A magazine includes a report on the energy costs per year for​32-inch liquid crystal display​ (LCD) televisions. The article states that 14 randomly selected​ 32-inch LCD televisions have a sample standard deviation of ​$3.42 . Assume the sample is taken from a normally distributed population. Construct 98 ​% confidence intervals for​ (a) the population variance sigma squared and​ (b) the population standard deviation sigma .
Interpret the results. ​(a) The confidence interval for the population variance is ​(___,___ ​). ​(Round to two decimal places as​ needed.)
Interpret the results. Select the correct choice below and fill in the answer​ box(es) to complete your choice. ​(Round to two decimal places as​ needed.)
A. With 98 ​% ​confidence, you can say that the population variance is less than ___.
B. With 2 ​% ​confidence, you can say that the population variance is between nothing and ___ .
C. With 98 ​% ​confidence, you can say that the population variance is between ___ and ___ .
D. With 2 ​% ​confidence, you can say that the population variance is greater than ___
​(b) The confidence interval for the population standard deviation is ​(n___,___​). ​(Round to two decimal places as​needed.)
Interpret the results. Select the correct choice below and fill in the answer​ box(es) to complete your choice. ​(Round to two decimal places as​ needed.)
A. With 98​% ​confidence, you can say that the population standard deviation is between ___ and ___ dollars per year.
B. With 98​% ​confidence, you can say that the population standard deviation is greater than ___ dollars per year.
C. With 2​% ​confidence, you can say that the population standard deviation is less than ___ dollars per year.
D. With 2​% ​confidence, you can say that the population standard deviation is between ___ and ___ dollars per year.

Answer:

a

The confidence interval for the population variance is ​[tex] 5.49 < \sigma^2 < 37.02 [/tex]  

The correct option is  C

b

The confidence interval for the population standard deviation is [tex] 2.34 < \sigma < 6.08 [/tex]

The correct option is  A

Step-by-step explanation:

From the question we are told that

    The sample size is  n =  14

    The standard deviation is  [tex]s = \$3.42[/tex]

Generally the degree of freedom is mathematically represented as

     [tex]df =  14 -1[/tex]

=>   [tex]df =  13 [/tex]    

Given that the confidence level is  98%  then that level of significance is  

     [tex]\alpha = (100 -98) \%[/tex]

=>  [tex]\alpha = 0.02[/tex]

Generally the 98% confidence interval for the variance  is mathematically represented as

     [tex]\frac{(n- 1) s^2}{x^2 _{[\frac{\alpha }{2} , df]}}  < \sigma^2 < \frac{(n- 1) s^2}{x^2 _{[1- \frac{\alpha }{2} , df]}}[/tex]

=> [tex]\frac{(14- 1) (3.42)^2}{x^2 _{[\frac{0.02 }{2} , 13]}}  < \sigma^2 < \frac{(14- 1) (3.42)^2}{x^2 _{[1- \frac{0.02 }{2} , 13]}}[/tex]

From the chi -distribution table  

     [tex]x^2 _{[1- \frac{0.02 }{2} , 13]} = 4.107[/tex]

and

    [tex]x^2 _{[\frac{0.02 }{2} , 13]} = 27.689[/tex]

So

=> [tex]\frac{(14- 1) (3.42)^2}{27.689}  < \sigma^2 < \frac{(14- 1) (3.42)^2}{4.107}[/tex]    

=> [tex] 5.49 < \sigma^2 < 37.02 [/tex]    

Generally the 98% confidence interval for the standard deviation   is mathematically represented as

   [tex]\sqrt{\frac{(n- 1) s^2}{x^2 _{[\frac{\alpha }{2} , df]}}}   < \sigma <\sqrt{ \frac{(n- 1) s^2}{x^2 _{[1- \frac{\alpha }{2} , df]}}}[/tex]

=> [tex]\sqrt{5.49}  < \sigma < \sqrt{37.02 }[/tex]

=> [tex] 2.34 < \sigma < 6.08 [/tex]

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