Download One Way ANOVA

Section 9–3:
ANOVA: Testing a claim that there is no difference
in 3 or more population means
Section 9–2 introduced the two-sample t-test method to test the hypothesis that the means of 2
( x1 − x 2 )
normal populations are equal. H 0: µ1 = µ2 using the two-sample test statistic: t =
2
2
( s1 ) + ( s2 )
n1
n2
ANOVA is a general technique to test the Null Hypothesis that the means of 3 or more
populations are equal. The populations must be normal and all have the same standard deviations.
H 0: µ1 = µ2 = µ3 = ....
µ1 = µ2 = µ2
H1: The population means ARE NOT ALL EQUAL
µ1
x
µ2
µ3
x
Example
An athletic trainer claims that the average resting heart rates of men in the age groups 15 to 24 years
old, 25 to 34 years old and 35 to 44 years old are the same rate. Does each group have the same
average heart rate?
We will call Population 1 the 15 to 24 years olds with a mean of µ1
We will call Population 2 the 25 to 34 years olds with a mean of µ2
We will call Population 3 the 35 to 44 years olds with a mean of µ3
The Null Hypotheses H 0 is a claim that all the age groups have the same average heart rate
H 0: µ1 = µ2 = µ3
versus the alternate Hypotheses H1
H1: At least one of theaverage heart rate is different from the others.
Section 9 –3
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Can we use Section 9-2 two-sample t-tests?
You could try to test all possible pairs of means by the use of two-sample t-tests.
Test 1
H 0: µ1 = µ 2
Test 2
H 0: µ1 = µ 3
Test 3
H 0: µ 2 = µ3
If a series of separate tests µ1 = µ2 and µ2 = µ3 and µ1 = µ3
all produced a DO NOT REJECT H 0
then you could conclude all the population means were equal µ1 = µ2 = µ3 .
If we use a σ = .05 vale to conduct one two-sample t-test. H 0: µ1 = µ2 then we would be able to say
with a 95% certainty that the means µ1 = µ2 are equal. To conclude that µ1 = µ2 = µ3 we would need
to conduct 3 two-sample t-tests, µ1 = µ2 and µ2 = µ3 and µ1 = µ3 , and have each test reach the
correct conclusion. If we conduct the three two-sample t-tests. µ1 = µ2 and µ2 = µ3 and µ1 = µ3 then
the certainty that all three are correct is .95 3 = .86 and the chances of a type I error becomes 1– .86 or
σ = .14
As the number of populations increases then the certainty of your conclusion decreases and the
chances of a type 1 error increases. The number of separate tests required also
increases as the number of populations increase. To address these problems ANOVA techniques
were developed. These techniques allow us to perform only one test and allows the certainty of the
conclusion to remain at 1− σ .
For these reasons, ANOVA techniques are preferred in comparing three or more means for equality.
This section will develop the techniques of One Way ANOVA where the populations are defined by
only one factor that is used to differentiate the populations.
Testing a Claim that the Means of Three or More Normal Populations are Equal
using One Way ANOVA
The computations required to perform ANOVA Tests are so tedious that the use of technology is the
only practical way to conduct the tests. The use of software like EXCEL, MINI TAB or the use of a TI
84 Calculator are very common. This section will introduce the use of EXCEL as the tool to conduct
ANOVA tests. You will NOT be required to use EXCELL. The reports table that is produced from
the data table will be given to you. You will be expected to find the numbers in the results table and
compare these numbers. That comparision will allow you to draw a conclusion about the claim the all
the means are equal to each other µ1 = µ2 = µ3
Section 9 –3
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One Way ANOVA Requirements
If the following requirements are met then the techniques of One Way ANOVA can be used to test if
H 0: µ1 = µ2 = µ3
versus
H1: At least one of the population mean is different form the others.
given k samples of size n and a desired level of significance.
1. k populations are defined with only one factor is used to differentiate the populations. All other
factors concerning the population must be the same ( or very similar) except for the one factor is
used to differentiate the populations.
2. The k populations must be normally distributed.
3. The k populations must have the same variance or standard deviation σ 1 = σ 2 = σ 3 ... . A rule of
thumb is that One Way ANOVA procedures can be used if the largest sample standard deviation
is no more than two times as large as the smallest sample standard deviation.
4. A random sample is taken form each of the k populations. The sample size for each sample should
be the same but this is not required. Each sample must be independent of the others. The
subjects in one sample cannot be related to the subjects in the others sample in any way.
The requirements for One Way ANOVA are robust. This means that you can be close to
meeting the requirements and the procedure will still provide a good test of the hypothesis. The
requirement of equal standard deviations can be relaxed if the sample size taken form each of the
separate populations are equal.
If the following requirements are met then the techniques of One Way ANOVA can be used to test
H 0: µ1 = µ2 = µ3
versus
H1: At least one of the population mean is different form the others.
given k samples of size n and a desired level of significance.
Conclusions
If the F test statistic is less than the critical value Do NOT Reject H 0 .
If the F test statistic is greater than the critical value Reject H 0 .
Section 9 –3
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© 2013 Eitel
Example 1
An athletic trainer claims that the resting heart rate of men in the age groups 15 to 24 years old, 25 to
34 years old and 35 to 44 years old are the same. It is known that the restring heart rates for the
populations are normally distributed. An independent random sample of 4 people in each of the 3
age groups was taken and their heart rates were measured and recorded. Test the hypothesis that
the populations have the same mean against the alternate hypothesis that at least one of the means is
different from the other means with a α = .05 level of significance .
Resting Heart Rates by Age Group in beats/min.
15 to 24
year olds
25 to 34
year olds
35 to 44
year olds
60
60
59
57
59
59
58
57
57
61
57
56
Sample Mean
59
58.25
57.75
Sample SD
1.825
1.500
1.500
H 0: µ1 = µ2 = µ3
H1: At least one of the population mean is different from the others.
Excel Results for α = .05
ANOVA
Source of
Variation
SS
df
MS
F
Test Stat
P
value
Critical
f value
Between
Groups
3.17
2
1.58
.0604
.566
4.265
Within
Groups
23.5
9
12.61
TOTAL
26.67
11
The F test statistic is F = .0604 and the F Critical Value is 4.265 The F test statistic is less than
the critical value and falls in the Do NoT Reject H 0 region.
Conclusion. DO Not Reject H 0: µ1 = µ2 = µ3 . I am 95% confident that the means of the three
populations are the same. The average resting heart rate for men in the age groups 15
to 24 years old, 25 to 34 years old and 35 to 44 years old
Section 9 –3
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© 2013 Eitel
Example 2
An student claims that the average test scores on test 1 are the same for the morning, afternoon and
night sections of Stat 300. It is known that the test scores for the populations are normally distributed.
An independent random sample of 4 test scores in each of the 3 section was taken and the scores are
recorded in the table below. Test the hypothesis that the populations have the same mean against
the alternate hypothesis that at least one of the means is different from the other means with a α = .05
level of significance .
Sample Test Scores for Test 1
Morning
Class
Afternoon
Class
Night
Class
72
72
86
74
76
76
73
75
79
74
68
83
Sample Mean
64
72.75
81
Sample SD
2.16
3.59
4.40
Excel Results for α = .05
ANOVA
Source of
Variation
SS
df
MS
F
Test Stat
P
value
Critical
f value
Between
Groups
158.16
2
79.08
6.427
.0184
4.256
Within
Groups
110.75
9
12.30
TOTAL
268.91
11
H 0: µ1 = µ2 = µ3
H1: At least one of the population mean is different from the others.
The F test statistic is F = 6.427 and the F Critical Value is 4.256 The F test statistic is more than the
critical value and falls in the Reject H 0 region.
Conclusion based on H 0 : Reject H 0: µ1 = µ2 = µ3
Conclusion based on the problem:
I am 95% confident that the means of the three populations ARE NOT the same. The average test
scores for morning, afternoon and night sections of STAT 300 ate NOT the same.
Section 9 –3
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Section 9 –3
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© 2013 Eitel