Apr 21, 2023 1

Chapter 13: Chapter 13: Comparing Several Means Comparing Several Means

(One-Way ANOVA)(One-Way ANOVA)

In Chapter 13:

13.1 Descriptive Statistics

13.2 The Problem of Multiple Comparisons

13.3 Analysis of Variance

13.4 Post Hoc Comparisons

13.5 The Equal Variance Assumption

13.6 Introduction to Nonparametric Tests

3

Illustrative Example: Data

Pets as moderators of a stress response. This chapter follows the analysis of data from a study in which heart rates (bpm) of participants were monitored after being exposed to a psychological stressor. Participants were randomized to one of three groups:

• Group 1 - monitored in presence of pet dog• Group 2 - monitored in the presence of human

friend• Group 3 - monitored with neither dog nor human

friend present

4

Illustrative Example: Data

5

SPSS Data Table

• Most computer programs require data in two columns

• One column is for the explanatory variable (group)

• One column is for the response variable (hrt_rate)

6

13.1 Descriptive Statistics

• Data are described and explored before moving to inferential calculations

• Here are summary statistics by group:

7

Exploring Group Differences

• John Tukey taught us the importance of exploratory data analysis (EDA)

• EDA techniques that apply:– Stemplots

– Boxplots

– Dotplots

John W. Tukey (1915 -2000)

8

Side-by-Side Stemplots

9

Side-by-Side Boxplots

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§13.2 The Problem of Multiple Comparisons

• Consider a comparison of three groups. There are three possible t tests when considering three groups:(1) H0: μ1 = μ2 versus Ha: μ1 ≠ μ2

(2) H0: μ1 = μ3 versus Ha: μ1 ≠ μ3

(3) H0: μ2 = μ3 versus Ha: μ2 ≠ μ3

• However, we do not perform separate t tests without modification → this would identify too many random differences

11

Problem of Multiple Comparisons

• Family-wise error rate = probability of at least one false rejection of H0

• Assume three null hypotheses are true:At α = 0.05, the Pr(retain all three H0s) = (1−0.05)3 = 0.857. Therefore, Pr(reject at least one) = 1−0.847 = 0.143 this is the family-wise error rate.

• The family-wise error rate is much greater than intended. This is “The Problem of Multiple Comparisons”

12

Problem of Multiple Comparisons

The more comparisons you make, the greater the family-wise error rate. This table demonstrates the magnitude of the problem

13

Two-step approach:

1. Test for overall significance using a technique called “Analysis of Variance”

2. Do post hoc comparison on individual groups

Mitigating the Problem of Multiple Comparisons

14

13.3 Analysis of Variance

• One-way ANalysis Of VAriance (ANOVA)– Categorical explanatory variable – Quantitative response variable– Test group means for a significant difference

• Statistical hypotheses

– H0: μ1 = μ2 = … = μk

– Ha: at least one of the μis differ

• Method: compare variability between groups to variability within groups (F statistic)

15

Analysis of Variance Overview, cont.

R. A. Fisher (1890-1962)

The F in the F statistic stands for “Fisher”

16

Variability Between Groups• Variability of group means around the grand

mean → provides a “signal” of group difference

• Based on a statistic called the Mean Square Between (MSB)

• NotationSSB ≡ sum of squares between

dfB ≡ degrees of freedom betweenk ≡ number of groupsx-bar ≡ grand mean

x-bari ≡ mean of group i

17

Mean Square Between: Formula

• Sum of Squares Between [Groups]

• Degrees of Freedom Between

• Mean Square Between

18

Mean Square Between: Graphically

19

Mean Square Between: Example

20

Variability Within Groups• Variability of data points within groups →

quantifies random “noise”

• Based on a statistic called the Mean Square Within (MSW)

• NotationSSW ≡ sum of squares withindfW ≡ degrees of freedom withinN ≡ sample size, all groups combinedni ≡ sample size, group Is2

i ≡ variance of group i

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Mean Square Within: Formula

• Mean Square Within

• Sum of Squares Within

• Degrees of Freedom Within

22

Mean Square Within: Graphically

23

Mean Square Within: Example

24

The F statistic and ANOVA table• Data are arranged to form an ANOVA

table

• F statistic is the ratio of the MSB to MSW

08.14793.84

843.1193

MSW

MSBFstat

Fstat “signal-to-noise” ratio

25

Fstat and P-value• The Fstat has numerator and denominator

degrees of freedom: df1 and df2 respectively (corresponding to dfB and dfW)

• Convert Fstat to P-value with a computer program or Table D

• The P-value corresponds to the area in the right tail beyond

26

Table D (“F Table”)• The F table has limited listings for df2.

• You often must round-down to the next available df2 (rounding down preferable for conservative estimate).

• Wedge the Fstat between listing to find the approximate P-value

df1 = 2, df2 = 42(Table D does not have df2 of 42;

next lowest df2 is 30).

Fstat = 14.02 falls below P of 0.001

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Fstat and P-value

P < 0.001

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ANOVA Example (Summary)A. Hypotheses: H0: μ1 = μ2 = μ3 vs. Ha: at least one

of the μis differ

B. Statistics: Fstat = 14.08 with 2 and 42 degrees of freedom

C. P-value = .000021 (via SPSS), providing highly significant evidence against the H0; conclude the heart rates (an indicator of the effects of stress) differed in the groups

D. Significance level (optional): Results are significantly at α = .00005

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ComputationBecause of the complexity of computations, ANOVA statistics are often calculated by computer

ANOVA

beats per minute

2387.685 2 1193.843 14.079 .000

3561.309 42 84.793

5948.994 44

Between Groups

Within Groups

Total

Sum ofSquares df Mean Square F Sig.

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ANOVA and the t test (Optional)

ANOVA for two groups is equivalent to the equal variance (pooled) t test (§12.4)

• Both address H0: μ1 = μ2

• dfW = df for t test = N – 2

• MSW = s2pooled

• Fstat = (tstat)2

• F1,df2,α = (tdf,1-α/2)2

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13.4 Post Hoc Comparisons

• ANOVA Ha says “at least one population mean differs” but does not delineate which differ.

• Post hoc comparisons are pursued after rejection of the ANOVA H0 to delineate differences

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SPSS Post Hoc Comparison Procedures

Many post hoc comparison procedures exist. We cover the LSD and Bonferroni methods.

33

Least Squares Difference Procedure

A. Hypotheses. H0: μi = μj vs. Ha: μi ≠ μj for each group i and j

B. Test statistic

C. P-value. Use t table or software

kNdfdf

nnMSWSE

SE

xxt

jixx

xx

ji

ji

Within

stat 11

where)(

21

Do after a significant ANOVA to protect against the problem of multiple comparisons

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LSD Procedure: Example

A. Hypotheses. H0: μ1 = μ2 against Ha: μ1 ≠ μ2

B. Test statistic.

C. P-value. P = 0.0000039; highly significant evidence of a difference.

42345 with 31.53623

3259148373

)(

362.315

1

15

1793.84

11

Within21

stat21

21

dfdf.

..

SE

xxt

nnMSWSE

xx

jixx

For the “pets” illustrative data, we test H0: μ1 = μ2 by hand. The other tests will be done by computer

35

LSD Procedure, SPSSResults for illustrative “pets” data.

H0: μ1 = μ2H0: μ1 = μ3

H0: μ2 = μ3

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95% Confidence Interval, Mean Difference, LSD Method

ji

kNji

ji

nnMSWtxx

μ

11)(

for CI %100)1(

21,

37

95% CI, LSD Method, Example

)0.11 ,6.24(

)362.3)(021.2(842.17

15

1

15

1793.84)325.91483.73(

11)(

for CI %95

975,.42

1,

21

2

t

nnMSWtxx

μ

jikNji

Comparing Group 1 to Group 2:

38

Bonferroni Procedure

A. Hypotheses. H0: μ1 = μ2 against Ha: μ1 ≠ μ2

B. Test statistic. Same as for the LSD method.

C. P-value. The LSD method produced P = .0000039 (two-tailed). Since there were three post hoc comparisons, PBonf = 3 × .0000039 = .000012.

The Bonferroni procedure is instituted by multiplying the P-value from the LSD procedure by the number of post hoc comparisons “c”.

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Bonferroni Confidence Interval

)4.9,3.26(

)362.3)(51.2(842.17

15

1

15

1793.84)()325.91483.73(

11)(for CI %95

9917,.42

1, 2

k

jikNjiji

t

nnMSWtxxμ

c

Let c represent the number of post hoc comparisons. Comparing Group 1 to Group 2:

40

Bonferroni Procedure, SPSSP-values from Bonferroni are higher and confidence intervals are broader than LSD method, reflecting its conservative approach

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§13.5. The Equal Variance Assumption

• Conditions for ANOVA:1. Sampling independence

2. Normal sampling distributions of mean

3. Equal variance within population groups

• Let us focus on condition 3, since conditions 1 and 2 are covered elsewhere.

• Equal variance is called homoscedasticity. (Unequal variance = heteroscedasticity).

• Homoscedasticity allows us to pool group variances to form the MSW

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Assessing “Equal Variance”

1. Graphical exploration. Compare spreads visually with side-by-side plots. 2. Descriptive statistics. If a group’s standard deviation is more than twice that of another, be alerted to possible heteroscedasticity

3. Test variances. A statistical test can be applied (next slide).

43

Levene’s Test of VariancesA.Hypotheses. H0: σ2

1 = σ22 = … = σ2

k

Ha: at least one σ2i differs

B. Test statistic. Test is performed by computer. The test statistic is a particular type of Fstat based on the rank transformed deviations (see p. 283 for details).

C. P-value. The Fstat is converted to a P-value by the computational program. Interpretation of P is routine small P evidence against H0, suggesting heteroscedasticity.

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Test of Homogeneity of Variances

beats per minute

.059 2 42 .943

LeveneStatistic df1 df2 Sig.

Levene’s Test – Example (“pets” data)

A. H0: σ21 = σ2

2 = σ23 versus Ha: at least one σ2

i differs

B. SPSS output (below). Fstat = 0.059 with 2 and 42 df

C. P = 0.943. Very weak evidence against H0 retain assumption of homoscedasticity

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Analyzing Groups with Unequal Variance

• Stay descriptive. Use summary statistics and EDA methods to compare groups.

• Remove outliers, if appropriate (p. 287).

• Mathematically transform the data to compensate for heteroscedasticity (e.g., a long right tail can be pulled in with a log transform).

• Use robust non-parametric methods.

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13.6 Intro to Nonparametric MethodsMany nonparametric procedures are based on rank transformed data (“rank tests”). Here are examples:

47

The Kruskal-Wallis Test• Let us explore the Kruskal-Wallis test as

an example of a non-parametric test

• The Kruskal-Wallis test is the non-parametric analogue of one-way ANOVA.

• It does not require Normality or Equal Variance conditions for inference.

• It is based on rank transformed data and seeing if the mean ranks in groups differ significantly.

48

Kruskal-Wallis Test• The K-W hypothesis can be stated in

terms of mean or median (depending on assumptions made about population shapes). Let us use the later.

• Let Mi ≡ the median of population i

• There are k groups

• H0: M1 = M2 = … = Mk

• Ha: at least one Mi differs

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Kruskal-Wallis, ExampleAlcohol and income. Data from a survey on alcohol consumption and income are presented.

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Kruskal-Wallis Test, ExampleWe wish to test whether the means differ significantly but find graphical and hypothesis testing evidence that the population variances are unequal.

Test of Homogeneity of Variances

Alcohol consumption

10.874 4 708 .000

LeveneStatistic df1 df2 Sig.

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Kruskal-Wallis Test, Example, cont.

A.Hypotheses.H0: M1 = M2 = M3 = M4 = M5 vs.Ha: at least one Mi differs

B.Test statistic. Some computer programs use chi-square statistic based upon a Normal approximation. SPSS derives Chi-square statistics = 7.793 with 4 df (next slide)

52

Kruskal-Wallis Test, Example, cont. Test Statisticsa,b

7.793

4

.099

Chi-Square

df

Asymp. Sig.

Alcoholconsumption

Kruskal Wallis Testa.

Grouping Variable: Incomeb.

Ranks

46 303.58

88 344.68

140 385.17

250 345.26

189 370.40

713

Income1

2

3

4

5

Total

Alcohol consumptionN Mean Rank

P = 0.099, providing marginally significant evidence against H0.