Comparing Means for Several Populations

When we wish to test for differences in means for only 1 or 2 populations, we use one- or two-sample t inference.

Testing for differences in more than 2 populations, or at several different levels (values) of a variable involves a different approach.

This is called Analysis of Variance, or ANOVA.

ANOVA partitions the total sum of squares into two parts:

1. within treatment variability

2. between treatment variability

Comparing Means for Several Populations

Example: Test 5 types of concrete for differences in moisture absorption.

The 5 types of concrete are the five levels of the treatment.

Within Variability – this seeks to quantify the variability in absorption for one particular type of concrete.

Between Variability – this seeks to quantify the differences between the types of concrete.

ANOVA seeks to answer the question “Are the differences between the 5 sample means what is expected purely from random variation alone?”

Definitions

• An experimental unit is an object, or subject, that produces a sample measurement.

• The experimental conditions that define the different populations in a completely randomized design are called treatments.

• Testing for differences in the treatments is equivalent to testing for differences in the population means.

Practice on Definitions

• See page 399 section 10.1 exercises.

Graphical demonstration:Employing two types of variability

Graphical demonstration:Employing two types of variability

20

25

30

1

7

Treatment 1 Treatment 2 Treatment 3

10

12

19

9

Treatment 1Treatment 2Treatment 3

20

161514

1110

9

10x1

15x2

20x3

10x1

15x2

20x3

The sample means are the same as before,but the larger within-sample variability makes it harder to draw a conclusionabout the population means.

A small variability withinthe samples makes it easierto draw a conclusion about the population means.

Assumptions for ANOVA

• 1. The samples are independent– Selection of objects from any one population is

unrelated to the selection of objects from any of the other populations. Selections are random.

– Examples• Different groups of people (no person in more than one

group)• Different types of music• Different concentrations of chemicals• Different models of automobiles

Assumptions for ANOVA

• 2. Each population has the same standard deviation, But the values of the population standard

deviations is not known before testing.

Assumptions for ANOVA

• 3. Each sample has a mean that can be calculated. This mean is somehow representative of the population mean for its population.

Assumptions for ANOVA

• 4. Each population is normally distributed– Quantitative data: sample size is at least 30

– However, we will assume normally distributed populations for all the problems we work.

Assumptions for ANOVA

The following assumptions are required for a 1-way ANOVA:• The k populations are independent.• Each population has common standard deviation, .

• Each population has a mean, i for i = 1, 2, …, k.

• Each population is normally distributed.

So we now are testing whether all the treatment means are equal.

H0: 1 = 2 = … = k

Ha: At least two of the population means are not equal

Test Statistic

• If the null hypothesis is true, we expect the k sample means to have reasonably similar values.

• In other words, if the population means are equal, we would expect the variability among the sample means to be relatively small.

• Variability among the sample means is one of the things we will be testing for.

Test Statistic

• If the null hypothesis is true, we do not expect the population means to be exactly the same, because there is a chance factor in our choice of sample experimental units.

• We need to take into account the variability due to chance among the sample means.

Test Statistic

• This method is called “analysis of variance” of ANOVA because we are comparing two sources of variance: the variance among the sample means and the variation expected by chance among the sample means when the null hypothesis is true.

Test Statistic

• Our test statistic is called F.

• F = Variability among the sample means Variability expected by chance

Degrees of freedom

• For a sample, (or group) (k) df = n – 1

• Total df = total number of units in the experiment – 1

• Error df = Total df – Group df – Or

• Error df = N - k

Minitab

• We will use Minitab to do our calculations.

• A typical Minitab display is on the next slide.

ANOVA Table: Tensile Strength for 6 Machines

Analysis of Variance for Tensile-StrengthSource DF SS MS F P

Machine 5 5.34 1.07 0.31 0.902

Error 18 62.64 3.48

Total 23 67.98

SSMachine = 5.34 (sample mean variability), k = 6 machines

SSError = 62.64 (variability due to chance)

Notice how much larger the “chance” variability is than the other.

There is little to no evidence that the machines differ in mean tensile-strength. Look at that HUGE p-value!

Another Minitab Example

• Example 102 page 369

• Sociologist and GPA college students

One-way ANOVA: GPA versus Group Source DF SS MS F PGroup 3 1.519 0.506 2.99 0.044Error 36 6.091 0.169Total 39 7.610

S = 0.4113 R-Sq = 19.96% R-Sq(adj) = 13.29%

Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ---+---------+---------+---------+------

Lower Middle 10 2.5240 0.4362 (--------*--------)Poor 10 2.2640 0.3161 (-------*--------)Upper Middle 10 2.7170 0.4125 (--------*-------)Well-to-do 10 2.7560 0.4653 (--------*--------) ---+---------+---------+---------+------ 2.10 2.40 2.70 3.00

Pooled StDev = 0.4113

Manual Calculation

• The formula for calculating F using the Mean Square Treatment is given on page 375.

Manual Calculation

• To determine the p value when the f value is known, we need to use a table.

• Table 5 is on pages VII, VIII, IX in the table appendix.

• In general, Table 5 will provide only approximate p-values. To find precise values, technology is needed.

ANOVA – What is expected from you?

Be able to complete each of the following exercises:

• State the two hypotheses. • What is the observed value of the test statistic?

(F = ?) • Is this valid? We will typically “assume” the

method is ok.• What is the p-value?• State a conclusion. • Using a table for comparisons, locate what

mean(s) are significantly different if you accepted the alternative hypothesis. (Sect 10.3)

Analysis of Variance results: Responses stored in Score. Factors stored in Hair Color. Factor means

Hair Color n Mean Std. Error

Dark Blond 6 39.5 3.3936214

Dark Brunette 6 32.666668 1.2560962

Light Blond 6 49.833332 3.5158372

Light Brunette 6 42.333332 3.4123957

ANOVA table

Source df SS MS F-Stat P-value

Treatments 3 908.8333 302.94446 5.4437456 0.0067

Error 20 1113 55.65

Total 23 2021.8334

Example Page 423 # 1

One-way ANOVA: Score versus Hair Color Source DF SS MS F PHair Color 3 908.8 302.9 5.44 0.007Error 20 1113.0 55.7Total 23 2021.8

H0: light_blond = dark_blond = … = dark_brunette

Ha: At least two population means are different.Accept Ha if p-value < 0.05

F = 5.44 p-value = 0.007

At the 0.05 level of significance, there is sufficient evidence to conclude that there is a difference among mean pain thresholds for people possessing these four hair colors.

10.3 Which means are different?Multiple Comparisons

• When an analysis of variance F-test indicates a significant difference among population means, (accept Ha), the next question is which means are different.

Which means are different?

• We need to test each of the following pairs of hypotheses.

• Pair 1: Ho: μ1-μ2=0 Ha: μ1-μ2≠0

• Pair 2: Ho: μ1-μ3=0 Ha: μ1-μ3≠0

• Pair 3: Ho: μ2-μ3=0 Ha: μ2-μ3≠0

Which means are different?

• To test each pair of hypothesis, we are only testing two means for a difference between them.

• This is the two-sample t-statistic that we used in section 9-2.

• However, we will substitute MSE(Mean Square Error) for s2

• See page 416 for entire equation.

Which mean is different?

• We can use StatCrunch to calculate the value of t and the p-value for each of the comparisons. We can then draw our conclusions based on the p-value for each pair (is it less than α? If so we accept the alternative hypothesis), and summarize our findings in a chart. This is how the revised section in the book does it.

• See example 10.4 p 418

• Let’s look further at the example on hair coloring.

Multiple Comparisons

Pair p-value t-value Interpretation

LB v DB +2.11 0.0606 NS

LB v LBr +1.53 0.1569 NS

LB v DBr +4.59 0.0033 LB > DBr

DB v LBr -0.59 0.5691 NS

DB v DBr +1.89 0.1052 NS

LBr v DBr +2.66 0.0357 LBr > DBr

Let’s look further at the example on hair coloring

Summary

• Ex 10.5 summarizes ideas from Chapter 10. See p 421

When should we use the multiple comparison method?

• The sample data are obtained from the k populations using a completely randomized design

• An analysis of variance F-test indicates that there are some differences among the k population means.

• The objective is to determine which of the k population means differ. It is usually of interest to determine which mean might be the largest (or smallest).