Interval Estimates - Confidence Interval for \sigma^2

Section 10.6 Interval Estimates - Confidence Interval for $σ^{2}$

Once again, you may need to approximate the population variance or standard deviation but only have the sample values available. One difference from the previous sections is that you are not dealing with an average of values (such as

\overset{―}{x}

\tilde{p}

) but with the average of the squares of values. The Central Limit Theorem does not directly help you in this case but the following result (presented without proof) provides a solution.

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Theorem 10.6.1. Relationship between Variance and $χ^{2}$ .

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S^{2}

is a random variable of possible sample variance values from a sample of size n, then

W = \frac{(n - 1) S^{2}}{σ^{2}}

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is approximately

χ^{2} (n - 1) .

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To create a confidence interval for

σ^{2}

first consider an interval of the form

E_{1} < σ^{2} < E_{2}

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and determine values for the boundaries so that the likelihood of this being true is high. For this case, since the chi-square distribution only has a positive domain and is not symmetrical, you will not expect to determine a symmetrical confidence interval. Therefore, consider

P (E_{1} < σ^{2} < E_{2}) = 1 - α

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and by playing around with algebra you get

P (\frac{E_{1}}{(n - 1) S^{2}} < \frac{σ^{2}}{(n - 1) S^{2}} < \frac{E_{2}}{(n - 1) S^{2}}) = 1 - α

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or by inverting the inequality yields

P (\frac{(n - 1) S^{2}}{E_{2}} < \frac{(n - 1) S^{2}}{σ^{2}} < \frac{(n - 1) S^{2}}{E_{1}}) = 1 - α .

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Using the previous theorem, note that the inside variable can be replaced with a chi-square variable. If F is the distribution function for chi-square, then you get

F (\frac{(n - 1) S^{2}}{E_{1}}) - F (\frac{(n - 1) S^{2}}{E_{2}}) = 1 - α .

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For a given value of

α

there are many possible choices but often one often utilized is one in which

F (χ_{1 - α / 2}^{2}) = F (\frac{(n - 1) S^{2}}{E_{1}}) = 1 - α / 2

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and

F (χ_{α / 2}^{2}) = F (\frac{(n - 1) S^{2}}{E_{2}}) = α / 2.

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Using the inverse chi-square gives values for the expression on the inside and algebra can be used to solve for each of

E_{1}, E_{2} .

Indeed,

E_{1} = \frac{(n - 1) S^{2}}{χ_{1 - α / 2}^{2}}

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and

E_{2} = \frac{(n - 1) S^{2}}{χ_{α / 2}^{2}}

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To determine appropriate values for

χ_{α / 2}^{2}

and

χ_{1 - α / 2}^{2}

with equal probabilities in each tail, consider using the interactive cell below:

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print("Confidence Interval for variance using equally-sized tails")
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# Chi-Square Calculator for confidence intervals with equal alpha/2 tails
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var('t')
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@interact(layout=dict(top=[['c'],['n']]))
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def _(c=input_box(0.95,width=10,label='Confidence Level = '),n=input_box(20,width=8,label='n =')):
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    alpha = 1-c
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    T = RealDistribution('chisquared', n)
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    a = T.cum_distribution_function_inv(alpha/2)
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    a1 = T.cum_distribution_function(a)
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    b = T.cum_distribution_function_inv(1-alpha/2)
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    b1 = T.cum_distribution_function(b)
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    print('From the Chi-Square distribution for X:')
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    print('P(',a,'< X < ',(b),') = ',c)
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    print('with')
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    print('P( X < ',a,') = ',a1)
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    print('P( X < ',b,') = ',b1)
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    f = x^(n/2-1)*e^(-x/2)/(gamma(n/2)*2^(n/2))
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    M = f(x=n).n()   # to scale by the maximum of f
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    G = plot(f,x,0,b+(b-a)/2)+plot(f,x,a,b,thickness=5,color='green',fill=True,fillcolor='yellow')
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    G += line([(a,0),(a,f(x=a))],color='green',thickness=5)
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    G += line([(b,0),(b,f(x=b))],color='green',thickness=5)
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    G += text(str(c.n(digits=5)),((a+b)/2,f(x=(a+b)/2)/3),color='green')
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    G += text(str((alpha/2).n(digits=5)),(a/2,0.01),color='red',fontsize=5)
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    G += text(str((alpha/2).n(digits=5)),(b+a/2,0.01),color='red',fontsize=5)
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    G += text("$\\chi^2_{\\alpha/2} $",(a,-M/5),color='red',fontsize=10)
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    G += text("$\\chi^2_{1-\\alpha/2} $",(b,-M/5),color='red',fontsize=10)
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    G.show(figsize=(5,3))

    
    
    
    
        
            
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        Messages

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Example 10.6.2. - Two-sided Confidence interval for $σ^{2}$ and $σ$ .

Given the data 570, 561, 546, 540, 609, 580, 550, 577, 585, determine a 95% confidence interval for

σ^{2} .

Using the computational forumaula (or your calculator) gives

s^{2} \approx 479.5 .

Also, notice for n=9, the resulting interval will use a Chi-square variable with 8 degrees of freedom. Using the symmetric option, gives

χ_{0.025}^{2} = 2.18

and

χ_{0.975}^{2} = 17.53 .

Therefore

E_{1} = \frac{8 \cdot 479.5}{17.53} \approx 221.095

and

E_{2} = \frac{8 \cdot 479.5}{2.18} \approx 1759.63 .

Hence, you are 95% certain that

221.095 < σ^{2} < 1759.63 .

By taking square roots you get

14.87 < σ < 41.95 .

Notice, this interval is relatively wide which is a result both of the number of data values being relatively small (n=9) and the actual data values being relatively large and spread out.

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Checkpoint 10.6.3. WebWork - Two-sided Confidence Interval with large n.

Find the critical values

χ_{L}^{2} = χ_{1 - α / 2}^{2}

and

χ_{R}^{2} = χ_{α / 2}^{2}

that correspond to

99

% degree of confidence and the sample size

n = 22.

χ_{L}^{2} =

χ_{R}^{2} =

Answer 1.

8.03365

Answer 2.

41.4011

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Suppose that you have n=400 data values and suppose you have computed from those a sample variance of

s^{2} = 479.5 .

Then, the only change in the calculation is the two chi-square statistic values. For 95% but now with 399 degrees of freedom

χ_{0.025}^{2} = 345.55

and

χ_{0.975}^{2} = 456.24 .

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Therefore

E_{1} = \frac{8 \cdot 479.5}{456.24} \approx 419.3

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and

E_{2} = \frac{8 \cdot 479.5}{345.55} \approx 553.7 .

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Hence, you are 95% certain that

419.24 < σ^{2} < 553.7 .

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By taking square roots you get

20.48 < σ < 23.53

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which is a relatively tight confidence interval. Notice, these are also completely contained in the confidence intervals from the previous small n example.

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Similar to above, another choice to estimate