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)*S^{2} / χ^{2}_{α}_{/2, n –
1} < σ^{2} < (n – 1)*S^{2} /
χ^{2}_{1 -}_{α/2, n}_{– 1}

Part a

We are given

Confidence level = 90%, α = 0.10

n = 20, df = n – 1 = 19

Sample variance = S^{2} = 12.5

χ^{2}_{α}_{/2, n – 1} = 30.1435

χ^{2}_{1 -}_{α/2, n}_{– 1} =
10.1170

(20 – 1)* 12.5/ 30.1435< σ^{2} < (20 – 1)* 12.5/
10.1170

7.8790 < σ^{2} < 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 = S^{2} = 12.5

χ^{2}_{α}_{/2, n – 1} = 42.5570

χ^{2}_{1 -}_{α/2, n}_{– 1} =
17.7084

(30 – 1)* 12.5/ 42.5570< σ^{2} < (30 – 1)* 12.5/
17.7084

8.5180< σ^{2} < 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 = S^{2} = 12.5

χ^{2}_{α}_{/2, n – 1} = 36.1909

χ^{2}_{1 -}_{α/2, n}_{– 1} =
7.6327

(20 – 1)* 12.5/ 36.1909 < σ^{2} < (20 – 1)* 12.5/
7.6327

6.5624< σ^{2} < 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)

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