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

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

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

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

From the chi -distribution table

`x^2 _{[1- (0.02 )/(2) , 13]} = 4.107`

and

`x^2 _{[(0.02 )/(2) , 13]} = 27.689`

So

=>
`((14- 1) (3.42)^2)/(27.689)  < \sigma^2 < ((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 _{[(\alpha )/(2) , df]}}}   < \sigma <\sqrt{ \frac{(n- 1) s^2}{x^2 _{[1- (\alpha )/(2) , df]}}}`

=>
`√(5.49)  < \sigma < √(37.02 )`

=>
`2.34 < \sigma < 6.08`