Statistics

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Statistics

Inferences About Population Variances

Contents Inference about a Population Variance Inferences about the Variances of Two

Populations

STATISTICS in PRACTICE The U.S. General Accounting Office (GAO) evaluators studied a Department of Interior program established to help clean up the nation’s rivers and lakes. The audits reviewed sample data on the

oxygen content, the pH level, and the amount of suspended solids in the effluent.

STATISTICS in PRACTICE

The hypothesis test was conducted about the variance in pH level for the population of effluent. The population variance in pH level expected at a properly functioning plant.

In this chapter you will learn how to conduct statistical inferences about the variances of one and two populations.

Inferences About a Population Variance

Chi-Square Distribution(2) Interval Estimation of 2

Hypothesis Testing

Chi-Square Distribution

The chi-square distribution is based on sampling from a normal population.

The chi-square distribution is the sum of squared standardized normal random variables such as

k

iizQ

1

2

(k)Q~χ 2


where

Mean: kVariance: 2k

Probability density function


We can use the chi-square distribution to develop interval estimates and conduct hypothesis tests about a population variance.

The sampling distribution of (n - 1)s2/2 has a chi-square distribution whenever a simplerandom sample of size n is selected from a normal population.

Examples of Sampling Distribution of (n - 1)s2/2

0

With 2 degrees of freedom

2

2

( 1)n s



2 2 2.975 .025


For example, there is a .95 probability of obtaining a (chi-square) value such that2

We will use the notation to denote the value for the chi-square distribution that provides an area of a to the right of the stated value.

2a

2a

Chi-Square Distribution A Chi-Square Distribution with 19 Degrees

of Freedom

Chi-Square Distribution Selected Values form the Chi-Square

Distribution Table

95% of thepossible 2 values

2

0

.025

2.025

.025

2.975

Interval Estimation of 2

22 2.975 .0252

( 1)n s


22 2(1 / 2) / 22

( 1)n sa a

Substituting (n – 1)s2/2 for the 2 we get

Performing algebraic manipulation we get

2)2/1(

22

22/

2 )1()1(

aa

snsn

There is a (1 – a) probability of obtaining a 2 value such that

22/

22)2/1( aa

Interval Estimate of a Population Variance


( ) ( )

/ ( / )

n s n s

1 12

22

22

1 22

a a

( ) ( )

/ ( / )

n s n s

1 12

22

22

1 22

a a

where the values are based on a chi-squaredistribution with n - 1 degrees of freedom andwhere 1 - a is the confidence coefficient.

Interval Estimation of Interval Estimate of a Population Standard

Deviation Taking the square root of the upper and lowerlimits of the variance interval provides the confidence interval for the population standard deviation.

2 2

2 2/ 2 (1 / 2)

( 1) ( 1)n s n s

a a

Buyer’s Digest rates thermostats manufactured for home temperature control. In a recent test, 10thermostats manufactured by The rmoRite wereselected and placed in a test room thatwas maintained at a temperature of 68oF. The temperature readings of the ten thermostats are shown on the next slide.


Example: Buyer’s Digest (A)


We will use the 10 readings below todevelop a 95% confidence intervalestimate of the population variance.

Example: Buyer’s Digest (A)

Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2

Thermostat 1 2 3 4 5 6 7 8 9 10

Degreesof Freedom .99 .975 .95 .90 .10 .05 .025 .01

5 0.554 0.831 1.145 1.610 9.236 11.070 12.832 15.0866 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.8127 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.4758 1.647 2.180 2.733 3.490 13.362 15.507 17.535 20.0909 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666

10 2.558 3.247 3.940 4.865 15.987 18.307 20.483 23.209

Area in Upper Tail

Selected Values from the Chi-Square Distribution Table

Our value

2.975

For n - 1 = 10 - 1 = 9 d.f. and a = .05


2

0

.025

2

2.0252

( 1)2.700 n s

Area inUpper Tail= .975

2.700

For n - 1 = 10 - 1 = 9 d.f. and a =.05



5 0.554 0.831 1.145 1.610 9.236 11.070 12.832 15.0866 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.8127 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.4758 1.647 2.180 2.733 3.490 13.362 15.507 17.535 20.0909 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666

10 2.558 3.247 3.940 4.865 15.987 18.307 20.483 23.209

Area in Upper Tail


For n - 1 = 10 - 1 = 9 d.f. and a = .05

Our value 2

.025


2

0

.025

2.700

n - 1 = 10 - 1 = 9 degrees of freedom and a = .05

2

2( 1)2.700 19.023n s

19.023

Area in UpperTail = .025


Sample variance s2 provides a point estimate of 2.


.33 < 2 < 2.33

70.93.6

1)( 2

2

nxx

s i

A 95% confidence interval for the population variance is given by:

70.270).110(

02.1970).110( 2

Left-Tailed Test

Hypothesis TestingAbout a Population Variance

22

021

( )n s

22

021 ( )n s

Test Statistic

Hypotheses 2 20 0: H

2 20: aH

where is the hypothesized valuefor the population variance

20

where is based on a chi-squaredistribution with n - 1 d.f.

2)1( a

Left-Tailed Test (continued)

Reject H0 if p-value < ap-Value approach:Critical value approach:

Rejection RuleReject H0 if 2 2

(1 )a


Right-Tailed TestH0

202: H0

202:

Ha : 202Ha : 202

22

021

( )n s

22

021 ( )n s

Test Statistic

Hypotheses


20


Reject H0 if2 2

a

Right-Tailed Test (continued)

Reject H0 if p-value < ap-Value approach:

Critical value approach: Rejection Rule

where is based on a chi-squaredistribution with n - 1 d.f.

2a


Two-Tailed Test

22

021 ( )n s

2

2

021 ( )n s

Test Statistic

HypothesesHa : 2

02Ha : 202

H02

02: H0

202:


20


Two-Tailed Test (continued)


Critical value approach: Rejection Rule

Reject H0 if 2 2 2 2(1 / 2) / 2 or a a

where and are based on achi-square distribution with n - 1 d.f.

2)2/1( a

22/a


Recall that Buyer’s Digest is ratingThermoRite thermostats. Buyer’s Digestgives an “acceptable” rating to a thermo-stat with a temperature variance of 0.5or less.

Example: Buyer’s Digest (B)

We will conduct a hypothesis test (witha = .10) to determine whether the ThermoRitethermostat’s temperature variance is “acceptable”.


Using the 10 readings, we willconduct a hypothesis test (with a = .10)to determine whether the ThermoRitethermostat’s temperature variance is“acceptable”.

Example: Buyer’s Digest (B)

Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2

Thermostat 1 2 3 4 5 6 7 8 9 10


Hypotheses2

0 : 0.5H 2: 0.5aH

Reject H0 if 2 > 14.684

Rejection Rule



5 0.554 0.831 1.145 1.610 9.236 11.070 12.832 15.0866 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.8127 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.4758 1.647 2.180 2.733 3.490 13.362 15.507 17.535 20.0909 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666

10 2.558 3.247 3.940 4.865 15.987 18.307 20.483 23.209

Area in Upper Tail


For n - 1 = 10 - 1 = 9 d.f. and a = .10

Our value 2.10


2

0 14.684

Area in UpperTail = .10

Rejection Region

2 22

2( 1) 9

.5n s s

Reject H0


Test Statistic

2 9(.7) 12.6.5

Because 2 = 12.6 is less than 14.684, we cannot reject H0. The sample variance s2 = .7 is insufficient evidence to conclude that the temperature variance for ThermoRite thermostats is unacceptable.

Conclusion

The sample variance s 2 = 0.7


• The rejection region for the ThermoRite thermostat example is in the upper tail; thus, the appropriate p-value is less than .90 (c 2 = 4.168) and greater than .10 (c 2 = 14.684).

Using Excel to Conduct a Hypothesis Testabout a Population Variance

Using the p-Value

• The sample variance of s 2 = .7 is insufficient evidence to conclude that the temperature variance is unacceptable (>.5).

• Because the p –value > a = .10, we cannot reject the null hypothesis.


Summary

One-Tailed Test

• Test Statistic

• Hypotheses

Hypothesis Testing About theVariances of Two Populations

Denote the population providing thelarger sample variance as population 1.

2 20 1 2: H

2 21 2: aH

21 2

2

sF s

Reject H0 if F > Fa

One-Tailed Test (continued)

Reject H0 if p-value < a

where the value of Fa is based on anF distribution with n1 - 1 (numerator)and n2 - 1 (denominator) d.f.

p-Value approach:

Critical value approach:• Rejection Rule


Two-Tailed Test

Test Statistic

Hypotheses H0 12

22: H0 1

222:

Ha : 12

22Ha : 1

222

Denote the population providing thelarger sample variance as population 1.

21 2

2

sF s


Two-Tailed Test (continued)


Critical value approach:• Rejection Rule

Reject H0 if F > Fa/2where the value of Fa/2 is based on anF distribution with n1 - 1 (numerator)and n2 - 1 (denominator) d.f.


F Distribution

F Distribution

Probability density function

where

F Distribution

mean for d2 > 2

Variance

for d2 > 2

F Distribution with (20, 20) Degrees of Freedom

F Distribution Selected Values From the F Distribution

Table

F Distribution Table

Buyer’s Digest has conducted thesame test, as was described earlier, onanother 10 thermostats, this timemanufactured by TempKing. Thetemperature readings of the tenthermostats are listed on the next slide.

Example: Buyer’s Digest (C)


Example: Buyer’s Digest (C) We will conduct a hypothesis test with a = .10 to see if the variances are equal for ThermoRite’s thermostats and TempKing’s thermostats.


Example: Buyer’s Digest (C)ThermoRite Sample

TempKing Sample

Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2

Thermostat 1 2 3 4 5 6 7 8 9 10

Temperature 67.7 66.4 69.2 70.1 69.5 69.7 68.1 66.6 67.3 67.5

Thermostat 1 2 3 4 5 6 7 8 9 10


The F distribution table (on next slide) shows that with with a = .10, 9 d.f. (numerator), and 9 d.f. (denominator), F.05 = 3.18.

HypothesesH0 1

222: H0 1

222:

Ha : 12

22Ha : 1

222

Reject H0 if F > 3.18

(Their variances are not equal)

(TempKing and ThermoRite thermostats have the same temperature variance)

Rejection Rule


Denominator Area inDegrees Upper

of Freedom Tail 7 8 9 10 158 .10 2.62 2.59 2.56 2.54 2.46

.05 3.50 3.44 3.39 3.35 3.22.025 4.53 4.43 4.36 4.30 4.10.01 6.18 6.03 5.91 5.81 5.52

9 .10 2.51 2.47 2.44 2.42 2.34.05 3.29 3.23 3.18 3.14 3.01

.025 4.20 4.10 4.03 3.96 3.77.01 5.61 5.47 5.35 5.26 4.96

Numerator Degrees of FreedomSelected Values from the F Distribution Table


Test StatisticTempKing’s sample variance is 1.768ThermoRite’s sample variance is .700

= 1.768/.700 = 2.5322

21

ss

F


We cannot reject H0. F = 2.53 < F.05 = 3.18.There is insufficient evidence to conclude thatthe population variances differ for the twothermostat brands.

Conclusion


Determining and Using the p-Value

• Because a = .10, we have p-value > a and therefore we cannot reject the null hypothesis.

• But this is a two-tailed test; after doubling the upper-tail area, the p-value is between .20 and .10.

• Because F = 2.53 is between 2.44 and 3.18, the area in the upper tail of the distribution is between .10 and .05.

Area in Upper Tail .10 .05 .025 .01F Value (df1 = 9, df2 = 9) 2.44 3.18 4.03 5.35



Summary

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