six easy steps for an anova
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Six Easy Steps for an ANOVA. 1) State the hypothesis 2) Find the F-critical value 3) Calculate the F-value 4) Decision 5) Create the summary table 6) Put answer into words. Example. - PowerPoint PPT PresentationTRANSCRIPT
Six Easy Steps for an ANOVA
• 1) State the hypothesis
• 2) Find the F-critical value
• 3) Calculate the F-value
• 4) Decision
• 5) Create the summary table
• 6) Put answer into words
Example
• Want to examine the effects of feedback on self-esteem. Three different conditions -- each have five subjects
• 1) Positive feedback
• 2) Negative feedback
• 3) Control
• Afterward all complete a measure of self-esteem that can range from 0 to 10.
Example:
• Question: Is the type of feedback a person receives significantly (.05) related their self-esteem?
Results
Positive Feedback
Negative Feedback
Control
8 5 2
7 6 4
9 7 5
10 4 3
6 3 6
Step 1: State the Hypothesis
• H1: The three population means are not all equal
• H0: pos = neg = cont
Step 2: Find F-Critical
• Step 2.1• Need to first find dfbetween and dfwithin
• Dfbetween = k - 1 (k = number of groups)
• dfwithin = N - k (N = total number of observations)
• dftotal = N - 1
• Check yourself
• dftotal = Dfbetween + dfwithin
Step 2: Find F-Critical
• Step 2.1• Need to first find dfbetween and dfwithin
• Dfbetween = 2 (k = number of groups)
• dfwithin = 12 (N = total number of observations)
• dftotal = 14
• Check yourself• 14 = 2 + 12
Step 2: Find F-Critical
• Step 2.2
• Look up F-critical using table F on pages 370 - 373.
• F (2,12) = 3.88
Step 3: Calculate the F-value
• Has 4 Sub-Steps
• 3.1) Calculate the needed ingredients
• 3.2) Calculate the SS
• 3.3) Calculate the MS
• 3.4) Calculate the F-value
Step 3.1: Ingredients
XX2
Tj2
• N
• n
Step 3.1: Ingredients
Positive Feedback
Negative Feedback
Control
8 5 2
7 6 4
9 7 5
10 4 3 6 3 6
X
PositiveFeedback
NegativeFeedback
Control
8 5 2
7 6 4
9 7 5
10 4 3
6 3 6
Xp = 40 Xn = 25 Xc = 20
X = 85
X2
PositiveFeedback
NegativeFeedback
Control
8 64 5 25 2 4
7 49 6 36 4 16
9 81 7 49 5 25
10 100 4 16 3 9
6 36 3 9 6 36
Xp = 40 Xn = 25 Xc = 20
X = 85
X2 = 555
X2p = 330 X2
n = 135 X2c = 90
T2 = (X)2 for each groupPositive
FeedbackNegativeFeedback
Control
8 64 5 25 2 4
7 49 6 36 4 25
9 81 7 49 5 25
10 100 4 16 3 9
6 36 3 9 6 36
Xp = 40 Xn = 25 Xc = 20
X = 85
X2 = 555
X2p = 330 X2
n = 135 X2c = 90
T2p = 1600 T2
n = 625 T2c = 400
Tj2
PositiveFeedback
NegativeFeedback
Control
8 64 5 25 2 4
7 49 6 36 4 25
9 81 7 49 5 25
10 100 4 16 3 9
6 36 3 9 6 36
Xp = 40 Xn = 25 Xc = 20
X = 85
X2 = 555
Tj2
= 2625
X2p = 330 X2
n = 135 X2c = 90
T2p = 1600 T2
n = 625 T2c = 400
NPositive
FeedbackNegativeFeedback
Control
8 64 5 25 2 4
7 49 6 36 4 25
9 81 7 49 5 25
10 100 4 16 3 9
6 36 3 9 6 36
Xp = 40 Xn = 25 Xc = 20
X = 85
X2 = 555
Tj2
= 2625
N = 15
X2p = 330 X2
n = 135 X2c = 90
T2p = 1600 T2
n = 625 T2c = 400
nPositive
FeedbackNegativeFeedback
Control
8 64 5 25 2 4
7 49 6 36 4 25
9 81 7 49 5 25
10 100 4 16 3 9
6 36 3 9 6 36
Xp = 40 Xn = 25 Xc = 20
X = 85
X2 = 555
Tj2
= 2625
N = 15
n = 5
X2p = 330 X2
n = 135 X2c = 90
T2p = 1600 T2
n = 625 T2c = 400
Step 3.2: Calculate SSX = 85
X2 = 555
Tj2
= 2625
N = 15
n = 5• SStotal
Step 3.2: Calculate SS
• SStotal
55585
1573.33
X = 85
X2 = 555
Tj2
= 2625
N = 15
n = 5
Step 3.2: Calculate SS
• SSWithin
X = 85
X2 = 555
Tj2
= 2625
N = 15
n = 5
Step 3.2: Calculate SS
• SSWithin
5552625
530
X = 85
X2 = 555
Tj2
= 2625
N = 15
n = 5
Step 3.2: Calculate SS
• SSBetween
X = 85
X2 = 555
Tj2
= 2625
N = 15
n = 5
Step 3.2: Calculate SS
• SSBetween
2625
5
85
15
43.33
X = 85
X2 = 555
Tj2
= 2625
N = 15
n = 5
Step 3.2: Calculate SS
• Check!
• SStotal = SSBetween + SSWithin
Step 3.2: Calculate SS
• Check!
• 73.33 = 43.33 + 30
Step 3.3: Calculate MS
Step 3.3: Calculate MS
43.33
221.67
Calculating this Variance Ratio
Step 3.3: Calculate MS
30
122.5
Step 3.4: Calculate the F value
21.67
2.58.67
Step 3.4: Calculate the F value
Step 4: Decision
• If F value > than F critical– Reject H0, and accept H1
• If F value < or = to F critical– Fail to reject H0
Step 4: Decision
• If F value > than F critical– Reject H0, and accept H1
• If F value < or = to F critical– Fail to reject H0
F value = 8.67
F crit = 3.88
Step 5: Create the Summary Table
Source SS df MS F
Between 43.33 2 21.67 8.67*
Within 30.00 12 2.5
Total 73.33 14
Step 6: Put answer into words
• Question: Is the type of feedback a person receives significantly (.05) related their self-esteem?
• H1: The three population means are not all equal
• The type of feedback a person receives is related to their self-esteem
SPSS
43.333 2 21.667 8.667 .005
30.000 12 2.500
73.333 14
BetweenGroups
WithinGroups
Total
ESTEEM
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Practice
• You are interested in comparing the performance of three models of cars. Random samples of five owners of each car were used. These owners were asked how many times their car had undergone major repairs in the last 2 years.
Results
VW Beetle
Ford Mustang
Geo Metro
2 5 9
1 4 6
2 3 3
3 4 7
2 4 5
Practice
• Is there a significant (.05) relationship between the model of car and repair records?
Step 1: State the Hypothesis
• H1: The three population means are not all equal
• H0: V = F = G
Step 2: Find F-Critical
• Step 2.1• Need to first find dfbetween and dfwithin
• Dfbetween = 2 (k = number of groups)
• dfwithin = 12 (N = total number of observations)
• dftotal = 14
• Check yourself• 14 = 2 + 12
Step 2: Find F-Critical
• Step 2.2
• Look up F-critical using table F on pages 370 - 373.
• F (2,12) = 3.88
Step 3.1: Ingredients
X = 60X2 = 304Tj
2 = 1400
• N = 15
• n = 5
Step 3.2: Calculate SSX = 60
X2 = 304
Tj2
= 1400
N = 15
n = 5• SStotal
Step 3.2: Calculate SS
• SStotal
30460
1564
X = 60
X2 = 304
Tj2
= 1400
N = 15
n = 5
Step 3.2: Calculate SS
• SSWithin
X = 60
X2 = 304
Tj2
= 1400
N = 15
n = 5
Step 3.2: Calculate SS
• SSWithin
3041400
524
X = 60
X2 = 304
Tj2
= 1400
N = 15
n = 5
Step 3.2: Calculate SS
• SSBetween
X = 60
X2 = 304
Tj2
= 1400
N = 15
n = 5
Step 3.2: Calculate SS
• SSBetween
1400
5
60
15
40
X = 60
X2 = 304
Tj2
= 1400
N = 15
n = 5
Step 3.2: Calculate SS
• Check!
• SStotal = SSBetween + SSWithin
Step 3.2: Calculate SS
• Check!
• 64 = 40 + 24
Step 3.3: Calculate MS
Step 3.3: Calculate MS
40
220
Calculating this Variance Ratio
Step 3.3: Calculate MS
24
122
Step 3.4: Calculate the F value
20
210
Step 3.4: Calculate the F value
Step 4: Decision
• If F value > than F critical– Reject H0, and accept H1
• If F value < or = to F critical– Fail to reject H0
Step 4: Decision
• If F value > than F critical– Reject H0, and accept H1
• If F value < or = to F critical– Fail to reject H0
F value = 10
F crit = 3.88
Step 5: Create the Summary Table
Source SS df MS F
Between 40 2 20 10*
Within 24 12 2
Total 64 14
Step 6: Put answer into words• Question: Is there a significant (.05) relationship
between the model of car and repair records?
• H1: The three population means are not all equal
• There is a significant relationship between the type of car a person drives and how often the car is repaired
A way to think about ANOVA
• Make no assumption about Ho
– The populations the data may or may not have equal means
A way to think about ANOVA
VW Beetle
Ford Mustang
Geo Metro
2 5 9
1 4 6
2 3 3
3 4 7
2 4 5
2 4 6
A way to think about ANOVA
• The samples can be used to estimate the variance of the population
• Assume that the populations the data are from have the same variance
• It is possible to use the same variances to estimate the variance of the populations
222222 ,, GEOGEOFordFordVWVW SSS
222GEOFordVW
k
S je
2
2
VW Beetle
Ford Mustang
Geo Metro
2 5 9
1 4 6
2 3 3
3 4 7
2 4 5
S2 = .50 S2 = .50 S2 = 5.0
222222 ,, GEOGEOFordFordVWVW SSS
222GEOFordVW
A way to think about ANOVA
00.23
)0.550.50(.2
e
40.000 2 20.000 10.000 .003
24.000 12 2.000
64.000 14
BetweenGroups
WithinGroups
Total
REP
Sum ofSquares df
MeanSquare F Sig.
ANOVA
A way to think about ANOVA
• Assume about Ho is true
– The population mean are not different from each other
• They are three samples from the same population– All have the same variance and the same
mean
VW Beetle
Ford Mustang
Geo Metro
2 5 9
1 4 6
2 3 3
3 4 7
2 4 5
2,1,2,3,2,5,4,3,4,4,9,6,3,7,5
Random A
Random B
Random C
2 5 9
1 4 6
2 3 3
3 4 7
2 4 5
2 4 6
2,1,2,3,2,5,4,3,4,4,9,6,3,7,5
A way to think about ANOVA
For any population of scores, regardless of form, the sampling distribution of the mean will approach a normal distribution a N (sample size) get larger. Furthermore, the sampling distribution of the mean will have a mean equal to and a standard deviation equal to / N
Central Limit Theorem
A way to think about ANOVA
For any population of scores, regardless of form, the sampling distribution of the mean will approach a normal distribution a N (sample size) get larger. Furthermore, the sampling distribution of the mean will have a mean equal to and a standard deviation equal to / N
Central Limit Theorem
A way to think about ANOVA
• Central Limit Theorem (remember)
• The variance of the means drawn from the same population equals the variance of the population divided by the sample size.
nS eX
22
A way to think about ANOVA
)( 22Xe Sn
nS eX
22
Can estimate population variance from the sample means with the formula
*This only works if the means are from the same population
A way to think about ANOVA
Random A
Random B
Random C
2 5 9
1 4 6
2 3 3
3 4 7
2 4 5
2 4 6 S2 = 4.00
A way to think about ANOVA
)( 22Xe Sn
)00.4(520
A way to think about ANOVA
40.000 2 20.000 10.000 .003
24.000 12 2.000
64.000 14
BetweenGroups
WithinGroups
Total
REP
Sum ofSquares df
MeanSquare F Sig.
ANOVA
)00.4(520 *Estimates population variance only if the three means are from the same population
A way to think about ANOVA
40.000 2 20.000 10.000 .003
24.000 12 2.000
64.000 14
BetweenGroups
WithinGroups
Total
REP
Sum ofSquares df
MeanSquare F Sig.
ANOVA
*Estimates population variance regardless if the three means are from the same population
What do all of these numbers mean?
40.000 2 20.000 10.000 .003
24.000 12 2.000
64.000 14
BetweenGroups
WithinGroups
Total
REP
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Why do we call it “sum of squares”?
• SStotal
• SSbetween
• SSwithin
• Sum of squares is the sum the squared deviations about the mean
2)( XX
Why do we use “sum of squares”?
1
)( 22
n
XXsx
2)( XX
SS are additive
Variances and MS are only additive if df are the same
Another way to think about ANOVA
• Think in “sums of squares”
2..)( XXSS ijtotal
Represents the SS of all observations, regardless of the treatment.
Another way to think about ANOVA
VW Beetle
Ford Mustang
Geo Metro
2 4.00 5 1.00 9 25.00
1 9.00 4 .00 6 4.00
2 4.00 3 1.00 3 1.00
3 1.00 4 .00 7 9.00
2 4.00 4 .00 5 1.00
Overall Mean= 4
64..)( 2 XX ij
Another way to think about ANOVA
2: ijtotal
total Sdf
SSNote
40.000 2 20.000 10.000 .003
24.000 12 2.000
64.000 14
BetweenGroups
WithinGroups
Total
REP
Sum ofSquares df
MeanSquare F Sig.
ANOVA
64..)( 2 XX ij
15 4.0000 4.571
15
VAR00001
Valid N(listwise)
N Mean Variance
Descriptive Statistics
Another way to think about ANOVA
• Think in “sums of squares”
2..)( XXnSS jbetween
Represents the SS deviations of the treatment means around the grand mean
Its multiplied by n to give an estimate of the population variance (Central limit theorem)
VW Beetle
Ford Mustang
Geo Metro
2 5 9
1 4 6
2 3 3
3 4 7
2 4 5
Overall Mean= 4
408)5(..)( 2 XXn j
2 4 6
Another way to think about ANOVA
40.000 2 20.000 10.000 .003
24.000 12 2.000
64.000 14
BetweenGroups
WithinGroups
Total
REP
Sum ofSquares df
MeanSquare F Sig.
ANOVA
408)5(..)( 2 XXn j
Another way to think about ANOVA
• Think in “sums of squares”
2)( jijwithin XXSS
Represents the SS deviations of the observations within each group
VW Beetle
Ford Mustang
Geo Metro
2 0 5 1 9 9
1 1 4 0 6 0
2 0 3 1 3 9
3 1 4 0 7 1
2 0 4 0 5 1
Overall Mean= 4
2 4 6
24)( 2 jijwithin XXSS
Another way to think about ANOVA
40.000 2 20.000 10.000 .003
24.000 12 2.000
64.000 14
BetweenGroups
WithinGroups
Total
REP
Sum ofSquares df
MeanSquare F Sig.
ANOVA
24)( 2 jijwithin XXSS
Sum of Squares
• SStotal
– The total deviation in the observed scores
• SSbetween
– The total deviation in the scores caused by the grouping variable and error
• SSwithin
– The total deviation in the scores not caused by the grouping variable (error)
Conceptual Understanding
Source SS df MS F
Between -- -- -- --
Within 152 -- - -
Total 182 --
Complete the above table for an ANOVA having 3 levels of the independent variable and n = 20. Test for significant at .05.
Conceptual UnderstandingSource SS df MS F
Between 30 2 15 5.62*
Within 152 57 2.67
Total 182 59
Fcrit = 3.18
Complete the above table for an ANOVA having 3 levels of the independent variable and n = 20. Test for significant at .05.
Fcrit (2, 57) = 3.15
Conceptual Understanding
• Distinguish between: Between-group variability and within-group variability
Conceptual Understanding
• Distinguish between: Between-group variability and within-group variability
• Between concerns the differences between the mean scores in various groups
• Within concerns the variability of scores within each group
Between and Within Group Variability
Between-group variability
Within-group variability
Between and Within Group Variability
sampling error + effect of variable
sampling error
Conceptual Understanding
• Under what circumstance will the F ratio, over the long run, approach 1.00? Under what circumstances will the F ratio be greater than 1.00?
Conceptual Understanding
• Under what circumstance will the F ratio, over the long run, approach 1.00? Under what circumstances will the F ratio be greater than 1.00?
• F ratio will approach 1.00 when the null hypothesis is true
• F ratio will be greater than 1.00 when the null hypothesis is not true
Conceptual Understanding
A B C
3 5 7
3 5 7
3 5 7
3 5 7
Without computing the SS within, what must its value be? Why?
Conceptual Understanding
A B C
3 5 7
3 5 7
3 5 7
3 5 7
The SS within is 0. All the scores within a group are the same (i.e., there is NO variability within groups)
Example
• Freshman, Sophomore, Junior, Senior
• Measure Happiness (1-100)
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
ANOVA
• Traditional F test just tells you not all the means are equal
• Does not tell you which means are different from other means
Why not
• Do t-tests for all pairs
• Fresh vs. Sophomore• Fresh vs. Junior• Fresh vs. Senior• Sophomore vs. Junior• Sophomore vs. Senior• Junior vs. Senior
Problem
• What if there were more than four groups?
• Probability of a Type 1 error increases.
• Maximum value = comparisons (.05)
• 6 (.05) = .30
Chapter 12
• A Priori and Post Hoc Comparisons
• Multiple t-tests• Linear Contrasts• Orthogonal Contrasts• Trend Analysis• Bonferroni t• Fisher Least Significance Difference• Studentized Range Statistic• Dunnett’s Test
Multiple t-tests
• Good if you have just a couple of planned comparisons
• Do a normal t-test, but use the other groups to help estimate your error term
• Helps increase you df
Remember
21
21
xxS
XXt
Note
nMS
XXt
within221
ProofCandy Gender
5.00 1.004.00 1.007.00 1.006.00 1.004.00 1.005.00 1.001.00 2.002.00 2.003.00 2.004.00 2.003.00 2.002.00 2.00
6 5.1667 1.1690 .4773
6 2.5000 1.0488 .4282
GENDER1.00
2.00
CANDYN Mean
Std.Deviation
Std. ErrorMean
Group Statistics
.027 .873 4.159 10 .002 2.6667 .6412 1.2380 4.0953
4.159 9.884 .002 2.6667 .6412 1.2358 4.0976
Equalvariancesassumed
Equalvariancesnotassumed
CANDYF Sig.
Levene's Test forEquality of Variances
t dfSig.
(2-tailed)Mean
DifferenceStd. ErrorDifference Lower Upper
95% ConfidenceInterval of the Mean
t-test for Equality of Means
Independent Samples Test
21.333 1 21.333 17.297 .002
12.333 10 1.233
33.667 11
BetweenGroups
WithinGroups
Total
CANDY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
.027 .873 4.159 10 .002 2.6667 .6412 1.2380 4.0953
4.159 9.884 .002 2.6667 .6412 1.2358 4.0976
Equalvariancesassumed
Equalvariancesnotassumed
CANDYF Sig.
Levene's Test forEquality of Variances
t dfSig.
(2-tailed)Mean
DifferenceStd. ErrorDifference Lower Upper
95% ConfidenceInterval of the Mean
t-test for Equality of Means
Independent Samples Test
21.333 1 21.333 17.297 .002
12.333 10 1.233
33.667 11
BetweenGroups
WithinGroups
Total
CANDY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
t = 2.667 / .641 = 4.16
.027 .873 4.159 10 .002 2.6667 .6412 1.2380 4.0953
4.159 9.884 .002 2.6667 .6412 1.2358 4.0976
Equalvariancesassumed
Equalvariancesnotassumed
CANDYF Sig.
Levene's Test forEquality of Variances
t dfSig.
(2-tailed)Mean
DifferenceStd. ErrorDifference Lower Upper
95% ConfidenceInterval of the Mean
t-test for Equality of Means
Independent Samples Test
21.333 1 21.333 17.297 .002
12.333 10 1.233
33.667 11
BetweenGroups
WithinGroups
Total
CANDY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
t = 2.667 / .641 = 4.16
nMS
XXt
within221
.027 .873 4.159 10 .002 2.6667 .6412 1.2380 4.0953
4.159 9.884 .002 2.6667 .6412 1.2358 4.0976
Equalvariancesassumed
Equalvariancesnotassumed
CANDYF Sig.
Levene's Test forEquality of Variances
t dfSig.
(2-tailed)Mean
DifferenceStd. ErrorDifference Lower Upper
95% ConfidenceInterval of the Mean
t-test for Equality of Means
Independent Samples Test
21.333 1 21.333 17.297 .002
12.333 10 1.233
33.667 11
BetweenGroups
WithinGroups
Total
CANDY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
t = 2.667 / .641 = 4.16
6)233.1(2
5.217.5 t
.027 .873 4.159 10 .002 2.6667 .6412 1.2380 4.0953
4.159 9.884 .002 2.6667 .6412 1.2358 4.0976
Equalvariancesassumed
Equalvariancesnotassumed
CANDYF Sig.
Levene's Test forEquality of Variances
t dfSig.
(2-tailed)Mean
DifferenceStd. ErrorDifference Lower Upper
95% ConfidenceInterval of the Mean
t-test for Equality of Means
Independent Samples Test
21.333 1 21.333 17.297 .002
12.333 10 1.233
33.667 11
BetweenGroups
WithinGroups
Total
CANDY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
t = 2.667 / .641 = 4.16
16.4641.
67.2t
.027 .873 4.159 10 .002 2.6667 .6412 1.2380 4.0953
4.159 9.884 .002 2.6667 .6412 1.2358 4.0976
Equalvariancesassumed
Equalvariancesnotassumed
CANDYF Sig.
Levene's Test forEquality of Variances
t dfSig.
(2-tailed)Mean
DifferenceStd. ErrorDifference Lower Upper
95% ConfidenceInterval of the Mean
t-test for Equality of Means
Independent Samples Test
21.333 1 21.333 17.297 .002
12.333 10 1.233
33.667 11
BetweenGroups
WithinGroups
Total
CANDY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Also, when F has 1 df between Ft 2tF
Within Variability
• Within variability of all the groups represents “error”
• You can therefore get a better estimate of error by using all of the groups in your ANOVA when computing a t-value
nMS
XXt
within221
Note: This formula is for equal n
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Hyp 1: Juniors and Seniors will have different levels of happiness
Hyp 2: Seniors and Freshman will have different levels of happiness
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Hyp 1: Juniors and Seniors will have different levels of happiness
nMS
XXt
within221
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Hyp 1: Juniors and Seniors will have different levels of happiness
6)90.100(2
8562 t
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Hyp 1: Juniors and Seniors will have different levels of happiness
80.5
2397.3
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Hyp 1: Juniors and Seniors will have different levels of happiness
80.5
2397.3
t crit (20 df) = 2.086
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Hyp 1: Juniors and Seniors will have different levels of happiness
80.5
2397.3
t crit (20 df) = 2.086
Juniors and seniors do have significantly different levels of happiness
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Hyp 2: Seniors and Freshman will have different levels of happiness
nMS
XXt
within221
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Hyp 2: Seniors and Freshman will have different levels of happiness
6)90.100(2
8576 t
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Hyp 2: Seniors and Freshman will have different levels of happiness
80.5
955.1
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Hyp 2: Seniors and Freshman will have different levels of happiness
t crit (20 df) = 2.086
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
80.5
955.1
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Hyp 2: Seniors and Freshman will have different levels of happiness
t crit (20 df) = 2.086
Freshman and seniors do not have significantly different levels of happiness
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
80.5
955.1
HAPPY76.0000
6
9.8793
72.0000
6
13.3417
62.0000
6
8.2219
85.0000
6
7.7717
73.7500
24
12.6052
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Mean
N
Std.Deviation
Fresh
Soph
Jun
Sen
Total
Report
1636.500 3 545.500 5.406 .007
2018.000 20 100.900
3654.500 23
BetweenGroups
WithinGroups
Total
HAPPY
Sum ofSquares df
MeanSquare F Sig.
ANOVA
Hyp 1: Juniors and Sophomores will have different levels of happiness
Hyp 2: Seniors and Sophomores will have different levels of happiness
PRACTICE!
Practice
• 11.1
• Figure out if 5 days is different than 35 days.
Practice
Source SS df MS F
Between 2100 2 1050 40.13*
Within 392.5 15 26.17
Total 2492.5 17
* p < .05