Transcript Analysis of Variance - LISA (Laboratory for

T-TESTS AND
ANALYSIS OF VARIANCE
Jennifer Kensler
ONE SAMPLE T-TEST
ONE SAMPLE T-TEST


Used to test whether the population mean is
different from a specified value.
Example: Is the mean amount of soda in a 20 oz.
bottle different from 20 oz?
STEP 1: FORMULATE THE HYPOTHESES



The population mean is not equal to a specified
value.
H0: μ = μ0
Ha: μ ≠ μ0
The population mean is greater than a specified
value.
H0: μ = μ0
Ha: μ > μ0
The population mean is less than a specified value.
H0: μ = μ0
Ha: μ < μ0
STEP 2: CHECK THE ASSUMPTIONS


The sample is random.
The population from which the sample is drawn
is either normal or the sample size is large.
STEPS 3-5

Step 3: Calculate the test statistic:
y  0
t
s/ n
n
Where


s
2


y

y
 i
i 1
n 1
Step 4: Calculate the p-value based on the
appropriate alternative hypothesis.
Step 5: Write a conclusion.
IRIS EXAMPLE
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A researcher would like to know whether the
mean sepal width of a variety of irises is different
from 3.5 cm.
The researcher randomly measures the sepal
width of 50 irises.
Step 1: Hypotheses
H0: μ = 3.5 cm
Ha: μ ≠ 3.5 cm
JMP

Steps 2-4:
JMP Demonstration
Analyze  Distribution
Y, Columns: Sepal Width
Test Mean
Specify Hypothesized Mean: 3.5
JMP OUTPUT
Step 5 Conclusion: The sepal width is not
significantly different from 3.5 cm.

TWO SAMPLE T-TEST
TWO SAMPLE T-TEST


Two sample t-tests are used to determine
whether the mean of one group is equal to, larger
than or smaller than the mean of another group.
Example: Is the mean cholesterol of people taking
drug A lower than the mean cholesterol of people
taking drug B?
STEP 1: FORMULATE THE HYPOTHESES



The population means of the two groups are not
equal.
H0: μ1 = μ2
Ha: μ1 ≠ μ2
The population mean of group 1 is greater than the
population mean of group 2.
H0: μ1 = μ2
Ha: μ1 > μ2
The population mean of group 1 is less than the
population mean of group 2.
H0: μ1 = μ2
Ha: μ1 < μ2
STEP 2: CHECK THE ASSUMPTIONS
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The two samples are random and independent.
The populations from which the samples are
drawn are either normal or the sample sizes are
large.
The populations have the same standard
deviation.
STEPS 3-5

Step 3: Calculate the test statistic
y1  y2
t
1 1
sp

n1 n2
where
(n1  1) s12  (n2  1) s22
sp 
n1  n2  2
Step 4: Calculate the appropriate p-value.
 Step 5: Write a Conclusion.

TWO SAMPLE EXAMPLE
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
A researcher would like to know whether the
mean sepal width of a setosa irises is different
from the mean sepal width of versicolor irises.
Step 1 Hypotheses:
H0: μsetosa = μversicolor
Ha: μsetosa ≠ μversicolor
JMP

Steps 2-4:
JMP Demonstration:
Analyze  Fit Y By X
Y, Response: Sepal Width
X, Factor: Species
JMP OUTPUT
Step 5 Conclusion: There is strong evidence (pvalue < 0.0001) that the mean sepal widths for
the two varieties are different.
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PAIRED T-TEST
PAIRED T-TEST
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
The paired t-test is used to compare the means of
two dependent samples.
Example:
A researcher would like to determine if
background noise causes people to take longer to
complete math problems. The researcher gives 20
subjects two math tests one with complete silence
and one with background noise and records the
time each subject takes to complete each test.
STEP 1: FORMULATE THE HYPOTHESES
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The population mean difference is not equal to zero.
H0: μdifference = 0
Ha: μdifference ≠ 0
The population mean difference is greater than
zero.
H0: μdifference = 0
Ha: μdifference > 0
The population mean difference is less than a zero.
H0: μdifference = 0
Ha: μdifference < 0
STEP 2: CHECK THE ASSUMPTIONS

The sample is random.
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The data is matched pairs.

The differences have a normal distribution or the
sample size is large.
STEPS 3-5

Step 3: Calculate the test Statistic:
d 0
t
sd / n
Where d bar is the mean of the differences and sd
is the standard deviations of the differences.

Step 4: Calculate the p-value.

Step 5: Write a conclusion.
PAIRED T-TEST EXAMPLE
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A researcher would like to determine whether a
fitness program increases flexibility. The
researcher measures the flexibility (in inches) of
12 randomly selected participants before and
after the fitness program.
Step 1: Formulate a Hypothesis
H0: μAfter - Before = 0
Ha: μ After - Before > 0
PAIRED T-TEST EXAMPLE
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Steps 2-4:
JMP Analysis:
Create a new column of After – Before
Analyze  Distribution
Y, Columns: After – Before
Test Mean
Specify Hypothesized Mean: 0
JMP OUTPUT
Step 5 Conclusion: There is not evidence that
the fitness program increases flexibility.
ONE-WAY ANALYSIS OF
VARIANCE
ONE-WAY ANOVA

ANOVA is used to determine whether three or
more populations have different distributions.
A
B
Medical Treatment
C
ANOVA STRATEGY
The
first step is to use the ANOVA F test to
determine if there are any significant differences
among means.

If the ANOVA F test shows that the means are
not all the same, then follow up tests can be
performed to see which pairs of means differ.
ONE-WAY ANOVA MODEL
yij  i   ij
Where
yij is theresponseof the jth trialon t heith factorlevel
i is themean of theith group
 ij ~ N (0,  2 )
i  1,  , r
j  1, , ni
In other words, for each group the observed
value is the group mean plus some random
variation.
ONE-WAY ANOVA HYPOTHESIS

Step 1: We test whether there is a difference in
the means.
H 0 : 1  2    r
H a : T hei are not all equal.
STEP 2: CHECK ANOVA ASSUMPTIONS
The samples are random and independent of each
other.
 The populations are normally distributed.
 The populations all have the same variance.
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The ANOVA F test is robust to the assumptions
of normality and equal variances.
STEP 3: ANOVA F TEST
A
B
C
A
B
C
Medical Treatment
Compare the variation within the samples to the
variation between the samples.
ANOVA TEST STATISTIC
F
Variationbetween Groups MSG

Variationwithin Groups
MSE
Variation within groups
small compared with
variation between groups
→ Large F
Variation within groups
large compared with
variation between groups
→ Small F
MSG
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The mean square for groups, MSG, measures the
variability of the sample averages.
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SSG stands for sums of squares groups.
SSG
MSG 
r -1
n1 ( y1  y ) 2  n 2 ( y2  y ) 2    n r ( y1  y ) 2

r -1
MSE
Mean square error, MSE, measures the variability
within the groups.
 SSE stands for sums of squares error.

SSE
n-r
(n1 - 1)s12  (n 2 - 1)s22    (n r - 1)s2r

n-r
Where
MSE 
ni
si 
(y
j 1
ij
 yi  )
ni  1
STEPS 4-5
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Step 4: Calculate the p-value.
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Step 5: Write a conclusion.
ANOVA EXAMPLE
A researcher would like to determine if three
drugs provide the same relief from pain.
 60 patients are randomly assigned to a treatment
(20 people in each treatment).
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Step 1: Formulate the Hypotheses
H0: μDrug A = μDrug B = μDrug C
Ha : The μi are not all equal.
STEPS 2-4
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JMP demonstration
Analyze  Fit Y By X
Y, Response: Pain
X, Factor: Drug
EXAMPLE 1: JMP OUTPUT AND
CONCLUSION
Step 5 Conclusion: There is strong evidence
that the drugs are not all the same.
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FOLLOW-UP TEST
The p-value of the overall F test indicates that
level of pain is not the same for patients taking
drugs A, B and C.
 We would like to know which pairs of treatments
are different.
 One method is to use Tukey’s HSD (honestly
significant differences).
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TUKEY TESTS
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Tukey’s test simultaneously tests
H 0 : i  i '
H a : i  i '
for all pairs of factor levels. Tukey’s HSD
controls the overall type I error.
JMP demonstration
Oneway Analysis of Pain By Drug 
Compare Means  All Pairs, Tukey HSD
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JMP OUTPUT
The JMP output shows that drugs A and C are
significantly different.
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ANALYSIS OF COVARIANCE
ANALYSIS OF COVARIANCE (ANCOVA)
Covariates are variables that may affect the
response but cannot be controlled.
 Covariates are not of primary interest to the
researcher.
 We will look at an example with two covariates,
the model is

yij  i  covariates  ij
ANCOVA EXAMPLE
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Consider the previous example where we tested
whether the patients receiving different drugs
reported different levels of pain. Perhaps age and
gender may influence the efficacy of the drug. We
can use age and gender as covariates.
JMP demonstration
Analyze  Fit Model
Y: Pain
Add: Drug
Age
Gender
JMP OUTPUT
CONCLUSION
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
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The one sample t-test allows us to test whether
the mean of a group is equal to a specified value.
The two sample t-test and paired t-test allows us
to determine if the means of two groups are
different.
ANOVA and ANCOVA methods allow us to
determine whether the means of several groups
are statistically different.
SAS AND SPSS

For information about using SAS and SPSS to do
ANOVA:
http://www.ats.ucla.edu/stat/sas/topics/anova.htm
http://www.ats.ucla.edu/stat/spss/topics/anova.htm
REFERENCES
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Fisher’s Irises Data (used in one sample and two
sample t-test examples).
Flexibility data (paired t-test example):
Michael Sullivan III. Statistics Informed
Decisions Using Data. Upper Saddle River, New
Jersey: Pearson Education, 2004: 602.