10.5 Testing Claims about the Population Standard Deviation

for testing claims about a population standard deviation or variance

1) The sample is a simple random sample.

2) The population has values that are normally distributed (a strict requirement).

Test Statistic

X 2 =(n - 1) s 2

2

n = sample size

s 2 = sample variance

2 = population variance (given in null hypothesis)

Test Statistic

X 2 =(n - 1) s 2

2

Found in Table VI

Degrees of freedom = n -1

All values of X2 are nonnegative, and the distribution is not symmetric.

There is a different distribution for each number of degrees of freedom.

The critical values are found in Table VI using n-1 degrees of freedom.

Figure 7-12

All values are nonnegative

Not symmetric

x2

Properties of the Chi-Square Distribution

0 5 10 15 20 25 30 35 40 45

Figure 7-13

df = 10

df = 20

Figure 7-12

Not symmetric

x2

There is a different distribution for each number of degrees of freedom.

Properties of the Chi-Square Distribution

Chi-Square Distribution for 10 and 20 Degrees of Freedom

All values are nonnegative

Claim: 43.7 H0: = 43.7

H1: 43.7

Example: Aircraft altimeters have measuring errors with a standard deviation of 43.7 ft. With new production equipment, 81 altimeters measure errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to test the claim that the new altimeters have a standard deviation different from the old value of 43.7 ft.

Claim: 43.7 H0: = 43.7

H1: 43.7

2= 0.025

0.0250.025

= 0.05

Example: Aircraft altimeters have measuring errors with a standard deviation of 43.7 ft. With new production equipment, 81 altimeters measure errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to test the claim that the new altimeters have a standard deviation different from the old value of 43.7 ft.

Claim: 43.7 H0: = 43.7

H1: 43.7

106.629

0.025

n = 81

df = 80

Table A-4

0.025

= 0.05

Example: Aircraft altimeters have measuring errors with a standard deviation of 43.7 ft. With new production equipment, 81 altimeters measure errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to test the claim that the new altimeters have a standard deviation different from the old value of 43.7 ft.

2= 0.025

Claim: 43.7 H0: = 43.7

H1: 43.7

57.153

0.025

n = 81

df = 80

Table A-4

0.975

0.025

= 0.05

Example: Aircraft altimeters have measuring errors with a standard deviation of 43.7 ft. With new production equipment, 81 altimeters measure errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to test the claim that the new altimeters have a standard deviation different from the old value of 43.7 ft.

106.629

2= 0.025

x2 = = 114.586(81 -1) (52.3)2(n -1)s2

2 43.72

x2 = 114.586

x2 = = 114.586(81 -1) (52.3)2(n -1)s2

2 43.72

Reject H0

57.153 106.629

x2 = 114.586

x2 = = 114.586(81 -1) (52.3)2(n -1)s2

2 43.72

Reject H0

57.153 106.629

The sample evidence supports the claim that the standard deviation is different from 43.7 ft.

x2 = 114.586

x2 = = 114.586(81 -1) (52.3)2(n -1)s2

2 43.72

Reject H0

57.153 106.629

The new production method appears to be worse than the old method. The data supports that there is more variation in the error readings than before.

Table VI includes only selected values of

Specific P-values usually cannot be found

Use Table to identify limits that contain the P-value

Some calculators and computer programs will find exact P-values

Use the Chi-squaredistribution with

(n -1)s2

2x2 =

St. Dev or

Variance 2Which

parameter does the claim

address ?

ProportionP

Use the normal distribution

where P = x/nz = P - Pˆpqn

ˆ

Yes

Mean (µ)

Isn > 30

?

Use the normal distribution with

(If Is unknown use s instead.)

z =x - µx

n

Start

YesIsn > 30

?

Use the normal distribution with

(If is unknown use s instead.)

z =x - µx

n

No

Yes

No

Yes

No

Is thedistribution of

the population essentiallynormal ? (Use a

histogram.)

Use nonparametric methodswhich don’t require a normal distribution. See Chapter 13.

Use the normal distribution with

z =x - µx

n

(This case is rare.)

Is known

?

Use the Student tdistribution with

t = x - µx

sn

Example: Tests in the author’s past statistics classes have scores with a standard deviation equal to 14.1. One of his current classes now has 27 test scores with a standard deviation of 9.3. Use a 0.01 significance level to test the claim that this current class has less variation that past classes. Does a lower standard deviation suggest that the current class is doing better?

Claim: <14.1 H0: > 14.1

H1: 14.1

Example: Tests in the author’s past statistics classes have scores with a standard deviation equal to 14.1. One of his current classes now has 27 test scores with a standard deviation of 9.3. Use a 0.01 significance level to test the claim that this current class has less variation that past classes. Does a lower standard deviation suggest that the current class is doing better?

0.01

= 0.01

Example: Tests in the author’s past statistics classes have scores with a standard deviation equal to 14.1. One of his current classes now has 27 test scores with a standard deviation of 9.3. Use a 0.01 significance level to test the claim that this current class has less variation than past classes. Does a lower standard deviation suggest that the current class is doing better?Claim: <14.1 H0: > 14.1

H1: 14.1

= 0.01

Example: Tests in the author’s past statistics classes have scores with a standard deviation equal to 14.1. One of his current classes now has 27 test scores with a standard deviation of 9.3. Use a 0.01 significance level to test the claim that this current class has less variation than past classes. Does a lower standard deviation suggest that the current class is doing better?Claim: <14.1 H0: > 14.1

H1: 14.1

n = 27

df = 26

Table A-412.198

0.01

0.99

x2 = = 11.311(27 -1) (9.3)2(n -1)s2

2 14.12

x2 = 11.311

x2 = = 11.311(27 -1) (9.3)2(n -1)s2

2 14.12

Reject H0

12.198

x2 = 11.311

x2 = = 11.311(27 -1) (9.3)2(n -1)s2

2 14.1

Reject H0

12.198

The sample evidence supports the claim that the standard deviation is less than previous classes. A lower standard deviation means there is less variance in their scores.

P 556 3 – 5, 9, 10, 17, 18