1 copyright © 2005 brooks/cole, a division of thomson learning, inc. chapter 15 the analysis of...

1 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.

Chapter 15

The Analysis of Variance


A study was done on the survival time of patients with advanced cancer of the stomach, bronchus, colon, ovary or breast when treated with ascorbate1. In this study, the authors wanted to determine if the survival times differ based on the affected organ.

1 Cameron, E. and Pauling, L. (1978) Supplemental ascorbate in the supportive treatment of cancer: re-evaluation of prolongation of survival time in terminal human cancer. Proceedings of the National Academy of Science, USA, 75, 4538-4542.

A Problem


A comparative dotplot of the survival times is shown below.

A Problem

3000200010000

Survival Time (in days)

Dotplot for Survival Time

Cancer Type

Breast

Bronchus

Colon

Ovary

Stomach


H0: µstomach = µbronchus = µcolon = µovary = µbreast

Ha: At least two of the µ’s are different

A Problem

The hypotheses used to answer the question of interest are

The question is similar to ones encountered in chapter 11 where we looked at tests for the difference of means of two different variables. In this case we are interested in looking a more than two variable.


A single-factor analysis of variance (ANOVA) problems involves a comparison of k population or treatment means µ1, µ2, … , µk.

The objective is to test the hypotheses:

H0: µ1 = µ2 = µ3 = … = µk

Ha: At least two of the µ’s are different

Single-factor Analysis of Variance (ANOVA)


The analysis is based on k independently selected samples, one from each population or for each treatment.

In the case of populations, a random sample from each population is selected independently of that from any other population.

When comparing treatments, the experimental units (subjects or objects) that receive any particular treatment are chosen at random from those available for the experiment.



A comparison of treatments based on independently selected experimental units is often referred to as a completely randomized design.



70

60

50

40

FertilizerY

ield

Dotplots of Yield by Fertilizer(group means are indicated by lines)

Type 1 Type 2 Type 3

Notice that in the above comparative dotplot, the differences in the treatment means is large relative to the variability within the samples.



Sta

tistic

s

Psy

cho

log

y

Eco

nom

ics

Bus

ine

ss

85

75

65

SubjectP

rice

Dotplots of Price by Subject(group means are indicated by lines)

Notice that in the above comparative dotplot, the differences in the treatment means is not easily understood relative to the sample variability.

ANOVA techniques will allow us to determined if those differences are significant.



ANOVA Notation

k = number of populations or treatments being compared

Population or treatment 1 2 … k

Population or treatment mean µ1 µ2 … µk

Sample mean …1x 2x kx

Population or treatment variance …21 2

2 2k

Sample variance …21s 2

2s 2ks

Sample size n1 n2 … nk


N = n1 + n2 + … + nk (Total number of observations in the data set)

ANOVA Notation

Tx grand mean

N

T = grand total = sum of all N observations

1 1 2 2 k kn x n x n x


Assumptions for ANOVA

1. Each of the k populations or treatments, the response distribution is normal.

2. 1 = 2 = … = k (The k normal distributions have identical standard deviations.

3. The observations in the sample from any particular one of the k populations or treatments are independent of one another.

4. When comparing population means, k random samples are selected independently of one another. When comparing treatment means, treatments are assigned at random to subjects or objects.


Definitions

2 2 2

1 1 2 2 k kSSTr n x x n x x n x x

A measure of disparity among the sample means is the treatment sum of squares, denoted by SSTr is given by

A measure of variation within the k samples, called error sum of squares and denoted by SSE is given by

2 2 21 1 2 2 k kSSE n 1 s n 1 s n 1 s


Definitions

A mean square is a sum of squares divided by its df. In particular,

The error df comes from adding the df’s associated with each of the sample variances:

(n1 - 1) + (n2 - 1) + …+ (nk - 1)

= n1 + n2 … + nk - 1 - 1 - … - 1 = N - k

mean square for

treatments = MSTr = SSTrk 1

mean square for error = MSE = SSEN k


ExampleThree filling machines are used by a bottler to fill 12 oz cans of soda. In an attempt to determine if the three machines are filling the cans to the same (mean) level, independent samples of cans filled by each were selected and the amounts of soda in the cans measured. The samples are given below.

Machine 112.033 11.985 12.009 12.00912.033 12.025 12.054 12.050

Machine 212.031 11.985 11.998 11.99211.985 12.027 11.987

Machine 312.034 12.021 12.038 12.05812.001 12.020 12.029 12.01112.021


Example

2 2 2

1 1 2 2 k k

2 2 2

SSTr n x x n x x n x x

8(0.0065833) 7(-0.0174524) 9(0.0077222)

0.000334672+0.00213210+0.00053669

0.00301552

1 1 1n 8, x 12.0248, s 0.02301

3 3 3n 9, x 12.0259, s 0.01650 2 2 2n 7, x 12.0007, s 0.01989

x 12.018167


Example

2 2 21 1 2 2 k k

2 2 2

SSE n 1 s n 1 s n 1 s

7(0.0230078) 6(0.0198890) 8(0.01649579)

0.0037055 0.0023734 0.0021769

0.00825582

1 1 1n 8, x 12.0248, s 0.02301

3 3 3n 9, x 12.0259, s 0.01650 2 2 2n 7, x 12.0007, s 0.01989

x 12.018167


Example

SSTrk 1

mean square for treatments = MSTr =

SSTr 0.00301552MSTr 0.0015078

k 1 3 1

mean square for error = MSE = SSEN k

SSE 0.0082579MSE 0.00039313

N k 24 3

1 1 1n 8, x 12.0248, s 0.02301

3 3 3n 9, x 12.0259, s 0.01650 2 2 2n 7, x 12.0007, s 0.01989

x 12.018167


Comments

Both MSTr and MSE are quantities that are calculated from sample data.

As such, both MSTr and MSE are statistics and have sampling distributions.

More specifically, when H0 is true, µMSTr = µMSE.

However, when H0 is false, µMSTr µMSE and the greater the differences among the ’s, the larger µMSTr will be relative to µMSE.


The Single-Factor ANOVA F Test

Null hypothesis: H0: µ1 = µ2 = µ3 = … = µk

Alternate hypothesis: At least two of the µ’s are different

Test Statistic: MSTrF

MSE


The Single-Factor ANOVA F Test

When H0 is true and the ANOVA assumptions are reasonable, F has an F distribution with df1 = k - 1 and df2 = N - k.

Values of F more contradictory to H0 than what was calculated are values even farther out in the upper tail, so the P-value is the area captured in the upper tail of the corresponding F curve.


Example

Consider the earlier example involving the three filling machines.

Machine 112.033 11.985 12.009 12.009 12.03312.025 12.054 12.050

Machine 212.031 11.985 11.998 11.992 11.98512.027 11.987

Machine 312.034 12.021 12.038 12.058 12.00112.020 12.029 12.011 12.021


Example

SSTr 0.00301552 SSE 0.00825582

MSTr 0.0015078 MSE 0.00039313

1 1 1n 8, x 12.0248, s 0.02301

3 3 3n 9, x 12.0259, s 0.01650 2 2 2n 7, x 12.0007, s 0.01989

x 12.018167


Example

1. Let µ1, µ2 and µ3 denote the true mean amount of soda in the cans filled by machines 1, 2 and 3, respectively.

2. H0: µ1 = µ2 = µ3

3. Ha: At least two among are µ1, µ2 and µ3 different

4. Significance level: = 0.01

5. Test statistic:MSTr

FMSE


Example

6. Looking at the comparative dotplot, it seems reasonable to assume that the distributions have the same ’s. We shall look at the normality assumption on the next slide.*

12.0612.0512.0412.0312.0212.0112.0011.99

Fill

Dotplot for FillMachine

Machine 1

Machine 2

Machine 3

*When the sample sizes are large, we can make judgments about both the equality of the standard deviations and the normality of the underlying populations with a comparative boxplot.


Example6. Looking at normal plots for the samples, it

certainly appears reasonable to assume that the samples from Machine’s 1 and 2 are samples from normal distributions. Unfortunately, the normal plot for the sample from Machine 2 does not appear to be a sample from a normal population. So as to have a computational example, we shall continue and finish the test, treating the result with a “grain of salt.”

P-Value: 0.692A-Squared: 0.235

Anderson-Darling Normality Test

N: 8StDev: 0.0230078Average: 12.0248

12.05512.04512.03512.02512.01512.00511.99511.985

.999

.99

.95

.80

.50

.20

.05

.01

.001

Pro

bab

ility

Machine 1

Normal Probability Plot



N: 7StDev: 0.0198890Average: 12.0007

12.0312.0212.0112.0011.99

.999

.99

.95

.80

.50

.20

.05

.01

.001

Pro

bab

ility

Machine 2




N: 9StDev: 0.0164958Average: 12.0259

12.0612.0512.0412.0312.0212.0112.00

.999

.99

.95

.80

.50

.20

.05

.01

.001

Pro

bab

ility

Machine 3



Example7. Computation:

SSTr 0.00301552 SSE 0.00825582

MSTr 0.0015078 MSE 0.00039313

1 1 1n 8, x 12.0248, s 0.02301

3 3 3n 9, x 12.0259, s 0.01650 2 2 2n 7, x 12.0007, s 0.01989

x 12.018167

1 2 3N n n n 8 7 9 24, k 3

1

2

MSTr 0.0015078F 3.835

MSE 0.00039313df treatment df k 1 3 1 2

df error df N k 24 3 21


Example8. P-value:

3.835

dfden / dfnum 2

21 0.100 2.570.050 3.470.025 4.420.010 5.780.001 9.77

From the F table with numerator df1 = 2 and denominator df2 = 21 we can see that

0.025 < P-value < 0.05

(Minitab reports this value to be 0.038

1

2

MSTr 0.0015078F 3.835



Recall

1

2

MSTr 0.0015078F 3.835



Recall


Example

9. Conclusion:

Since P-value > = 0.01, we fail to reject H0. We are unable to show that the mean fills are different and conclude that the differences in the mean fills of the machines show no statistically significant differences at the 1% level of significance.


Total Sum of Squares

The relationship between the three sums of squares is SSTo = SSTr + SSEwhich is often called the fundamental identity for single-factor ANOVA.

Informally this relation is expressed as

Total variation = Explained variation + Unexplained variation

Total sum of squares, denoted by SSTo, is given by

with associated df = N - 1.all N obs.

2SSTo (x x)


Single-factor ANOVA Table

The following is a fairly standard way of presenting the important calculations from an single-factor ANOVA. The output from most statistical packages will contain an additional column giving the P-value.


Single-factor ANOVA Table

The ANOVA table supplied by Minitab

One-way ANOVA: Fills versus Machine

Analysis of Variance for Fills Source DF SS MS F PMachine 2 0.003016 0.001508 3.84 0.038Error 21 0.008256 0.000393Total 23 0.011271


Another Example

A food company produces 4 different brands of salsa. In order to determine if the four brands had the same sodium levels, 10 bottles of each Brand were randomly (and independently) obtained and the sodium content in milligrams (mg) per tablespoon serving was measured.

The sample data are given on the next slide.

Use the data to perform an appropriate hypothesis test at the 0.05 level of significance.


Another Example

Brand A43.85 44.30 45.69 47.13 43.3545.59 45.92 44.89 43.69 44.59

Brand B42.50 45.63 44.98 43.74 44.9542.99 44.95 45.93 45.54 44.70

Brand C45.84 48.74 49.25 47.30 46.4146.35 46.31 46.93 48.30 45.13

Brand D43.81 44.77 43.52 44.63 44.8446.30 46.68 47.55 44.24 45.46


Another Example

1. Let µ1, µ2 , µ3 and µ4 denote the true mean sodium content per tablespoon in each of the brands respectively.

2. H0: µ1 = µ2 = µ3 = µ4

3. Ha: At least two among are µ1, µ2, µ3 and µ4 are different

4. Significance level: = 0.05

5. Test statistic:MSTr

FMSE


6. Looking at the following comparative boxplot, it seems reasonable to assume that the distributions have the equal ’s as well as the samples being samples from normal distributions.

Another Example

Bra

nd

D

Bra

nd

C

Bra

nd

B

Bra

nd

A

49

48

47

46

45

44

43

42

Boxplots of Brand A - Brand D(means are indicated by solid circles)


Example

Treatment df = k - 1 = 4 - 1 = 3

7. Computation:Brand k si

Brand A 10 44.9001.180Brand B 10 44.5911.148Brand C 10 47.0561.331Brand D 10 45.1801.304

xi

2 2 2 21 1 2 2 3 3 4 4

2 2

2 2

SSTr n (x x) n (x x) n (x x) n (x x)

10(44.900 45.432) 10(44.591 45.432)

10(47.056 45.432) 10(45.180 45.432)

36.912

x 45.432


Example7. Computation (continued):

Error df = N - k = 40 - 4 = 36

2 2 2 21 1 2 2 3 3 4 4

2 2 2 2

SSE n 1 s n 1 s n 1 s n 1 s

9(1.180) 9(1.148) 9(1.331) 9(1.304)

55.627

SSTr

SSE

SSTr 36.912MSTr 12.304df 3F 7.963

SSE 55.627MSE 1.5452df 36


Example8.P-value:

F = 7.96 with dfnumerator= 3 and dfdenominator= 36

Using df = 30 we find

P-value < 0.001

7.96


Example

9. Conclusion:

Since P-value < = 0.001, we reject H0. We can conclude that the mean sodium content is different for at least two of the Brands.

We need to learn how to interpret the results and will spend some time on developing techniques to describe the differences among the µ’s.


Multiple Comparisons

A multiple comparison procedure is a method for identifying differences among the µ’s once the hypothesis of overall equality (H0) has been rejected.

The technique we will present is based on computing confidence intervals for difference of means for the pairs.

Specifically, if k populations or treatments are studied, we would create k(k-1)/2 differences. (i.e., with 3 treatments one would generate confidence intervals for µ1 - µ2, µ1 - µ3 and µ2 - µ3.) Notice that it is only necessary to look at a confidence interval for µ1 - µ2 to see if µ1 and µ2 differ.


The Tukey-Kramer Multiple Comparison Procedure

When there are k populations or treatments being compared, k(k-1)/2 confidence intervals must be computed. If we denote the relevant Studentized range critical value by q, the intervals are as follows:

For i - j:

Two means are judged to differ significantly if the corresponding interval does not include zero.

i ji j

MSE 1 1( ) q

2 n n


The Tukey-Kramer Multiple Comparison Procedure

When all of the sample sizes are the same, we denote n by n = n1 = n2 = n3 = … = nk, and the confidence intervals (for µi - µj) simplify to

i j

MSE( ) q

n


Example (continued)

Continuing with example dealing with the sodium content for the four Brands of salsa we shall compute the Tukey-Kramer 95% Tukey-Kramer confidence intervals for µA - µB, µA - µC, µA - µD, µB - µC, µB - µD and µC - µD.

A B C D

55.627MSE 1.5452, n n n n n 10

36Interpolating from the table

q 3.81 i.e. 60% of the way from 3.85 to 3.79

MSE 1.5452q 3.81 1.498

n 10


Example (continued)

Difference95% Confidence

Limits95% Confidence

Interval

A - B 0.309 ± 1.498 (-1.189, 1.807)

A - C -2.156 ± 1.498 (-3.654, -0.658)

A - D -0.280 ± 1.498 (-1.778, 1.218)

B - C -2.465 ± 1.498 (-3.963, -0.967)

B - D -0.589 ± 1.498 (-2.087, 0.909)

C - D 1.876 ± 1.498 (0.378, 3.374)

Notice that the confidence intervals for µA – µB, µA – µC

and µC – µD do not contain 0 so we can infer that the mean sodium content for Brands C is different from Brands A, B and D.


Example (continued)

We also illustrate the differences with the following listing of the sample means in increasing order with lines underneath those blocks of means that are indistinguishable.

Brand B Brand A Brand D Brand C

44.591 44.900 45.180 47.056

Notice that the confidence interval for µA – µC, µB – µC, and µC – µD do not contain 0 so we can infer that the mean sodium content for Brand C and all others differ.


Minitab Output for Example

One-way ANOVA: Sodium versus Brand

Analysis of Variance for Sodium Source DF SS MS F PBrand 3 36.91 12.30 7.96 0.000Error 36 55.63 1.55Total 39 92.54 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ------+---------+---------+---------+Brand A 10 44.900 1.180 (-----*------) Brand B 10 44.591 1.148 (------*-----) Brand C 10 47.056 1.331 (------*------) Brand D 10 45.180 1.304 (------*-----) ------+---------+---------+---------+Pooled StDev = 1.243 44.4 45.6 46.8 48.0


Minitab Output for Example

Tukey's pairwise comparisons

Family error rate = 0.0500Individual error rate = 0.0107

Critical value = 3.81

Intervals for (column level mean) - (row level mean)

Brand A Brand B Brand C

Brand B -1.189 1.807

Brand C -3.654 -3.963 -0.658 -0.967

Brand D -1.778 -2.087 0.378 1.218 0.909 3.374


Simultaneous Confidence Level

The Tukey-Kramer intervals are created in a manner that controls the simultaneous confidence level.

For example at the 95% level, if the procedure is used repeatedly on many different data sets, in the long run only about 5% of the time would at least one of the intervals not include that value of what it is estimating.

We then talk about the family error rate being 5% which is the maximum probability of one or more of the confidence intervals of the differences of mean not containing the true difference of mean.

1 copyright © 2005 brooks/cole, a division of thomson learning, inc. chapter 15 the analysis of...

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