Question

A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample​ variance,

s squareds2​,

is determined to be

12.512.5.

Complete parts​ (a) through​ (c).

​(a) Construct a​ 90% confidence interval for

sigma squaredσ2

if the sample​ size, n, is 20.The lower bound is

nothing.

​(Round to two decimal places as​ needed.)The upper bound is

nothing.

​(Round to two decimal places as​ needed.)​(b) Construct a​ 90% confidence interval for

sigma squaredσ2

if the sample​ size, n, is 30.The lower bound is

nothing.

​(Round to two decimal places as​ needed.)The upper bound is

nothing.

​(Round to two decimal places as​ needed.)

How does increasing the sample size affect the width of the​ interval?

The width increases

The width does not change

The width decreases

​(c) Construct a​ 98% confidence interval for

sigma squaredσ2

if the sample​ size, n, is 20.The lower bound is

nothing.

​(Round to two decimal places as​ needed.)The upper bound is

nothing.

​(Round to two decimal places as​ needed.)

Compare the results with those obtained in part​ (a). How does increasing the level of confidence affect the confidence​ interval?

The width decreases

The width increases

The width does not change

Answer 1

Solution:

Confidence interval for population variance formula is given as below:

(n – 1)*S² / χ²_{α/2, n – 1} < σ² < (n – 1)*S² / χ²_{1 -α/2, n– 1}

Part a

We are given

Confidence level = 90%, α = 0.10

n = 20, df = n – 1 = 19

Sample variance = S² = 12.5

χ²_{α/2, n – 1} = 30.1435

χ²_{1 -α/2, n– 1} = 10.1170

(20 – 1)* 12.5/ 30.1435< σ² < (20 – 1)* 12.5/ 10.1170

7.8790 < σ² < 23.4753

Lower bound = 7.88

Upper bound = 23.48

Part b

We are given

Confidence level = 90%, α = 0.10

n = 30, df = n – 1 = 29

Sample variance = S² = 12.5

χ²_{α/2, n – 1} = 42.5570

χ²_{1 -α/2, n– 1} = 17.7084

(30 – 1)* 12.5/ 42.5570< σ² < (30 – 1)* 12.5/ 17.7084

8.5180< σ² < 20.4706

Lower bound = 8.52

Upper bound = 20.47

Part c

We are given

Confidence level = 98%, α = 0.02

n = 20, df = n – 1 = 19

Sample variance = S² = 12.5

χ²_{α/2, n – 1} = 36.1909

χ²_{1 -α/2, n– 1} = 7.6327

(20 – 1)* 12.5/ 36.1909 < σ² < (20 – 1)* 12.5/ 7.6327

6.5624< σ² < 31.1160

Lower bound = 6.56

Upper bound = 31.12

How does increasing the level of confidence affect the confidence ​interval?

The width increases

(By comparing part a and c)

(All critical values are taken from Chi square table)