not symmetric around the point estimate ${\displaystyle s^{2}}$, since the chi-square distribution is not a symmetric distribution.
• The length of the confidence interval ${\displaystyle (n-1)s^{2}\left({\frac {1}{\chi _{{\frac {\alpha }{2}};n-1}^{2}}}-{\frac {1}{\chi _{1-{\frac {\alpha }{2}};n-1}^{2}}}\right)}$ depends on the sampled values ${\displaystyle x_{1},\dots ,x_{n}}$ and is a random variable. The length of the interval also depends on the sample size ${\displaystyle n}$ and on the confidence level ${\displaystyle 1-\alpha }$.

The following variables have been measured for a population of ${\displaystyle N=3000}$ households:${\displaystyle X1}$ = monthly net income${\displaystyle X2}$ = monthly expenditure on rent${\displaystyle X3}$ = monthly expenditure on the automobile The expectation ${\displaystyle \mu }$ and the variance ${\displaystyle \sigma ^{2}}$ of the variables are unknown. We assume the distributions to be normal. Find point and interval estimates for the unknown variance ${\displaystyle \sigma ^{2}}$. One can examine the effect of the confidence level and sample size on the length of the confidence interval. We recommend that only one feature be altered at a time. Please select

• the variable to be analyzed
• the sample size ${\displaystyle n}$
• the confidence level ${\displaystyle 1-\alpha }$ (as a decimal e.g. 0.95)

Result:This interactive examples produces

1. a boxplot

If you choose the same variable repeatedly, but enter different confidence levels/sample sizes, the previous results are displayed for comparison purposes. Click on your mouse to start the interactive example.Launching can take several seconds. Some time may elapse before the results are displayed. For a population of ${\displaystyle N=2000}$ households let ${\displaystyle X}$ denote net household income . We assume that ${\displaystyle X}$ is approximately normally distributed ${\displaystyle X\sim N(\mu ;\sigma ^{2})}$;  the two parameters ${\displaystyle \mu }$ and the variance ${\displaystyle \sigma ^{2}}$ are unknown. Construction of confidence intervals for the unknown mean ${\displaystyle \mu }$ has been studied in chapter confidence intervals for the expectation. Here, we want to focus on the unknown variance ${\displaystyle \sigma ^{2}}$, for which we will construct a confidence interval with confidence level ${\displaystyle 1-\alpha =0.95}$. A random sample of size ${\displaystyle n=20}$ yields the following realizations (ordered by size):

${\displaystyle i}$ Household net income (DM) ${\displaystyle x_{i}}$ ${\displaystyle i}$ Household net income (DM) ${\displaystyle x_{i}}$
1 800 11 2500
2 1200 12 2500
3 1400 13 2500
4 1500 14 2700
5 1500 15 2850
6 1500 16 3300
7 1800 17 3650
8 1800 18 3700
9 2300 19 4100
10 2400 20 4300

Mean household income of the sample is ${\displaystyle {\bar {x}}=48\,300/20=2\,415\,.}$ Our point estimate for the unknown variance ${\displaystyle \sigma ^{2}}$ is given by ${\displaystyle s^{2}=1\,002\,131.58}$ Using chi-square tables we find ${\displaystyle \chi _{\alpha /2;n-1}^{2}=\chi _{0.025;19}^{2}=8.91\ \,\,and\,\,\ \chi _{1-\alpha /2;n-1}^{2}=\chi _{0.975;19}^{2}=32.85}$ Hence the confidence interval is given by ${\displaystyle \left[{\frac {19\cdot 1002131.58}{32.85}},\,\,\,\,{\frac {19\cdot 1002131.58}{8.91}}\right]=[579619.48,\,\,\,\,\,2136980.92]\,.}$