© buddy freeman, 2015 h 0 : h 1 : α = decision rule: if then do not reject h 0, otherwise reject h...
TRANSCRIPT
© Buddy Freeman, 2015
H0:
H1:
α =
Decision Rule:
If
then do not reject H0, otherwise reject H0.
Test Statistic:
Decision:
Conclusion: We have found ________________ evidence at the _____ level of significance that
© Buddy Freeman, 2015
H0:
H1:
α =
Decision Rule:
If
then do not reject H0, otherwise reject H0.
Test Statistic:
Decision:
Conclusion: We have found ________________ evidence at the _____ level of significance that
.05
.05
© Buddy Freeman, 2015
H0:
H1:
α =
Decision Rule:
If
then do not reject H0, otherwise reject H0.
Test Statistic:
Decision:
Conclusion: We have found ________________ evidence at the _____ level of significance that the means of the number of parts produced per hour using methods: A, B, and C are not all equal.
.05
.05
not all the means are equal
μ1 = μ2 = μ3
A B C
© Buddy Freeman, 2014
# of groups?
Parameter?
Can we make all
fe > 5?
Normalpopulations
?
Hartley’s Fmax *(not in text)
Resample and try again.
yes
no
yes
no
chi-squaredf = (R-1)(C-1)pp. 368-374
yes
no Kruskal-Wallis*pp. 621-625
1-way ANOVApp. 386-395ANOVA
OK?
meanor
median
proportion
varianceor
standard deviation
morethan
2
Parameter?
RelatedSamples
?
meanor
median
proportionvariance
orstandard deviation Normal
populations?
yes
no
Levine-Brown-Forsythe
F = S12/S2
2
pp. 344-354
Z for proportionspp. 322-328
yes
no unequal-variances t-testp. 307-315
pooled-variances t-testpp. 307-315
Wilcoxon Rank Sum*pp. 616-621
no
yesNormalpopulations
?
Normalpopulations
?
yes
noyes
non1 > 30
andn2 > 30
?
Z for means with σ1 & σ2 pp. 307-315
yes
no
σ1 and σ2
both known?
no
Normalpopulations
?
yes
noyes
at leastinterval
level data?
yes
noSign Test*pp. 631-634.
Wilcoxon Signed-Ranks*pp. 614-616
paired-difference t-testpp. 315-322
2
chi-square goodness-of-fit test pp. for the Multinomial Experiment 362-368and the Normal Distribution 374-376
2 Groups and > 2 Groups FlowchartSpearman Rank Correlation testpp. 625-630 1
2
3
4
5
7
8
9
10
11
12
13
14
15
yes
no
n1 > 30and
n2 > 30?
6
σ1 = σ2
?
Levine-Brown-Forsythe
Jaggiaand Kelly
(1st edition)
Default case
* means coverage is different from text.
© Buddy Freeman, 2015
ANOVAOK?
There are three major assumptions for doing an Analysis of Variance (ANOVA) Test:
1. Normality: All the populations are normal.2. Equality of Variances: The variances of all the populations are equal.3. Independence of Error: The deviation of each value from the mean of the group containing that value should be independent of any other such deviation.
© Buddy Freeman, 2015
For this part of the course you may assume thethird one (Independence of Error). Care takenwhen obtaining the sample results should minimize the possibility of a violation of thisassumption. However, when data is collectedover a lengthy interval of time, this assumption should be checked. We will discuss this furtherwhen we get into regression analysis.
You are expected to check the othertwo assumptions:
1. Normality2. Equality of Variances
© Buddy Freeman, 2015
This problem has the normality assumption as a given.Later on in the course, we will cover a procedure totest a population to see if it is normally distributed.
How do we determine if all the population variances are equal ?
© Buddy Freeman, 2015
We can use the following hypothesis test:
23
22
21 H0:
H1: not all variances are equal
© Buddy Freeman, 2015
If we do not reject the null hypothesiswe can perform an ANOVA test tocompare the means, but if we do rejectthe null hypothesis, the ANOVA testis NOT appropriate.
If you have one or more of yourpopulations that are definitely NOTnormal the ANOVA test is NOTappropriate.
© Buddy Freeman, 2015
The flowchart can be used todetermine the test statisticneeded to test:
23
22
21 H0:
H1: not all variances are equal
© Buddy Freeman, 2014
# of groups?
Parameter?
Can we make all
fe > 5?
Normalpopulations
?
Hartley’s Fmax *(not in text)
Resample and try again.
yes
no
yes
no
chi-squaredf = (R-1)(C-1)pp. 368-374
yes
no Kruskal-Wallis*pp. 621-625
1-way ANOVApp. 386-395ANOVA
OK?
meanor
median
proportion
varianceor
standard deviation
morethan
2
Parameter?
RelatedSamples
?
meanor
median
proportionvariance
orstandard deviation Normal
populations?
yes
no
Levine-Brown-Forsythe
F = S12/S2
2
pp. 344-354
Z for proportionspp. 322-328
yes
no unequal-variances t-testp. 307-315
pooled-variances t-testpp. 307-315
Wilcoxon Rank Sum*pp. 616-621
no
yesNormalpopulations
?
Normalpopulations
?
yes
noyes
non1 > 30
andn2 > 30
?
Z for means with σ1 & σ2 pp. 307-315
yes
no
σ1 and σ2
both known?
no
Normalpopulations
?
yes
noyes
at leastinterval
level data?
yes
noSign Test*pp. 631-634.
Wilcoxon Signed-Ranks*pp. 614-616
paired-difference t-testpp. 315-322
2
chi-square goodness-of-fit test pp. for the Multinomial Experiment 362-368and the Normal Distribution 374-376
2 Groups and > 2 Groups FlowchartSpearman Rank Correlation testpp. 625-630 1
2
3
4
5
7
8
9
10
11
12
13
14
15
yes
no
n1 > 30and
n2 > 30?
6
σ1 = σ2
?
Levine-Brown-Forsythe
Jaggiaand Kelly
(1st edition)
Default case
* means coverage is different from text.
© Buddy Freeman, 2015
H0:
H1: not all the variances are equal
α =
Decision Rule:
If
then do not reject H0, otherwise reject H0.
Test Statistic:
Decision:
Conclusion: We have found ________________ evidence at the _____ level of significance that that the variances of the number of parts produced per hour using methods: A, B, and C are not all equal.
.05
.05
23
22
21
A B C
2min
2max
,max S
SF
c
© Buddy Freeman, 2015
H0:
H1: not all the variances are equal
α =
Decision Rule:
If
then do not reject H0, otherwise reject H0.
Test Statistic:
Decision:
Conclusion: We have found ________________ evidence at the _____ level of significance that that the variances of the number of parts produced per hour using methods: A, B, and C are not all equal.
.05
.05
23
22
21
A B C
2min
2max
,max S
SF
c
More than 2 groupsis an upper-tail test.
© Buddy Freeman, 2015
Do not reject H0 Reject H0
.05
Do not reject H0 Reject H0
.05
H0:
H1: not all the variances are equal
α =
Decision Rule:
If
then do not reject H0, otherwise reject H0.
Test Statistic:
Decision:
Conclusion: We have found ________________ evidence at the _____ level of significance that that the variances of the number of parts produced per hour using methods: A, B, and C are not all equal.
.05
.05
23
22
21
A B C
2min
2max
,max S
SF
c
More than 2 groupsis an upper-tail test.
Fmax = 5.34
c = number of groups = 39 1 10 1 n
© Buddy Freeman, 2015
Do not reject H0 Reject H0
.05
Do not reject H0 Reject H0
.05
H0:
H1: not all the variances are equal
α =
Decision Rule:
If Fmaxcomputed < 5.34
then do not reject H0, otherwise reject H0.
Test Statistic:
Decision:
Conclusion: We have found ________________ evidence at the _____ level of significance that that the variances of the number of parts produced per hour using methods: A, B, and C are not all equal.
.05
.05
23
22
21
A B CMore than 2 groupsis an upper-tail test.
Fmax = 5.34
c = number of groups = 39 1 10 1 n
2min
2max
,max S
SF
c
© Buddy Freeman, 2015
Calculation of Variations(and Variances)
© Buddy Freeman, 2015
Method Method Method A B C n 10 10 10 X = 900 840 810 X2 = 81,882 72,076 67,048
Calculation of Variations(and Variances)
© Buddy Freeman, 2015
)1(2
ndf
ssssS
Method Method Method A B C n 10 10 10 X = 900 840 810 X2 = 81,882 72,076 67,048
n
xxSS
22 )(
Calculation of Variations(and Variances)
© Buddy Freeman, 2015
9
882
)1(2
ndf
ssssS
Method Method Method A B C n 10 10 10 X = 900 840 810 X2 = 81,882 72,076 67,048
n
xxSS
22 )(
Method A:
88210
90081882
2
SS
Calculation of Variations(and Variances)
© Buddy Freeman, 2015
9
516,1
)1(2
ndf
ssssS
Method Method Method A B C n 10 10 10 X = 900 840 810 X2 = 81,882 72,076 67,048
n
xxSS
22 )(
Method B:
151610
84072076
2
SS
Calculation of Variations(and Variances)
© Buddy Freeman, 2015
9
438,1
)1(2
ndf
ssssS
Method Method Method A B C n 10 10 10 X = 900 840 810 X2 = 81,882 72,076 67,048
n
xxSS
22 )(
Method C:
143810
81067048
2
SS
Calculation of Variations(and Variances)
© Buddy Freeman, 2015
Method Method Method A B C n 10 10 10 X = 900 840 810 X2 = 81,882 72,076 67,048
SS = 882 1,516 1,438
S2 = 882/9 1,516/9 1,438/9
Calculation of Variations(and Variances)
Largestvariance
Smallestvariance
© Buddy Freeman, 2015
Do not reject H0 Reject H0
.05
Do not reject H0 Reject H0
.05
72.1
9
8829
1516
2min
2max
,max
S
SF
c
H0:
H1: not all the variances are equal
α =
Decision Rule:
If Fmaxcomputed < 5.34
then do not reject H0, otherwise reject H0.
Test Statistic:
Decision:
Conclusion: We have found ________________ evidence at the _____ level of significance that that the variances of the number of parts produced per hour using methods: A, B, and C are not all equal.
.05
.05
23
22
21
A B CMore than 2 groupsis an upper-tail test.
Fmax = 5.34
c = number of groups = 39 1 10 1 n
Do not reject H0
Do not reject H0.
insufficient
© Buddy Freeman, 2015
We have found no evidence of any difference in the variancesso the ANOVA test is appropriateto use to test the means.
We now resume our interruptedhypothesis test of the means.
© Buddy Freeman, 2015
H0:
H1:
α =
Decision Rule:
If
then do not reject H0, otherwise reject H0.
Test Statistic:
Decision:
Conclusion: We have found ________________ evidence at the _____ level of significance that the means of the number of parts produced per hour using methods: A, B, and C are not all equal.
.05
.05
not all the means are equal
μ1 = μ2 = μ3
A B C
where k = # of groupsand nT = n1 + n2 + … + nk
2
2
,1W
Bknk S
SF
T
)( kn
SSW
T
)1(k
SSB
More than 2 groupsis an upper-tail test.
© Buddy Freeman, 2015
Do not reject H0 Reject H0
.050
Do not reject H0 Reject H0
.050
H0:
H1:
α =
Decision Rule:
If
then do not reject H0, otherwise reject H0.
Test Statistic:
Decision:
Conclusion: We have found ________________ evidence at the _____ level of significance that the means of the number of parts produced per hour using methods: A, B, and C are not all equal.
.05
.05
not all the means are equal
μ1 = μ2 = μ3
A B C
where k = # of groupsand nT = n1 + n2 + … + nk
2
2
,1W
Bknk S
SF
T
)( kn
SSW
T
)1(k
SSB
dfd = nT - k = 30 - 3 = 27
dfn = k - 1 = 3 - 1 = 2
F = 3.354
More than 2 groupsis an upper-tail test.
© Buddy Freeman, 2015
Do not reject H0 Reject H0
.050
Do not reject H0 Reject H0
.050
H0:
H1:
α =
Decision Rule:
If Fcomputed < 3.354
then do not reject H0, otherwise reject H0.
Test Statistic:
Decision:
Conclusion: We have found ________________ evidence at the _____ level of significance that the means of the number of parts produced per hour using methods: A, B, and C are not all equal.
.05
.05
not all the means are equal
μ1 = μ2 = μ3
A B C
where k = # of groupsand nT = n1 + n2 + … + nk
2
2
,1W
Bknk S
SF
T
)( kn
SSW
T
)1(k
SSB
dfd = nT - k = 30 - 3 = 27
dfn = k - 1 = 3 - 1 = 2
F = 3.354
More than 2 groupsis an upper-tail test.
© Buddy Freeman, 2015
Calculation of the ANOVA test statistic: The Denominator Variance
The denominator is the variance within the groups,but remember variance = variation/df.
Therefore, (SW)2 = SSW/dfW.
SSW is the variation WITHIN the groups =variation WITHIN group 1 + variation WITHIN group 2+ … + variation WITHIN group k = SS1 + SS2 + … + SSk.Here k = 3, so SSW = SS1 + SS2 + SS3 = 882 + 1516 + 1438 = 3836.
dfW = df WITHIN group 1 + df WITHIN group 2 + … +df WITHIN group k = (n1 - 1) + (n2 - 1) + (n3 - 1) = (nT - k) = 27.
© Buddy Freeman, 2015
Calculation of the ANOVA test statistic: The Numerator Variance
The numerator is the variance between the groups,but remember variance = variation/df. Therefore, (SB)2 = SSB/dfB.
THE CONCEPTHere the group is the ‘entity.’ If you have k groups then you have(k-1) degrees of freedom, so dfB = (k-1).
Recall that the variation of a sample is defined as: SS = å(X-X)2. We will expand this concept to groups. The X represents the group, so we will replace X with a value to represent the group, the mean of the group, X. In the sample variation the Xrepresents the mean of all the X values, so we will replace it withthe mean of all values in all the groups, X. The catch is that youmust remember to multiply each square by the sample size of the group to account all the values of each group appropriately.
© Buddy Freeman, 2015
The definition formula for SSB is given by:
The means of each group are:
SSB = 10(90-85)2 + 10(84-85)2 + 10(81-85)2 = 420
The next slide shows an alternative way to calculate SSB.
21
XXnSSB j
k
jj
85
30
2550
101010
810840900
Here, groups. theallin values theall ofmean theis
X
X
8110
810 84
10
840 90
10
900321 XXX
Calculating SSB by Definition
© Buddy Freeman, 2015
SSB + SSW = SST, the total variation.Therefore, SSB = SST – SSW = SST – 3,836.To calculate SST, just add a column with the totals.
Method Method Method A B C Total n 10 10 10 30 X = 900 840 810 2,550 X2 = 81,882 72,076 67,048 221,006
n
xxSS
22 )(
Total:
256,430
550,2006,221
2
SST
Hence, SSB = SST – SSW = 4,256 – 3,836 = 420.With SSB, SSW, and the dfs, the test statistic may be computed.
Alternative Way to Calculate SSB
© Buddy Freeman, 2015
478.1074.142
210
273836
2420
12
2
,1
knSSW
kSSB
S
SF
T
W
Bknk T
Do not reject H0 Reject H0
.050
Do not reject H0 Reject H0
.050
H0:
H1:
α =
Decision Rule:
If Fcomputed < 3.354
then do not reject H0, otherwise reject H0.
Test Statistic:
Decision:
Conclusion: We have found ________________ evidence at the _____ level of significance that the means of the number of parts produced per hour using methods: A, B, and C are not all equal.
.05
.05
not all the means are equal
μ1 = μ2 = μ3
A B C
dfd = nT - k = 30 - 3 = 27
dfn = k - 1 = 3 - 1 = 2
F = 3.354
More than 2 groupsis an upper-tail test.
Do not reject H0
Do not reject H0.
insufficient