10.4: The One Variance Chi-Squared Procedure
- Page ID
- 105857
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)To create the One Mean Z and T Procedures and One Proportion Z Procedure, we used the Central Limit Theorem for sample means and proportions. While there is no Central Limit Theorem for sample variances, we can still construct a confidence interval using the fact that \(\dfrac{s^2}{\sigma^2/(n-1)}\) follows a well-studied \(\chi^2\)-distribution with \(n-1\) degrees of freedom, i.e.
\(\dfrac{s^2}{\sigma^2/(n-1)}\sim \chi^2_{df=n-1}\)
The following is the equivalent of the Generalized Empirical Rule:
\(P\left(\chi^2_{1-\alpha/2}<\dfrac{s^2}{\sigma^2/(n-1)}<\chi^2_{\alpha/2}\right)=1-\alpha\)
After manipulating the inequality inside the parenthesis, we can conclude that:
\(P\left(\dfrac{\sigma^2\chi^2_{1-\alpha/2}}{n-1}<s^2<\dfrac{\sigma^2\chi^2_{\alpha/2}}{n-1}\right)=1-\alpha\)
Therefore, we have the following two interpretations:
We are \((1-\alpha)\cdot100\%\) confident that
(1) \(s^2_i\) is between \(\dfrac{\sigma^2\chi^2_{1-\alpha/2}}{n-1}\) and \(\dfrac{\sigma^2\chi^2_{\alpha/2}}{n-1}\)
OR
(2) \(\sigma^2\) is between \(\dfrac{s^2_i(n-1)}{\chi^2_{\alpha/2}}\) and \(\dfrac{s^2_i(n-1)}{\chi^2_{1-\alpha/2}}\).
Hence, we are \((1-\alpha)\cdot100\%\) confident that \(\sigma^2\) is between \(\dfrac{s^2_i\chi^2_{1-\alpha/2}}{n-1}\) and \(\dfrac{s^2_i\chi^2_{\alpha/2}}{n-1}\).
Unlike the previous procedures, the One Variance Chi-Square Procedure is based on a distribution that is not symmetric. As a result, the confidence interval is not symmetric. While the sample variance, \(s_i^2\), is still an unbiased estimate of the population variance it is not the center of the interval. From the practical point of view, this means that the interval can be found by computing its lower and upper bounds directly with the following formulas:
\(LB=\dfrac{s^2_i(n-1)}{\chi^2_{\alpha/2}}\) and \(RB=\dfrac{s^2_i(n-1)}{\chi^2_{1-\alpha/2}}\)
Note that we can always turn a confidence interval for the population variance into a confidence interval for the population standard deviation by taking the square root of the boundaries!
\(LB=\sqrt{\dfrac{s^2_i(n-1)}{\chi^2_{\alpha/2}}}\) and \(RB=\sqrt{\dfrac{s^2_i(n-1)}{\chi^2_{1-\alpha/2}}}\)
As a result, the following template will be used to construct and interpret the confidence intervals:
|
Unknown Parameter |
\(\sigma^2\), population variance |
|---|---|
|
Point Estimate |
\(s_i^2\), the sample variance |
|
Confidence Level (of the point estimate) |
0% |
|
Confidence Level (for a new interval estimate) |
\((1-\alpha)100\%\) \(\alpha/2\) \(1-\alpha/2\) |
| Critical Value(s) | \(\chi^2_{\alpha/2}\) and \(\chi^2_{1-\alpha/2}\) for \(df=n-1\) |
|
Margin of Error |
n/a |
|
Lower Bound |
\(LB=\dfrac{s^2_i(n-1)}{\chi^2_{\alpha/2}}\) |
|
Upper Bound |
\(RB=\dfrac{s^2_i(n-1)}{\chi^2_{1-\alpha/2}}\) |
| Interpretation | We are \((1-\alpha)100\%\) confident that the unknown population variance \(\sigma^2\) is between LB and UB. |
For the procedure to work the following assumptions must be made:
- The sample is simple random.
- The population has a normal distribution.
Consider the following example.
For a certain drug, based on standards set by the United States Pharmacopeia (USP) - an official public standards-setting authority for all prescription and over-the-counter medicines and other health care products manufactured or sold in the United States, a standard deviation of capsule weights of less than 2 mg is acceptable. A sample of 10 capsules was taken and the weights are provided below:
99.8, 100.6, 101.7, 100.8, 99.5, 98.9, 101.7, 100.8, 99.8, 98.2
Construct and interpret a 90% confidence interval for the standard deviation of the weights of all capsules. You may assume that the weights of capsules are normally distributed.
Solution
First, let’s check if all necessary assumptions are satisfied:
- The sample is simple random.
- The population has a normal distribution.
Also, note that the sample mean and variance are respectively:
\(\bar{x}_i=100.18\)
\(s^2_i=1.32\)
Once the assumptions are verified, we may apply the procedure by using the template.
| Example | Template | |
|---|---|---|
|
Unknown Parameter |
\(\sigma^2\), population variance of capsule weights |
\(\sigma^2\), population variance |
|
Point Estimate |
\(s_i^2=1.32\), the sample variance of a sample of size \(n=10\) |
\(s_i^2\), the sample variance |
|
Confidence Level (of the point estimate) |
0% |
0% |
|
Confidence Level (for a new interval estimate) |
\(CL=90\%\) \(\alpha/2=0.05\) \(1-\alpha/2=0.95\) |
\((1-\alpha)100\%\) \(\alpha/2\) \(1-\alpha/2\) |
| Critical Value(s) | When \(df=9\): \(\chi^2_{0.95}=3.325\) and \(\chi^2_{0.05}=16.919\) | \(\chi^2_{\alpha/2}\) and \(\chi^2_{1-\alpha/2}\) for \(df=n-1\) |
|
Margin of Error |
n/a |
n/a |
|
Lower Bound |
\(LB=\dfrac{1.32\cdot9}{16.919}=0.70\) | \(LB=\dfrac{s^2_i(n-1)}{\chi^2_{\alpha/2}}\) |
|
Upper Bound |
\(RB=\dfrac{1.32\cdot9}{3.325}=3.57\) | \(RB=\dfrac{s^2_i(n-1)}{\chi^2_{1-\alpha/2}}\) |
| Interpretation | We are 90% confident that the variance of capsule weights is between 0.70 and 3.57. | We are \((1-\alpha)100\%\) confident that the unknown population variance \(\sigma^2\) is between LB and UB. |
Now we can interpret the results: We are 90% confident that the variance of capsule weights is between 0.70 and 3.57.
Remember that we can always convert a confidence interval for the population variance to a confidence interval for the population standard deviation by taking the square root of the boundaries!
In our example, we are \(90\%\) confident that the weight variance of capsules is between 0.70 and 3.57, therefore we are \(90\%\) confident that the weight standard deviation of capsules is between 0.83 and 1.89 mg - well within the standard 2 mg.