Question 1040410
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Let's set up some notation. 


XL = left chi-square critical value
XR = right chi-square critical value


The X will stand for the greek letter chi (pronounced "kai", rhymes with "sky"). R for "right" and L for "left. 


XL and XR are the chi-square critical values such that P(XL < chi-square < XR) = 0.95
basically the area between XL and XR is 0.95


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Use a <a href="https://people.richland.edu/james/lecture/m170/tbl-chi.html">chi-square table</a> of values. Highlight the df = 10 row (n = 11, so df = n-1 = 11-1 = 10). 
This row is marked in red. 


Keep in mind that if the confidence level is 95%, then alpha = 1-CL = 1-0.95 = 0.05. Cut this in half to get 0.05/2 = 0.025


Mark the column that has 0.025 at the top. I marked in blue. The blue and red boxes intersect at the value 20.483. Let's call this XR, so XR = 20.483 approximately


Then we compute 1 - 0.025 = 0.975


Mark the column that has 0.975 at the top. This is marked in green. The green and red boxes intersect at 3.247


<img width=600 src = "http://i.imgur.com/fejkHg9.png">


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So we just found that
XL = 3.247
XR = 20.483
both are approximate. These are the chi-square critical values.


we will plug in 
n = 11 (sample size)
s = 12 (standard deviation


Use the values to compute the confidence interval


*[Tex \LARGE \sqrt{\frac{(n-1)*s^2}{XR}} < \sigma < \sqrt{\frac{(n-1)*s^2}{XL}}]


*[Tex \LARGE \sqrt{\frac{(11-1)*12^2}{20.483}} < \sigma < \sqrt{\frac{(11-1)*12^2}{3.247}}]


*[Tex \LARGE \sqrt{\frac{10*144}{20.483}} < \sigma < \sqrt{\frac{10*144}{3.247}}]


*[Tex \LARGE \sqrt{\frac{1440}{20.483}} < \sigma < \sqrt{\frac{1440}{3.247}}]


*[Tex \LARGE \sqrt{70.3022018} < \sigma < \sqrt{443.486295}]


*[Tex \LARGE 8.3846408 < \sigma < 21.05911429]


*[Tex \LARGE 8.4 < \sigma < 21.1]


where *[Tex \LARGE \sigma] is the greek letter sigma and it is the population standard deviation



So the answer is <font color=red size=5>choice C</font>
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