Question

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 1

Answer:

a

The confidence interval for the population variance is ​$5.49 < \sigma^2 < 37.02$  

The correct option is  C

b

The confidence interval for the population standard deviation is $2.34 < \sigma < 6.08$

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  $s = \ 3.42$

Generally the degree of freedom is mathematically represented as

     $df =  14 -1$

=>   $df =  13$    

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

     $\alpha = (100 -98) \%$

=>  $\alpha = 0.02$

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

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

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

From the chi -distribution table  

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

and

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

So

=> $\frac{(14- 1) (3.42)^2}{27.689}  < \sigma^2 < \frac{(14- 1) (3.42)^2}{4.107}$    

=> $5.49 < \sigma^2 < 37.02$    

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

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

=> $\sqrt{5.49}  < \sigma < \sqrt{37.02 }$

=> $2.34 < \sigma < 6.08$