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1 1 Slide Slide
Chapter 11Chapter 11 Inferences About Population Variances Inferences About Population Variances
Inference about a Population VarianceInference about a Population Variance Inferences about Two Population VariancesInferences about Two Population Variances
2 2 Slide Slide
Inferences About a Population VarianceInferences About a Population Variance
Chi-Square DistributionChi-Square Distribution Interval EstimationInterval Estimation Hypothesis TestingHypothesis Testing
3 3 Slide Slide
Chi-Square DistributionChi-Square Distribution
We can use the chi-square distribution to developWe can use the chi-square distribution to develop interval estimates and conduct hypothesis testsinterval estimates and conduct hypothesis tests about a population variance.about a population variance.
The sampling distribution of (The sampling distribution of (nn - 1) - 1)ss22//22 has a has a chi-chi- square distribution whenever a simple random square distribution whenever a simple random samplesample of size of size nn is selected from a normal population. is selected from a normal population.
The chi-square distribution is based on samplingThe chi-square distribution is based on sampling from a normal population.from a normal population.
The The chi-square distributionchi-square distribution is the sum of is the sum of squaredsquared standardized normal random variables standardized normal random variables such assuch as
((zz11))22+(+(zz22))22+(+(zz33))22 and so on. and so on.
4 4 Slide Slide
Examples of Sampling Distribution of (Examples of Sampling Distribution of (nn - - 1)1)ss22//22
00
With 2 degreesWith 2 degrees of freedomof freedomWith 2 degreesWith 2 degrees of freedomof freedom
2
2
( 1)n s
With 5 degreesWith 5 degrees of freedomof freedomWith 5 degreesWith 5 degrees of freedomof freedom
With 10 degreesWith 10 degrees of freedomof freedomWith 10 degreesWith 10 degrees of freedomof freedom
5 5 Slide Slide
6 6 Slide Slide
2 2 2.975 .025 2 2 2.975 .025
Chi-Square DistributionChi-Square Distribution
For example, there is a .95 probability of For example, there is a .95 probability of obtaining a obtaining a 22 (chi-square) value such that (chi-square) value such that
We will use the notation to denote the We will use the notation to denote the value for the chi-square distribution that value for the chi-square distribution that provides an area of provides an area of to the right of the stated to the right of the stated value. value.
2 2
2 2
7 7 Slide Slide
95% of thepossible 2 values 95% of thepossible 2 values
22
00
.025.025
2.025 2.025
.025.025
2.975 2.975
Interval Estimation of Interval Estimation of 22
22 2.975 .0252
( 1)n s
2
2 2.975 .0252
( 1)n s
8 8 Slide Slide
Interval Estimation of Interval Estimation of 22
( ) ( )
/ ( / )
n s n s
1 12
22
22
1 22
( ) ( )
/ ( / )
n s n s
1 12
22
22
1 22
2 2 2(1 / 2) / 2 2 2 2(1 / 2) / 2
22 2(1 / 2) / 22
( 1)n s
2
2 2(1 / 2) / 22
( 1)n s
Substituting (Substituting (nn – 1) – 1)ss22//22 for the for the 22 we get we get
Performing algebraic manipulation we getPerforming algebraic manipulation we get
There is a (1 – There is a (1 – ) probability of obtaining a ) probability of obtaining a 22 valuevalue
such thatsuch that
9 9 Slide Slide
Interval Estimate of a Population VarianceInterval Estimate of a Population Variance
Interval Estimation of Interval Estimation of 22
( ) ( )
/ ( / )
n s n s
1 12
22
22
1 22
( ) ( )
/ ( / )
n s n s
1 12
22
22
1 22
where the where the values are based on a chi-squarevalues are based on a chi-square
distribution with distribution with nn - 1 degrees of freedom and - 1 degrees of freedom and
where 1 - where 1 - is the confidence coefficient. is the confidence coefficient.
10 10 Slide Slide
Interval Estimation of Interval Estimation of
Interval Estimate of a Population Standard DeviationInterval Estimate of a Population Standard Deviation
Taking the square root of the upper and lowerTaking the square root of the upper and lower
limits of the variance interval provides the confidencelimits of the variance interval provides the confidence
interval for the population standard deviation.interval for the population standard deviation.
2 2
2 2/ 2 (1 / 2)
( 1) ( 1)n s n s
2 2
2 2/ 2 (1 / 2)
( 1) ( 1)n s n s
11 11 Slide Slide
Buyer’s Digest rates thermostatsBuyer’s Digest rates thermostats
manufactured for home temperaturemanufactured for home temperature
control. In a recent test, 10 thermostatscontrol. In a recent test, 10 thermostats
manufactured by ThermoRite weremanufactured by ThermoRite were
selected and placed in a test room thatselected and placed in a test room that
was maintained at a temperature of 68was maintained at a temperature of 68ooF.F.
The temperature readings of the ten thermostats The temperature readings of the ten thermostats areare
shown on the next slide. shown on the next slide.
Interval Estimation of Interval Estimation of 22
Example: Buyer’s Digest (A)Example: Buyer’s Digest (A)
12 12 Slide Slide
Interval Estimation of Interval Estimation of 22
We will use the 10 readings below toWe will use the 10 readings below todevelop a 95% confidence intervaldevelop a 95% confidence intervalestimate of the population variance.estimate of the population variance.
Example: Buyer’s Digest (A)Example: Buyer’s Digest (A)
Temperature Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.267.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2
ThermostatThermostat 1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10
13 13 Slide Slide
Degrees
of 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 TailDegrees
of 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
Interval Estimation of Interval Estimation of 22
Selected Values from the Chi-Square Distribution TableSelected Values from the Chi-Square Distribution Table
Our Our value value
Our Our value value
2.9752.975
For For nn - 1 = 10 - 1 = 9 d.f. and - 1 = 10 - 1 = 9 d.f. and = .05= .05
14 14 Slide Slide
Interval Estimation of Interval Estimation of 22
22
00
.025.025
2
2.0252
( 1)2.700
n s
2
2.0252
( 1)2.700
n s
Area inArea inUpper TailUpper Tail
= .975= .975
2.7002.700
For For nn - 1 = 10 - 1 = 9 d.f. and - 1 = 10 - 1 = 9 d.f. and = .05= .05
15 15 Slide Slide
Degrees
of 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 TailDegrees
of 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
Interval Estimation of Interval Estimation of 22
Selected Values from the Chi-Square Distribution TableSelected Values from the Chi-Square Distribution Table
For For nn - 1 = 10 - 1 = 9 d.f. and - 1 = 10 - 1 = 9 d.f. and = .05= .05
Our Our value value
Our Our value value 2
.025 2
.025
16 16 Slide Slide
22
00
.025.025
2.7002.700
Interval Estimation of Interval Estimation of 22
nn - 1 = 10 - 1 = 9 degrees of freedom and - 1 = 10 - 1 = 9 degrees of freedom and = .05= .05
2
2
( 1)2.700 19.023
n s
2
2
( 1)2.700 19.023
n s
19.02319.023
Area in UpperArea in UpperTail = .025Tail = .025Area in UpperArea in UpperTail = .025Tail = .025
17 17 Slide Slide
Sample variance Sample variance ss22 provides a point estimate of provides a point estimate of 22..
sx xni2
2
16 39
70
( ) .
.sx xni2
2
16 39
70
( ) .
.
( )..
( )..
10 1 7019 02
10 1 702 70
2 ( ).
.( ).
.10 1 70
19 0210 1 70
2 702
Interval Estimation of Interval Estimation of 22
.33 .33 << 2 2 << 2.33 2.33
A 95% confidence interval for the population A 95% confidence interval for the population variance is given by:variance is given by:
18 18 Slide Slide
Hypothesis TestingHypothesis TestingAbout a Population VarianceAbout a Population Variance
Lower-tail test:
H0: σ2 σ02
H1: σ2 < σ02
Upper-tail test:
H0: σ2 ≤ σ02
H1: σ2 > σ02
Two-tail test:
H0: σ2 = σ02
H1: σ2 ≠ σ02
/2 /2
Reject H0 ifReject H0 if Reject H0 if
or
2, 1n χ
2,1 1n χ
2,1 1n 2/χ
2, 1n 2/χ
2,1 1n
21n χχ
2, 1n
21n χχ
2, 1n
21n 2/ χχ
2,1 1n
21n 2/ χχ
22
02
1
( )n s
22
02
1
( )n s
19 19 Slide Slide
Recall that Buyer’s Digest is ratingRecall that Buyer’s Digest is rating
ThermoRite thermostats. Buyer’s DigestThermoRite thermostats. Buyer’s Digest
gives an “acceptable” rating to a thermo-gives an “acceptable” rating to a thermo-
stat with a temperature variance of 0.5stat with a temperature variance of 0.5
or less.or less.
Hypothesis TestingHypothesis TestingAbout a Population VarianceAbout a Population Variance
Example: Buyer’s Digest (B)Example: Buyer’s Digest (B)
We will conduct a hypothesis test (withWe will conduct a hypothesis test (with
= .10) to determine whether the ThermoRite= .10) to determine whether the ThermoRite
thermostat’s temperature variance is “acceptable”.thermostat’s temperature variance is “acceptable”.
20 20 Slide Slide
Hypothesis TestingHypothesis TestingAbout a Population VarianceAbout a Population Variance
Using the 10 readings, we willUsing the 10 readings, we will
conduct a hypothesis test (with conduct a hypothesis test (with = .10) = .10)
to determine whether the ThermoRiteto determine whether the ThermoRite
thermostat’s temperature variance isthermostat’s temperature variance is
““acceptable”.acceptable”.
Example: Buyer’s Digest (B)Example: Buyer’s Digest (B)
Temperature Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.267.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2
ThermostatThermostat 1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10
21 21 Slide Slide
HypothesesHypotheses2
0 : 0.5H 20 : 0.5H
2: 0.5aH 2: 0.5aH
Hypothesis TestingHypothesis TestingAbout a Population VarianceAbout a Population Variance
Reject Reject HH00 if if 22 >> 14.684 14.684
Rejection RuleRejection Rule
22 22 Slide Slide
Degrees
of 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 TailDegrees
of 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 TailSelected Values from the Chi-Square Distribution TableSelected Values from the Chi-Square Distribution Table
For For nn - 1 = 10 - 1 = 9 d.f. and - 1 = 10 - 1 = 9 d.f. and = .10= .10
Hypothesis TestingHypothesis TestingAbout a Population VarianceAbout a Population Variance
Our Our value value Our Our value value 2
.10 2
.10
23 23 Slide Slide
22
00 14.68414.684
Area in UpperArea in UpperTail = .10Tail = .10Area in UpperArea in UpperTail = .10Tail = .10
Hypothesis TestingHypothesis TestingAbout a Population VarianceAbout a Population Variance
Rejection RegionRejection Region
2 22
2
( 1) 9.5
n s s
2 2
22
( 1) 9.5
n s s
Reject Reject HH00Reject Reject HH00
24 24 Slide Slide
Test StatisticTest Statistic
2 9(.7)12.6
.5 2 9(.7)
12.6.5
Hypothesis TestingHypothesis TestingAbout a Population VarianceAbout a Population Variance
Because Because 22 = 12.6 is less than 14.684, we cannot = 12.6 is less than 14.684, we cannot
reject reject HH00. The sample variance . The sample variance ss22 = .7 is insufficient = .7 is insufficient
evidence to conclude that the temperature varianceevidence to conclude that the temperature variance
for ThermoRite thermostats is unacceptable.for ThermoRite thermostats is unacceptable.
ConclusionConclusion
The sample variance The sample variance s s 22 = 0.7 = 0.7
25 25 Slide Slide
Using the Using the pp-Value-Value
• The sample variance of The sample variance of s s 22 = .7 is = .7 is insufficient evidence to conclude that theinsufficient evidence to conclude that the temperature variance is unacceptable (>.5).temperature variance is unacceptable (>.5).
• Because the Because the pp –value > –value > = .10, we = .10, we cannot reject the null hypothesis.cannot reject the null hypothesis.
• The rejection region for the ThermoRite The rejection region for the ThermoRite thermostat example is in the upper tail; thus, thethermostat example is in the upper tail; thus, the appropriate appropriate pp-value is less than .90 (-value is less than .90 (22 = 4.168) = 4.168) and greater than .10 (and greater than .10 (22 = 14.684). = 14.684).
Hypothesis TestingHypothesis TestingAbout a Population VarianceAbout a Population Variance
A precise A precise pp-value-valuecan be found usingcan be found usingMinitab or Excel.Minitab or Excel.
A precise A precise pp-value-valuecan be found usingcan be found usingMinitab or Excel.Minitab or Excel.
26 26 Slide Slide
27 27 Slide Slide
28 28 Slide Slide
Hypothesis Tests for Two VariancesHypothesis Tests for Two Variances
Tests for TwoPopulation Variances
F test statistic
H0: σx2 = σy
2
H1: σx2 ≠ σy
2Two-tail test
Lower-tail test
Upper-tail test
H0: σx2 σy
2
H1: σx2 < σy
2
H0: σx2 ≤ σy
2
H1: σx2 > σy
2
Goal: Test hypotheses about two population variances
The two populations are assumed to be independent and normally distributed
29 29 Slide Slide
FF 分布分布
FF 分布定义为分布定义为 :: 设设 XX 、、 YY 为两个独立的随机变量为两个独立的随机变量,, XX 服从自由度为服从自由度为 mm 的卡方分布,的卡方分布, YY 服从自由服从自由度为度为 nn 的卡方分布,这的卡方分布,这 2 2 个独立的卡方分布被个独立的卡方分布被各自的自由度除以后的比率这一统计量的分布即各自的自由度除以后的比率这一统计量的分布即::
F=F= (( x/mx/m )) /(y/n)/(y/n)
服从自由度为(服从自由度为( m,n)m,n) 的的 F-F- 分布, 上式分布, 上式 FF 服从服从第一自由度为第一自由度为 mm ,第二自由度为,第二自由度为 nn 的的 FF 分布 分布
30 30 Slide Slide
FF 分布分布
FF 分布的性质 分布的性质 11 、它是一种非对称分布; 、它是一种非对称分布; 22 、它有两个自由度,即、它有两个自由度,即 n1 -1n1 -1 和和 n2-1n2-1 ,相应的,相应的
分布记为分布记为 FF ( ( n1 –1n1 –1 , , n2-1n2-1 ), ), n1 –1n1 –1 通常通常称为分子自由度, 称为分子自由度, n2-1n2-1 通常称为分母自由度; 通常称为分母自由度;
33 、、 FF 分布是一个以自由度分布是一个以自由度 n1 –1n1 –1 和和 n2-1n2-1 为参数为参数的分布族,不同的自由度决定了的分布族,不同的自由度决定了 F F 分布的形状。 分布的形状。
44 、、 FF 分布的倒数性质:分布的倒数性质: FFα,df1,df2α,df1,df2=1/F=1/F1-1-α,df2,df1α,df2,df1
31 31 Slide Slide
FF 分布分布
32 32 Slide Slide
Hypothesis Tests for Two VariancesHypothesis Tests for Two Variances
Tests for TwoPopulation Variances
F test statistic
2y
2y
2x
2x
/σs
/σsF
The random variable
Has an F distribution with (nx – 1) numerator degrees of freedom and (ny – 1) denominator degrees of freedom
Denote an F value with 1 numerator and 2
denominator degrees of freedom by
(continued)
33 33 Slide Slide
Test StatisticTest Statistic
Tests for TwoPopulation Variances
F test statistic 2y
2x
s
sF
The critical value for a hypothesis test about two population variances is
where F has (nx – 1) numerator degrees of freedom and (ny – 1) denominator degrees of freedom
34 34 Slide Slide
One-Tailed TestOne-Tailed Test
•Test StatisticTest Statistic
•HypothesesHypotheses
Hypothesis Testing About theHypothesis Testing About theVariances of Two PopulationsVariances of Two Populations
Denote the population providing theDenote the population providing thelarger sample variance as population 1.larger sample variance as population 1.
2 20 1 2: H 2 20 1 2: H
2 21 2: aH 2 21 2: aH
21
22
sFs
21
22
sFs
35 35 Slide Slide
One-Tailed Test (continued)One-Tailed Test (continued)
Reject Reject HH00 if if pp-value -value <<
where the value of where the value of FF is based on anis based on an
F F distribution with distribution with nn11 - 1 (numerator) - 1 (numerator)
and and nn2 2 - 1 (denominator) d.f.- 1 (denominator) d.f.
pp-Value approach:-Value approach:
Critical value approach:Critical value approach:•Rejection RuleRejection Rule
Hypothesis Testing About theHypothesis Testing About theVariances of Two PopulationsVariances of Two Populations
Reject Reject HH00 if if FF >> FF
36 36 Slide Slide
Two-Tailed TestTwo-Tailed Test
•Test StatisticTest Statistic
•HypothesesHypotheses
Hypothesis Testing About theHypothesis Testing About theVariances of Two PopulationsVariances of Two Populations
H0 12
22: H0 1
222:
Ha : 12
22Ha : 1
222
Denote the population providing theDenote the population providing thelarger sample variance as population 1.larger sample variance as population 1.
21
22
sFs
21
22
sFs
37 37 Slide Slide
虽然我们可以通过以上公式求左侧虽然我们可以通过以上公式求左侧 FF 值,但通常在进行假设检验值,但通常在进行假设检验计算时,只需知道右侧计算时,只需知道右侧 FF 值。进行 假设检验时,可以随意值。进行 假设检验时,可以随意规定哪个总体是总体规定哪个总体是总体 11 或总体或总体 22 ,我们用总体,我们用总体 11 来代表方差较大来代表方差较大的总体时,只有在右侧才有可能出现的总体时,只有在右侧才有可能出现 HH00 被拒绝的情况。虽然左被拒绝的情况。虽然左侧临界值仍存在,但我们不需要知道它值,因为使用方差较大的侧临界值仍存在,但我们不需要知道它值,因为使用方差较大的总体作为总体总体作为总体 11 ,, ss11
22/s/s2222 往往出现在右侧。往往出现在右侧。
38 38 Slide Slide
Two-Tailed Test (continued)Two-Tailed Test (continued)
Reject Reject HH00 if if pp-value -value << pp-Value approach:-Value approach:
Critical value approach:Critical value approach:•Rejection RuleRejection Rule
Hypothesis Testing About theHypothesis Testing About theVariances of Two PopulationsVariances of Two Populations
Reject Reject HH00 if if FF >> FF/2/2
where the value of where the value of FF/2 /2 is based on anis based on an
F F distribution with distribution with nn11 - 1 (numerator) - 1 (numerator)
and and nn2 2 - 1 (denominator) d.f.- 1 (denominator) d.f.
39 39 Slide Slide
Decision Rules: Two VariancesDecision Rules: Two Variances
rejection region for a two-tail test is:
F 0
Reject H0Do not reject H0
F 0 /2
Reject H0Do not reject H0
H0: σx2 = σy
2
H1: σx2 ≠ σy
2H0: σx2 ≤ σy
2
H1: σx2 > σy
2
Use sx2 to denote the larger variance.
α1,n1,n yxF
2/α1,n1,n0 yxFF if H Reject
2/α1,n1,n yxF
where sx2 is the larger of the
two sample variances
α1,n1,n0 yxFF if H Reject
40 40 Slide Slide
Buyer’s Digest has conducted theBuyer’s Digest has conducted the
same test, as was described earlier, onsame test, as was described earlier, on
another 10 thermostats, this timeanother 10 thermostats, this time
manufactured by TempKing. Themanufactured by TempKing. The
temperature readings of the tentemperature readings of the ten
thermostats are listed on the next slide. thermostats are listed on the next slide.
Hypothesis Testing About theHypothesis Testing About theVariances of Two PopulationsVariances of Two Populations
Example: Buyer’s Digest (C)Example: Buyer’s Digest (C)
We will conduct a hypothesis test with We will conduct a hypothesis test with = .10 = .10 to seeto seeif the variances are equal for ThermoRite’s if the variances are equal for ThermoRite’s thermostatsthermostatsand TempKing’s thermostats.and TempKing’s thermostats.
41 41 Slide Slide
Hypothesis Testing About theHypothesis Testing About theVariances of Two PopulationsVariances of Two Populations
Example: Buyer’s Digest (C)Example: Buyer’s Digest (C)
ThermoRite SampleThermoRite Sample
TempKing SampleTempKing Sample
Temperature Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.267.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2
ThermostatThermostat 1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10
Temperature Temperature 67.7 66.4 69.2 70.1 69.5 69.7 68.1 66.6 67.3 67.567.7 66.4 69.2 70.1 69.5 69.7 68.1 66.6 67.3 67.5
ThermostatThermostat 1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10
42 42 Slide Slide
HypothesesHypotheses
H0 12
22: H0 1
222:
Ha : 12
22Ha : 1
222
Hypothesis Testing About theHypothesis Testing About theVariances of Two PopulationsVariances of Two Populations
Reject Reject HH00 if if FF >> 3.18 3.18
The The FF distribution table (on next slide) shows distribution table (on next slide) shows that withthat withwith with = .10, 9 d.f. (numerator), and 9 d.f. = .10, 9 d.f. (numerator), and 9 d.f. (denominator),(denominator),
FF.05.05 = 3.18. = 3.18.
(Their variances are not equal)(Their variances are not equal)
(TempKing and ThermoRite thermostats(TempKing and ThermoRite thermostatshave the same temperature variance)have the same temperature variance)
Rejection RuleRejection Rule
43 43 Slide Slide
Denominator Area in
Degrees 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 Freedom
Denominator Area in
Degrees 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 Freedom
Selected Values from the Selected Values from the FF Distribution Table Distribution Table
Hypothesis Testing About theHypothesis Testing About theVariances of Two PopulationsVariances of Two Populations
44 44 Slide Slide
Test StatisticTest Statistic
Hypothesis Testing About theHypothesis Testing About theVariances of Two PopulationsVariances of Two Populations
We We cannotcannot reject reject HH00. . FF = 2.53 < = 2.53 < FF.05.05 = 3.18. = 3.18.There is insufficient evidence to conclude thatThere is insufficient evidence to conclude thatthe population variances differ for the twothe population variances differ for the twothermostat brands.thermostat brands.
ConclusionConclusion
21
22
sFs
21
22
sFs
= 1.768/.700 = 2.53= 1.768/.700 = 2.53
TempKing’s sample variance is 1.768TempKing’s sample variance is 1.768
ThermoRite’s sample variance is .700ThermoRite’s sample variance is .700
45 45 Slide Slide
Determining and Using the Determining and Using the pp-Value-Value
Hypothesis Testing About theHypothesis Testing About theVariances of Two PopulationsVariances of Two Populations
• Because Because = .10, we have = .10, we have pp-value > -value > and therefore and therefore we cannot reject the null hypothesis.we cannot reject the null hypothesis.
• But this is a two-tailed test; after doubling the But this is a two-tailed test; after doubling the upper-upper- tail area, the tail area, the pp-value is between .20 and .10. -value is between .20 and .10. (A precise(A precise pp-value can be found using Minitab or Excel.)-value can be found using Minitab or Excel.)
• Because Because FF = 2.53 is between 2.44 and 3.18, the area = 2.53 is between 2.44 and 3.18, the area in the upper tail of the distribution is between .10in the upper tail of the distribution is between .10 and .05.and .05.
Area in Upper TailArea in Upper Tail .10 .05 .025 .01 .10 .05 .025 .01
FF Value (df Value (df11 = 9, df = 9, df22 = 9) 2.44 3.18 4.03 5.35 = 9) 2.44 3.18 4.03 5.35
46 46 Slide Slide
47 47 Slide Slide
48 48 Slide Slide
End of Chapter 11End of Chapter 11