of 32 /32
© Buddy Freeman, 2015 H 0 : H 1 : α = Decision Rule: If then do not reject H 0 , otherwise reject H 0 . Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that
• Author

oliver-nichols
• Category

## Documents

• view

31

2

Embed Size (px)

description

H 0 : H 1 : α = Decision Rule: If then do not reject H 0 , otherwise reject H 0 . Test Statistic: Decision: Conclusion: We have found ________________ evidence at the _____ level of significance that. H 0 : H 1 : α = Decision Rule: If - PowerPoint PPT Presentation

### Transcript of H 0 : H 1 : α = Decision Rule: If then do not reject H 0 , otherwise reject H 0 . Test Statistic:

Segment 2 - HT-6

H0:H1: =

Decision Rule:Ifthen do not reject H0, otherwise reject H0.Test Statistic:

Decision:Conclusion: We have found ________________ evidence at the _____ level of significance that Buddy Freeman, 2015

1H0:H1: =

Decision Rule:Ifthen 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

2H0:H1: =

Decision Rule:Ifthen 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..05.05not all the means are equal1 = 2 = 3 A B C Buddy Freeman, 2015

# of groups?Parameter?Can we make allfe > 5?Normalpopulations?Hartleys Fmax *(not in text)Resample and try again.yesnoyesnochi-squaredf = (R-1)(C-1)pp. 368-374yesnoKruskal-Wallis*pp. 621-6251-way ANOVApp. 386-395ANOVAOK?meanormedianproportionvarianceorstandard deviationmorethan2Parameter?RelatedSamples?meanormedianproportionvarianceorstandard deviationNormalpopulations?yesnoLevine-Brown-ForsytheF = S12/S22pp. 344-354Z for proportionspp. 322-328yesnounequal-variances t-testp. 307-315pooled-variances t-testpp. 307-315Wilcoxon Rank Sum*pp. 616-621noyesNormalpopulations?Normalpopulations?yesnoyesnon1 > 30andn2 > 30?Z for means with 1 & 2 pp. 307-315yesno1 and 2both known?noNormalpopulations?yesnoyesat leastinterval level data?yesnoSign Test*pp. 631-634.Wilcoxon Signed-Ranks*pp. 614-616paired-difference t-testpp. 315-3222chi-square goodness-of-fit test pp. for the Multinomial Experiment 362-368and the Normal Distribution 374-3762 Groups and > 2 Groups FlowchartSpearman Rank Correlation testpp. 625-63012345789101112131415yesnon1 > 30andn2 > 30?61 = 2?Levine-Brown-ForsytheJaggiaand Kelly(1st edition)Default case* means coverage is different from text. Buddy Freeman, 20144ANOVAOK?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

5For 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

6This 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

7We can use the following hypothesis test:

H0:H1: not all variances are equal Buddy Freeman, 2015

8If 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

9The flowchart can be used todetermine the test statisticneeded to test:

H0:H1: not all variances are equal Buddy Freeman, 2015

10# of groups?Parameter?Can we make allfe > 5?Normalpopulations?Hartleys Fmax *(not in text)Resample and try again.yesnoyesnochi-squaredf = (R-1)(C-1)pp. 368-374yesnoKruskal-Wallis*pp. 621-6251-way ANOVApp. 386-395ANOVAOK?meanormedianproportionvarianceorstandard deviationmorethan2Parameter?RelatedSamples?meanormedianproportionvarianceorstandard deviationNormalpopulations?yesnoLevine-Brown-ForsytheF = S12/S22pp. 344-354Z for proportionspp. 322-328yesnounequal-variances t-testp. 307-315pooled-variances t-testpp. 307-315Wilcoxon Rank Sum*pp. 616-621noyesNormalpopulations?Normalpopulations?yesnoyesnon1 > 30andn2 > 30?Z for means with 1 & 2 pp. 307-315yesno1 and 2both known?noNormalpopulations?yesnoyesat leastinterval level data?yesnoSign Test*pp. 631-634.Wilcoxon Signed-Ranks*pp. 614-616paired-difference t-testpp. 315-3222chi-square goodness-of-fit test pp. for the Multinomial Experiment 362-368and the Normal Distribution 374-3762 Groups and > 2 Groups FlowchartSpearman Rank Correlation testpp. 625-63012345789101112131415yesnon1 > 30andn2 > 30?61 = 2?Levine-Brown-ForsytheJaggiaand Kelly(1st edition)Default case* means coverage is different from text. Buddy Freeman, 201411H0:H1: not all the variances are equal =

Decision Rule:Ifthen 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

A B C

Buddy Freeman, 2015

12H0:H1: not all the variances are equal =

Decision Rule:Ifthen 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

A B C

More than 2 groupsis an upper-tail test. Buddy Freeman, 2015

13

H0:H1: not all the variances are equal =

Decision Rule:Ifthen 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

A B C

More than 2 groupsis an upper-tail test.Fmax = 5.34c = number of groups = 3

Buddy Freeman, 2015

14

H0:H1: not all the variances are equal =

Decision Rule:If Fmaxcomputed < 5.34then 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

A B CMore than 2 groupsis an upper-tail test.Fmax = 5.34c = number of groups = 3

Buddy Freeman, 2015

15Calculation of Variations(and Variances) Buddy Freeman, 2015

16 Method Method Method A B C n 10 10 10 X = 900 840 810 X2 = 81,882 72,076 67,048Calculation of Variations(and Variances) Buddy Freeman, 2015

17

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

18

Method Method Method A B C n 10 10 10 X = 900 840 810 X2 = 81,882 72,076 67,048

Method A:

Calculation of Variations(and Variances) Buddy Freeman, 2015

19

Method Method Method A B C n 10 10 10 X = 900 840 810 X2 = 81,882 72,076 67,048

Method B:

Calculation of Variations(and Variances) Buddy Freeman, 2015

20

Method Method Method A B C n 10 10 10 X = 900 840 810 X2 = 81,882 72,076 67,048

Method C:

Calculation of Variations(and Variances) Buddy Freeman, 2015

21 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 1,438/9Calculation of Variations(and Variances)LargestvarianceSmallestvariance Buddy Freeman, 2015

22

H0:H1: not all the variances are equal =

Decision Rule:If Fmaxcomputed < 5.34then 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

A B CMore than 2 groupsis an upper-tail test.Fmax = 5.34c = number of groups = 3

Do not reject H0 Do not reject H0.insufficient Buddy Freeman, 2015

23We 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

24H0:H1: =

Decision Rule:Ifthen 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..05.05not all the means are equal1 = 2 = 3 A B Cwhere k = # of groupsand nT = n1 + n2 + + nk

More than 2 groupsis an upper-tail test. Buddy Freeman, 2015

H0:H1: =

Decision Rule:Ifthen 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..05.05not all the means are equal1 = 2 = 3 A B Cwhere k = # of groupsand nT = n1 + n2 + + nk

dfd = nT - k = 30 - 3 = 27dfn = k - 1 = 3 - 1 = 2F = 3.354More than 2 groupsis an upper-tail test. Buddy Freeman, 2015

H0:H1: =

Decision Rule:If Fcomputed < 3.354then 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..05.05not all the means are equal1 = 2 = 3 A B Cwhere k = # of groupsand nT = n1 + n2 + + nk

dfd = nT - k = 30 - 3 = 27dfn = k - 1 = 3 - 1 = 2F = 3.354More than 2 groupsis an upper-tail test. Buddy Freeman, 2015

Calculation of the ANOVA test statistic: The Denominator VarianceThe 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

28Calculation of the ANOVA test statistic: The Numerator VarianceThe 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 = (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

29The 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.

Calculating SSB by Definition Buddy Freeman, 2015

30SSB + 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

Total:

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

H0:H1: =

Decision Rule:If Fcomputed < 3.354then 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..05.05not all the means are equal1 = 2 = 3 A B Cdfd = nT - k = 30 - 3 = 27dfn = k - 1 = 3 - 1 = 2F = 3.354More than 2 groupsis an upper-tail test.Do not reject H0 Do not reject H0.insufficient Buddy Freeman, 2015