1. Hypothesis test flow chart
START HERE
number of
correlations
number of
variables
χ2 test for
independence
(19.9) Table I
Test H0: r=0
(17.2)
Table G 1
correlation (r)
frequency
data 2
Measurement
scale
Means 2 1
basic χ2 test
(19.5)
Table I
Test H0: r1= r2
(17.4)
Tables H and A
Do you know s?
number of
means
number of
factors
z -test
(13.1)
Table A
Yes 1
More than 2 2
2-way ANOVA
Ch 21
Table E No 2 1
independent
samples?
1-way ANOVA
Ch 20
Table E
t -test
(13.14)
Table D
Yes No
Test H0: m1= m2
(15.6)
Table D
Test H0: D=0
(16.4)
Table D

2. Chapter 18: Testing for difference among three or more groups: One way Analysis of Variance (ANOVA)B C
Suppose you wanted to compare the results of three tests (A, B and C) to see if there was any differences difficulty. To test this, you randomly sample these ten scores from each of the three populations of test scores.
How would you test to see if there was any difference across the mean scores for these three tests?
The first thing is obvious – calculate the mean for each of the three samples of 10 scores.
But then what? You could run three two-sample t-tests on each of the pairs (A vs. B, A vs. C and B vs. C). 84 85 62 78 62 79 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75
Means

3. There are two problems with this: C
You could run an two-sample t-test on each of the pairs (A vs. B, A vs. C and B vs. C).
There are two problems with this:
The three tests wouldn’t be truly independent of each other, since they contain common values, and
We run into the problem of making multiple comparisons: If we use an a value of .05, the probability of obtaining at least one significant comparison by chance is 1-(1-.05)3, or about .14 84 85 62 78 62 79 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75
Means
So how do we test the null hypothesis: H0: mA = mB = mC ?

4. So how do we test the null hypothesis: H0: mA = mB = mC ?
In the 1920’s Sir Ronald Fisher developed a method called ‘Analysis of Variance’ or ANOVA to test hypotheses like this.
The trick is to look at the amount of variability between the means.
So far in this class, we’ve usually talked about variability in terms of standard deviations. ANOVA’s focus on variances instead, which (of course) is the square of the standard deviation. The intuition is the same.
The variance of these three mean scores (81, 72 and 75) is 22.5
Intuitively, you can see that if the variance of the means scores is ‘large’, then we should reject H0.
But what do we compare this number 22.5 to? A B C 84 85 62 78 62 79 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75
Means

5. So how do we test the null hypothesis: H0: mA = mB = mC ?
The variance of these three mean scores (81, 72 and 75) is 22.5
How ‘large’ is 22.5?
Suppose we knew the standard deviation of the population of scores (s).
If the null hypothesis is true, then all scores across all three columns are drawn from a population with standard deviation s.
It follows that the mean of n scores should be drawn from a population with standard deviation:
This means multiplying the variance of the means by n gives us an estimate of the variance of the population. A B C 84 85 62 78 62 79 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75
Means
With a little algebra:

6. The variance of these three mean scores (81, 72 and 75) is 22.5
Multiplying the variance of the means by n gives us an estimate of the variance of the population.
For our example, A B C 84 85 62 78 62 79
We typically don’t know what s2 is. But like we do for t-tests, we can use the variance within our samples to estimate it. The variance of the 10 numbers in each column (61, 94, and 55) should each provide an estimate of s2.
We can combine these three estimates of s2 by taking their average, which is 70. 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75
Means
n x Variance of means
225 61 94 55
Variances
Mean of variances 70

7. One is n times the variance of the means of each column.
If H0: mA = mB = mC is true, we now have two separate estimates of the variance of the population (s2).
One is n times the variance of the means of each column.
The other is the mean of the variances of each column.
If H0 is true, then these two numbers should be, on average, the same, since they’re both estimates of the same thing (s2).
For our example, these two numbers (225 and 70) seem quite different.
Remember our intuition that a large variance of the means should be evidence against H0. Now we have something to compare it to seems large compared to 70. A B C 84 85 62 78 62 79 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75
Means
n x Variance of means
225
Variances
Mean of variances 61 94 55 70

8. Fcrit for a = .05 and df’s of 2 and 27 is 3.35.
When conducting an ANOVA, we compute the ratio of these two estimates of s2. This ratio is called the ‘F statistic’. For our example, 225/70 =
If H0 is true, then the value of F should be around 1. If H0 is not true, then F should be significantly greater than 1.
We determine how large F should be for rejecting H0 by looking up Fcrit in Table E. F distributions depend on two separate degrees of freedom – one for the numerator and one for the denominator.
df for the numerator is k-1, where k is the number of columns or ‘treatments’. For our example, df is 3-1 =2.
df for the denominator is N-k, where N is the total number of scores. In our case, df is 30-3 = 27. A B C 84 85 62 78 62 79 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75
Means
n x Variance of means
Fcrit for a = .05 and df’s of 2 and 27 is 3.35.
Since Fobs = 3.23 is less than Fcrit, we fail to reject H0. We cannot conclude that the exam scores come from populations with different means.
225
Ratio (F)
3.23
Variances
Mean of variances 61 94 55 70

9. Fcrit for a = .05 and df’s of 2 and 27 is 3.35.
Since Fobs = 3.23 is less than Fcrit, we fail to reject H0. We cannot conclude that the exam scores come from populations with different means.
Instead of finding Fcrit in Table E, we could have calculated the p-value using our F-calculator. Reporting p-values is standard.
Our p-value for F=3.23 with 2 and 27 degrees of freedom is p=.0552
Since our p-value is greater then .05, we fail to reject H0

10. Our resulting F statistic is 15.32.
Example: Consider the following n=12 samples drawn from k=5 groups. Use an ANOVA to test the hypothesis that the means of the populations that these 5 groups were drawn from are different.
Answer: The 5 means and variances are calculated below, along with n x variance of means, and the mean of variances.
Our resulting F statistic is
Our two dfs are k-1=4 (numerator) and 60-5 = 55(denominator). Table E shows that Fcrit for 4 and 55 is 2.54.
Fobs > Fcrit so we reject H0. A B C D E 87 99 78 91 81 69
105 70 51 76
104 75 64 79 71
108 86 68 74 83 92 62 72 66 93 85
111 65
107 84 67 61 97 90 82 78 96 74
100 97 76 46 75
149
Means
Variances
Mean of variances 93
n x Variance of means
1429
Ratio (F)
15.32

11. What does the probability distribution F(dfbet,dfw) look like?1 2 3 4 5
F(2,5)
F(2,10)
F(2,50)
F(2,100)
F(10,5)
F(10,10)
F(10,50)
F(10,100)
F(50,5)
F(50,10)
F(50,50)
F(50,100)

12. For a typical ANOVA, the number of samples in each group may be different, but the intuition is the same - compute F which is the ratio of the variance of the means over the mean of the variances.
Formally, the variance is divided up the following way:
Given a table of k groups, each containing ni scores (i= 1,2, …, k), we can represent the deviation of a given score, X from the mean of all scores, called the grand mean as:
Deviation of X from the grand mean
Deviation of X from the mean of the group
Deviation of the mean of the group from the grand mean

13. The total sums of squares can be partitioned into two numbers:
Total sum of squares:
SStotal
Within-groups sum of squares:
SSw
Between-groups sum of squares:
SSbet
SSbet is a measure of the variability between groups.
It is used as the numerator in our F-tests
The variance between groups, called ‘mean squared error’, or MSbet, is calculated by dividing SSbet by its degrees of freedom dfbet = k-1
MSbet=SSbet/dfbet and is another estimate of s2 if H0 is true. This is essentially n times the variance of the means.
If H0 is not true, then s2bet is an estimate of s2 plus any ‘treatment effect’ that would add to a difference between the means. .

14. The total sums of squares can be partitioned into two numbers:
Total sum of squares:
SStotal
Within-groups sum of squares:
SSw
Between-groups sum of squares:
SSbet
SSw is a measure of the variability within each group.
It is used as the denominator in all F-tests.
The variance within each group, MSw is calculated by dividing SSw by its degrees of freedom dfw = ntotal – k
MSw=SSw/dfw This is an estimate of s2 This is essentially the mean of the variances within each group. (It is exactly the mean of variances if our sample sizes are all the same.)

15. The F ratio is calculated by dividing up the sums of squares and df into ‘between’ and ‘within’
SStotal = SSw + SSbet
Variances are then calculated by dividing SS by df
MSbet=SSbet/dfbet
SStotal
MSw=SSw/dfw
SSw
SSbet
F is the ratio of variances between and within
F= 𝑀𝑆 𝑏𝑒𝑡 𝑀𝑆 𝑤
dftotal = dfw + dfbet
dftotal =ntotal-1
dfw =ntotal-k
dfbet =k-1

16. F= 𝑀𝑆 𝑏𝑒𝑡 𝑀𝑆 𝑤 Finally, the F ratio is the ratio of MSbet and MSw
We can write all these calculated values in a summary table like this:
Source SS df MS F
Between
k-1
MSbet=SSbet/dfbet
Within
ntotal-k
MSw=SSw/dfw
Total
ntotal-1
𝑀𝑆 𝑏𝑒𝑡 𝑀𝑆 𝑤
(k is the number of groups)

17. Calculating SStotal grand mean: Source SS df MS F Between Within Total87 99 78 91 81 69
105 70 51 76
104 75 64 79 71
108 86 68 74 83 92 62 72 66 93 85
111 65
107 84 67 61 97 90 82
Calculating SStotal
grand mean: 78 96 74
100 97 76 46 75
149
Means
Variances
Mean of variances 93
n x Variance of means
1429
Ratio (F)
15.32
Source SS df MS F
Between
Within
Total
10847 59

18. 𝑀𝑆 𝑏𝑒𝑡 = 𝑆𝑆 𝑏𝑒𝑡 𝑑𝑓 𝑏𝑒𝑡 = 5717 4 =1429 Calculating SSbet and MSbet87 99 78 91 81 69
105 70 51 76
104 75 64 79 71
108 86 68 74 83 92 62 72 66 93 85
111 65
107 84 67 61 97 90 82
Calculating SSbet and MSbet
𝑀𝑆 𝑏𝑒𝑡 = 𝑆𝑆 𝑏𝑒𝑡 𝑑𝑓 𝑏𝑒𝑡 = =1429 78 96 74
100 97 76 46 75
149
Means
Variances
Mean of variances 93
n x Variance of means
1429
Ratio (F)
15.32
Source SS df MS F
Between
5717
5-1=4
1429
Within
Total
10847 59

19. 𝑀𝑆 𝑤 = 𝑆𝑆 𝑤 𝑑𝑓 𝑤 = 5130 55 =93 Calculating SSw and MSw Source SS df MSB C D E 87 99 78 91 81 69
105 70 51 76
104 75 64 79 71
108 86 68 74 83 92 62 72 66 93 85
111 65
107 84 67 61 97 90 82
Calculating SSw and MSw
𝑀𝑆 𝑤 = 𝑆𝑆 𝑤 𝑑𝑓 𝑤 = =93 78 96 74
100 97 76 46 75
149
Means
Variances
Mean of variances 93
n x Variance of means
1429
Ratio (F)
15.32
Source SS df MS F
Between
5717
5-1=4
1429
Within
5130
12x5-5=55 93
Total
10847 59

20. A B C D E 87 99 78 91 81 69
105 70 51 76
104 75 64 79 71
108 86 68 74 83 92 62 72 66 93 85
111 65
107 84 67 61 97 90 82
Fcrit with dfs of 4 and 55 and a = .05 is 2.54
Our decision is to reject H0 since > 2.54
Calculating F
F= 𝑀𝑆 𝑏𝑒𝑡 𝑀𝑆 𝑤 = =15.32 78 96 74
100 97 76 46 75
149
Means
Variances
Mean of variances 93
n x Variance of means
1429
Ratio (F)
15.32
Source SS df MS F
Between
5717
5-1=4
1429
15.32
Within
5130
12x5-5=55 93
Total
10847 59