Download analysis of variance and experimental design

Basic concept
Measures of central tendency
Measures of dispersion & variability
Measures of tendency central
Arithmetic mean (= simple average)
• Best estimate of population mean is
the sample mean, X
n
summation
X 
X
i 1
n
measurement in
population
i
index of
measurement
sample size
Measures of variability

All describe how “spread out” the data
are.
1. Sum of squares,
sum of squared deviations from
the mean
• For a sample,
SS   ( X i  X )
2
2.
Average or mean sum of squares
= variance, s2:
• For a sample,
s
2
(X


i
 X)
n 1
2
Why?
s

2
(X


i
 X)
2
n 1
n – 1 represents the degrees of
freedom, , or number of independent
quantities in the estimate s2.
Greek
letter “nu”
n
 (X
i
 X)  0
i 1
• therefore, once n – 1 of all deviations are
specified, the last deviation is already
determined.
• Variance has squared measurement
units – to regain original units, take the
square root …
3. Standard deviation, s
• For a sample, s 
 (X
i
 X)
n 1
2
4. Standard error of the mean
2
• For a sample,

s
sX 
n
Standard error of the mean is a
measure of variability among the
means of repeated samples from a
population.
A Population of Values
14
16
13
14
15
14
12
14 16 14
13
14
14 12
13
14
13
13
16 15
14
13
14
15
16
13
14
14
Means of repeated random samples, each
with sample size, n = 5 values …
X  14
X  15
X  13

For a large enough number of large
samples, the frequency distribution of
the sample means (= sampling
distribution), approaches a normal
distribution.
Frequency
Normal distribution: bell-shaped curve
Sample mean
Testing
statistical hypotheses between 2 means
1. State the research question in
terms of statistical hypotheses.
• We always start with a statement that
hypothesizes “no difference”, called
the null hypothesis = H0.
E.g., H0: Mean bill length of female
hummingbirds is equal to mean bill
length of male hummingbirds, µ=µ .

Then we formulate a statement
that must be true if the null
hypothesis is false, called the
alternate hypothesis = HA .
E.g., HA: Mean bill length of female
hummingbirds is not equal to mean bill
length of male hummingbirds, µµ .
• If we reject H0 as a result of sample
evidence, then we conclude that HA is
true.
2. Choose an appropriate statistical
test that would allow you to reject
H0 if H0 were false.
E.g., Student’s t test for
hypotheses about means
William Sealey Gosset
(a.k.a. “Student”)
Is the difference
between sample
means bigger than
we would expect,
given the variability
in the sampled
populations?
t Statistic,
Standard error of the
difference between the
sample means
X1  X 2
t
s X 1  X 2 
Mean of
sample 1
Mean of
sample 2
• To estimate s(X1—X2), we must first know
the relation between both populations.
Relation between populations


Dependent population
Independent population
1. Identical (homogenous ) variance
2. Not identical (heterogeneous) variance
Independent Population with
homogenous variances

Pooled variance:
• Then,
s X 1  X 2  
SS1  SS2
s 
1   2
2
p
s
2
p
n1

s
2
p
n2
3. Select the level of significance for
the statistical test.
• Level of significance = alpha value =
 = the probability of incorrectly
rejecting the null hypothesis when it
is, in fact, true.

Traditionally, researchers choose 
= 0.05.
• 5 percent of the time, or 1 time out of
20, the statistical test will reject H0
when it is true.
• Note: the choice of 0.05 is arbitrary!
4. Determine the critical value the
test statistic must attain to be
declared significant.
Frequency
• Most test statistics have a frequency
distribution …
Test statistic

When sample sizes are small, the
sampling distribution is described
better by the t distribution than by
the standard normal (Z)
distribution.
• Shape of t distribution depends on
degrees of freedom,  = n – 1.
Z = t(=)
t(=25)
t(=5)
t
t(=1)

The distribution of a test statistic is
divided into an area of acceptance
and an area of rejection.
For  = 0.05
0.025
Area of
Rejection
Lower
critical
value
0.95
Area of
Acceptance
0
t
0.025
Area of
Rejection
Upper
critical
value
5. Perform the statistical test.
E.g., Mean bill length from a sample of
5 female hummingbirds, X1 = 15.75;
Mean bill length from a sample of 5
male hummingbirds, X2 = 14.25;
 s X 1  X 2   0.5
X 1  X 2 15.75  14.25
t

 3.00
sX 1 X 2 
0.5
6. Draw and state the conclusions.
• Compare the calculated test
statistic with the critical test
statistic at the chosen .
• Reject or fail to reject H0.
• Obtain the P-value = probability for
the test statistic.
• Critical t for a test about
equality = t(2),
E.g., to test H0: µ = µ, HA: µ 
µ at  = 0.05 using n = 5, n = 5,
t(2), = t0.05(2),8 = 2.306.
, if |t|  2.306, reject H0.
Since calculated t > t0.05(2),8
(because 3.000 > 2.306), reject
H0.
Conclude that hummingbird bill
length is sexually size-dimorphic.
What is the probability, P, of
observing by chance a difference as
large as we saw between female and
male hummingbird bill lengths?
 0.01 < P < 0.02
Analysis of Variance
(ANOVA)
What is ANOVA?


ANOVA (Analysis of Variance) is a procedure designed to
determine if the manipulation of one or more independent
variables in an experiment has a statistically significant
influence on the value of the dependent variable.
It is assumed
• Each independent variable is categorical (nominal scale).
Independent variables are called Factors and their values are called
levels.
• The dependent variable is numerical (ratio scale)

The basic idea is that the “variance” of the dependent
variable given the influence of one or more independent
variables {Expected Sum of Squares for a Factor} is checked
to see if it is significantly greater than the “variance” of
the dependent variable (assuming no influence of the
independent variables) {also known as the Mean-SquareError(MSE)}.
Analysis of Variance




Analysis of Variance (ANOVA) can be used to test for the
equality of three or more population means using data
obtained from observational or experimental studies.
We want to use the sample results to test the following
hypotheses.
H0: 1 = 2 = 3 = . . . = k
Ha: Not all population means are equal
If H0 is rejected, we cannot conclude that all population
means are different.
Rejecting H0 means that at least two population means
have different values.
Assumptions for Analysis of
Variance




For each population, the response variable is
normally distributed.
The variance of the response variable, denoted 2,
is the same for all of the populations.
The effect of independent variable is additive
The observations must be independent.
Analysis of Variance:
Testing for the Equality of K Population
Means




Between-Treatments Estimate of
Population Variance
Within-Treatments Estimate of Population
Variance
Comparing the Variance Estimates: The F
Test
ANOVA Table
Between-Treatments Estimate
of Population Variance

A between-treatments estimate of σ2 is called the
mean square due to treatments (MSTR).
k
MSTR 
2
n
(
x

x
)
 j j
j 1
k1
 The numerator of MSTR is called the sum of squares due
to treatments (SSTR).
 The denominator of MSTR represents the degrees of
freedom associated with SSTR.
Within-Treatments Estimate
of Population Variance

The estimate of 2 based on the variation of the
sample observations within each treatment is called
the mean square due to error (MSE).
k
MSE 
2
(
n

1)
s
 j
j
j 1
nT  k
 The numerator of MSE is called the sum of squares due
to error (SSE).
 The denominator of MSE represents the degrees of
freedom associated with SSE.
Comparing the Variance Estimates:
The F Test



If the null hypothesis is true and the ANOVA
assumptions are valid, the sampling distribution of
MSTR/MSE is an F distribution with MSTR d.f. equal
to k - 1 and MSE d.f. equal to nT - k.
If the means of the k populations are not equal, the
value of MSTR/MSE will be inflated because MSTR
overestimates σ 2.
Hence, we will reject H0 if the resulting value of
MSTR/MSE appears to be too large to have been
selected at random from the appropriate F
distribution.
Test for the Equality of k
Population Means

Hypotheses
H0: 1 = 2 = 3 = . . . = k
Ha: Not all population means are equal

Test Statistic
F = MSTR/MSE
Test for the Equality of k
Population Means

Rejection Rule
Using test statistic:
Using p-value:
Reject H0 if F > Fa
Reject H0 if p-value < a
where the value of Fa is based on an F distribution
with k - 1 numerator degrees of freedom and nT - k
denominator degrees of freedom
Sampling Distribution of MSTR/MSE

The figure below shows the rejection region
associated with a level of significance equal to 
where F denotes the critical value.
Do Not Reject H0
Reject H0
F
Critical Value
MSTR/MSE
ANOVA Table
Source of
Sum of
Variation
Squares
Treatment SSTR
Error
SSE
Total
SST
Degrees of
Freedom
k-1
nT - k
nT - 1
Mean
Squares
F
MSTR MSTR/MSE
MSE
SST divided by its degrees of freedom nT - 1 is simply
the overall sample variance that would be obtained if
we treated the entire nT observations as one data set.
k
nj
SST   ( xij  x) 2  SSTR  SSE
j 1 i 1
Example: Reed Manufacturing
Analysis of Variance
J. R. Reed would like to know if the mean number
of hours worked per week is the same for the
department managers at her three manufacturing
plants (Buffalo, Pittsburgh, and Detroit).
A simple random sample of 5 managers from each of
the three plants was taken and the number of hours
worked by each manager for the previous week is
shown on the next slide.

Example: Reed Manufacturing

Sample Data
Observation
1
2
3
4
5
Sample Mean
Sample Variance
Plant 1
Buffalo
48
54
57
54
62
55
26.0
Plant 2
Pittsburgh
73
63
66
64
74
Plant 3
Detroit
51
63
61
54
56
68
26.5
57
24.5
Example: Reed Manufacturing

Hypotheses
H 0:  1 =  2 =  3
Ha: Not all the means are equal
where:
 1 = mean number of hours worked per
week by the managers at Plant 1
 2 = mean number of hours worked per
week by the managers at Plant 2
 3 = mean number of hours worked per
week by the managers at Plant 3
Example: Reed Manufacturing

Mean Square Due to Treatments
Since the= sample sizes are all equal
μ= (55 + 68 + 57)/3 = 60
SSTR = 5(55 - 60)2 + 5(68 - 60)2 + 5(57 - 60)2 = 490
MSTR = 490/(3 - 1) = 245
 Mean Square Due to Error
SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308
MSE = 308/(15 - 3) = 25.667
Example: Reed Manufacturing

F - Test
If H0 is true, the ratio MSTR/MSE should be
near 1 because both MSTR and MSE are
estimating 2.
If Ha is true, the ratio should be significantly
larger than 1 because MSTR tends to
overestimate
2.
Example: Reed Manufacturing

Rejection Rule
Using test statistic:
Reject H0 if F > 3.89
Using p-value:
Reject H0 if p-value < .05
where F.05 = 3.89 is based on an F distribution
with 2 numerator degrees of freedom and 12
denominator degrees of freedom
Example: Reed Manufacturing


Test Statistic
F = MSTR/MSE = 245/25.667 = 9.55
Conclusion
F = 9.55 > F.05 = 3.89, so we reject H0. The
mean number of hours worked per week by
department managers is not the same at each
plant.

Example: Reed Manufacturing
ANOVA Table
Source of
Variation
Treatments
Error
Total
Sum of
Squares
490
308
798
Degrees of
Freedom
2
12
14
Mean
Square
245
25.667
F
9.55
Using Excel’s Anova:
Single Factor Tool



Step 1
Step 2
Step 3
Select the Tools pull-down menu
Choose the Data Analysis option
Choose Anova: Single Factor
from the list of Analysis Tools
… continued
Using Excel’s Anova:
Single Factor Tool

Step 4 When the Anova: Single Factor dialog
box appears:
Enter B1:D6 in the Input Range box
Select Grouped By Columns
Select Labels in First Row
Enter .05 in the Alpha box
Select Output Range
Enter A8 (your choice) in the Output
Range box
Click OK
Using Excel’s Anova:
Single Factor Tool

Value Worksheet (top portion)
1
2
3
4
5
6
A
Observation
1
2
3
4
5
B
Buffalo
48
54
57
54
62
C
Pittsburgh
73
63
66
64
74
D
Detroit
51
63
61
54
56
E
Using Excel’s Anova:
Single Factor Tool
Value Worksheet (bottom portion)
A

8 Anova: Single Factor
9
10 SUMMARY
11
Groups
12 Buffalo
13 Pittsburgh
14 Detroit
15
16
17 ANOVA
18 Source of Variation
19 Between Groups
20 Within Groups
21
22 Total
B
C
Count
5
5
5
SS
490
308
798
D
E
F
G
Sum Average Variance
275
55
26
340
68
26.5
285
57
24.5
df
MS
F
P-value F crit
2
245 9.54545 0.00331 3.88529
12 25.6667
14
Using Excel’s Anova:
Single Factor Tool

Using the p-Value
• The value worksheet shows that the p-value is
.00331
• The rejection rule is “Reject H0 if p-value <
.05”
• Thus, we reject H0 because the p-value =
.00331 <  = .05
• We conclude that the mean number of hours
worked per week by the managers differ
among the three plants