Chapter Outline: 6.4 Confidence Intervals for the Standard Deviation 7.5 Hypothesis testing for Standard Deviation

Chapter Additional:

Standard Deviation

and Chi-Square

Section 6.4 Objectives

Interpret the chi-square distribution and use a chi-square distribution table

Use the chi-square distribution to construct a confidence interval for the variance and standard deviation

The Chi-Square Distribution

The point estimate for σ2 is s2

The point estimate for σ is s

s2 is the most unbiased estimate for σ2

Estimate Population

Parameter…

with Sample

Statistic

Variance: σ2 s2

Standard deviation: σ s

The Chi-Square Distribution

You can use the chi-square distribution to construct a confidence interval for the variance and standard deviation.

If the random variable x has a normal distribution, then the distribution of

forms a chi-square distribution for samples of any size n > 1.

22

2

( 1)n s

σ

Properties of The Chi-Square Distribution

1. All chi-square values χ2 are greater than or equal to zero.

2. The chi-square distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for σ2, use the χ2-distribution with degrees of freedom equal to one less than the sample size.

• d.f. = n – 1 Degrees of freedom

3. The area under each curve of the chi-square distribution equals one.

Properties of The Chi-Square Distribution

4. Chi-square distributions are positively skewed.

Chi-square Distributions

There are two critical values for each level of confidence.

The value χ2R represents the right-tail critical value

The value χ2L represents the left-tail critical value.

Critical Values for χ2

The area between

the left and right

critical values is c. χ2

c

12

c

12

c

2

L2

R

Example: Finding Critical Values for χ2

Find the critical values and for a 95% confidence

interval when the sample size is 18.

Solution:

• d.f. = n – 1 = 18 – 1 = 17 d.f.

• Area to the right of χ2R =

1 0.950.025

21

2c

• Area to the right of χ2L =

1 0.950.975

21

2c

2

L2

R

• Each area in the table represents the region under the

chi-square curve to the right of the critical value.

Solution: Finding Critical Values for χ2

Table 6: χ2-Distribution

2

R 2

L

95% of the area under the curve lies between 7.564 and

30.191.

30.191 7.564

Confidence Interval for σ:

•

Confidence Intervals for σ2 and σ

2 2

2 2

( 1) ( 1)

R L

n s n s

2σ

• The probability that the confidence intervals contain

σ2 or σ is c.

Confidence Interval for σ2:

•

2 2

2 2

( 1) ( 1)

R L

n s n s

σ

Confidence Intervals for σ2 and σ

1. Verify that the population has a

normal distribution.

2. Identify the sample statistic n and

the degrees of freedom.

3. Find the point estimate s2.

4. Find the critical values χ2R and χ2

L

that correspond to the given level

of confidence c.

Use Table 6 in

Appendix B.

22 )

1x x

sn

(

d.f. = n – 1

In Words In Symbols

Confidence Intervals for 2 and

2 2

2 2

( 1) ( 1)

R L

n s n s

2σ

5. Find the left and right

endpoints and form the

confidence interval for the

population variance.

6. Find the confidence

interval for the population

standard deviation by

taking the square root of

each endpoint.

2 2

2 2

( 1) ( 1)

R L

n s n s

σ

In Words In Symbols

Example: Constructing a Confidence Interval

You randomly select and weigh 30 samples of an allergy medicine. The sample standard deviation is 1.20 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance and standard deviation.

Solution:

• d.f. = n – 1 = 30 – 1 = 29 d.f.

Solution: Constructing a Confidence Interval

• The critical values are

χ2R = 52.336 and χ2

L = 13.121

• Area to the right of χ2R =

1 0.990.005

21

2c

• Area to the right of χ2L =

1 0.990.995

21

2c

Solution: Constructing a Confidence Interval

2

22 (30 1)(1.20)0.80

52.336( 1)

R

n s

Confidence Interval for σ2:

2

22 (30 1)(1.20)3.18

13.121( 1)

L

n s

Left endpoint:

Right endpoint:

0.80 < σ2 < 3.18

With 99% confidence, you can say that the population

variance is between 0.80 and 3.18.

Solution: Constructing a Confidence Interval

2 2(30 1)(1.20) (30 1)(1.20)52.336 13.121

Confidence Interval for σ :

0.89 < σ < 1.78

With 99% confidence, you can say that the population

standard deviation is between 0.89 and 1.78 milligrams.

2 2

2 2

( 1) ( 1)

R L

n s n s

σ

Section 6.4 Summary

Interpreted the chi-square distribution and used a chi-square distribution table

Used the chi-square distribution to construct a confidence interval for the variance and standard deviation

Section 7.5 Objectives

Find critical values for a χ2-test

Use the χ2-test to test a variance or a standard deviation

Finding Critical Values for the χ2-Test

a. Specify the level of significance α.

b. Determine the degrees of freedom d.f. = n – 1.

c. The critical values for the χ2-distribution are found in Table 6 in Appendix B. To find the critical value(s) for a

a. right-tailed test, use the value that corresponds to d.f. and α.

b. left-tailed test, use the value that corresponds to d.f. and 1 – α.

c. two-tailed test, use the values that corresponds to d.f. and ½α, and d.f. and 1 – ½α.

Finding Critical Values for the χ2-Test

2

0

1 –

Right-tailed

Two-tailed

1 – 12

2

L2

R

12

1

2

0

Left-tailed

Example: Finding Critical Values for χ2

Find the critical χ2-value for a left-tailed test when n = 11 and α = 0.01.

Solution:

• Degrees of freedom: n – 1 = 11 – 1 = 10 d.f.

• The area to the right of the critical value is

1 – α = 1 – 0.01 = 0.99.

From Table 6, the critical value is . 2

0 2.558

0.01

χ0 = 2.558

Example: Finding Critical Values for χ2

Find the critical χ2-value for a two-tailed test when n = 9 and α = 0.05.

Solution:

• Degrees of freedom: n – 1 = 9 – 1 = 8 d.f.

• The areas to the right of the critical values are

From Table 6, the critical values are and

02

251

.0

01

1 . 7 .52

9

2 2.180L 2 17.535R

The Chi-Square Test

χ2-Test for a Variance or Standard Deviation

A statistical test for a population variance or standard deviation.

Can be used when the population is normal.

The test statistic is s2.

The standardized test statistic

follows a chi-square distribution with degrees of freedom d.f. = n – 1.

22

2

( 1)n s

Using the χ2-Test for a Variance or Standard Deviation

1. State the claim mathematically

and verbally. Identify the null

and alternative hypotheses.

2. Specify the level of significance.

3. Determine the degrees of

freedom.

4. Determine the critical value(s).

State H0 and Ha.

Identify α.

Use Table 6 in

Appendix B.

d.f. = n – 1

In Words In Symbols

Using the χ2-Test for a Variance or Standard Deviation

22

2

( 1)n s

If χ2 is in the rejection

region, reject H0.

Otherwise, fail to

reject H0.

5. Determine the rejection region(s).

6. Find the standardized test statistic

and sketch the sampling

distribution.

7. Make a decision to reject or fail

to reject the null hypothesis.

8. Interpret the decision in the

context of the original claim.

In Words In Symbols

Example: Hypothesis Test for the Population Variance

A dairy processing company claims that the variance of the amount of fat in the whole milk processed by the company is no more than 0.25. You suspect this is wrong and find that a random sample of 41 milk containers has a variance of 0.27. At α = 0.05, is there enough evidence to reject the company’s claim? Assume the population is normally distributed.

Solution: Hypothesis Test for the Population Variance

• H0:

• Ha:

• α =

• df =

• Rejection Region:

• Test Statistic:

• Decision:

σ2 ≤ 0.25 (Claim)

σ2 > 0.25

0.05

41 – 1 = 40

22

2

( 1) (41 1)(0.27)

0.25

43.2

n s

Fail to Reject H0 .

At the 5% level of significance,

there is not enough evidence to

reject the company’s claim that the

variance of the amount of fat in the

whole milk is no more than 0.25.

0.05

2

0 55.758 2 43.2

Example: Hypothesis Test for the Standard Deviation

A company claims that the standard deviation of the lengths of time it takes an incoming telephone call to be transferred to the correct office is less than 1.4 minutes. A random sample of 25 incoming telephone calls has a standard deviation of 1.1 minutes. At α = 0.10, is there enough evidence to support the company’s claim? Assume the population is normally distributed.

Solution: Hypothesis Test for the Standard Deviation

• H0:

• Ha:

• α =

• df =

• Rejection Region:

• Test Statistic:

• Decision:

σ ≥ 1.4 min.

σ < 1.4 min. (Claim)

0.10

25 – 1 = 24

Reject H0 . At the 10% level of significance, there

is enough evidence to support the

claim that the standard deviation of the

lengths of time it takes an incoming

telephone call to be transferred to the

correct office is less than 1.4 minutes.

2 22

2 2

( 1) (25 1)(1.1)

1.4

14.816

n s

Example: Hypothesis Test for the Population Variance

A sporting goods manufacturer claims that the variance of the strengths of a certain fishing line is 15.9. A random sample of 15 fishing line spools has a variance of 21.8. At α = 0.05, is there enough evidence to reject the manufacturer’s claim? Assume the population is normally distributed.

Solution: Hypothesis Test for the Population Variance

• H0:

• Ha:

• α =

• df =

• Rejection Region:

• Test Statistic:

• Decision:

σ2 = 15.9 (Claim)

σ2 ≠ 15.9

0.05

15 – 1 = 14

2 (n 1)s2

2

(151)(21.8)

15.9

19.195

Fail to Reject H0

At the 5% level of significance,

there is not enough evidence to

reject the claim that the variance in

the strengths of the fishing line is

15.9.

10.025

2

10.025

2

2 5.629L 2 26.119R 19.195

Section 7.5 Summary

Found critical values for a χ2-test

Used the χ2-test to test a variance or a standard deviation