anova by hazilah mohd amin
DESCRIPTION
Nota kuliah Pensampelan dan analisis dataTRANSCRIPT
![Page 1: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/1.jpg)
Analysis of Variance (ANOVA)
Hazilah Mohd Amin
![Page 2: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/2.jpg)
Goals
After completing, you should be able to:
• Recognize situations in which to use analysis of variance (ANOVA)
• Perform a single-factor hypothesis test for Comparing More Than Two Means and interpret results
![Page 3: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/3.jpg)
The F - Distribution
• Analysis-of-variance procedures rely on F-distribution.
• There are infinitely many F-distributions, and we identify an F-distribution (and F-curve) by its number of degrees of freedom. • F-distribution has two numbers of degrees of freedom.
![Page 4: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/4.jpg)
Key Fact F distribuition curve:
![Page 5: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/5.jpg)
Find Critical Value: Example
• Find the F value for 8 df for numerator, 14 df for denominator, and 0.05 area in the right tail of the F distribuition curve.
Critical value: F, df numerator,df denominator = F, 8,14 = ?
![Page 6: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/6.jpg)
Table 12.1 (p. 534) Critical value: F, 8,14 = 2.70
![Page 7: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/7.jpg)
Hypotheses of One-Way ANOVA
• – All population means are equal – i.e., no treatment effect (no variation in means among groups)
• – At least one population mean is different – i.e., there is a treatment effect – Does not mean that all population means are different (some pairs
may be the same)
k3210 μμμμ:H
same the are means population the of all Not:HA
The analysis of variance is a procedure that tests to determine whether differences exits between two or more population means.
![Page 8: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/8.jpg)
One-Factor ANOVA
All Means are the same:The Null Hypothesis is True
(No Treatment Effect)
k3210 μμμμ:H
same the are μ all Not:H iA
321 μμμ
![Page 9: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/9.jpg)
One-Factor ANOVA
At least one mean is different:The Null Hypothesis is NOT true
(Treatment Effect is present)
k3210 μμμμ:H
same the are μ all Not:H iA
321 μμμ 321 μμμ
or
![Page 10: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/10.jpg)
One-Way Analysis of Variance
![Page 11: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/11.jpg)
![Page 12: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/12.jpg)
![Page 13: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/13.jpg)
One-Factor ANOVA F Test: Example 1
You want to see if three different golf clubs yield different distances. You randomly select five measurements from trials on an automated driving machine for each club. At the .05 significance level, is there a difference in mean distance?
Club 1 Club 2 Club 3254 234 200263 218 222241 235 197237 227 206251 216 204
![Page 14: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/14.jpg)
Solution of Example 1– The data are interval– The problem objective is to compare mean
distances in three type of golf club.– We hypothesize that the three population
means are equal
One Way Analysis of Variance
![Page 15: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/15.jpg)
H0: 1 = 2= 3
H1: At least two means differ
• Solution
Defining the Hypotheses
![Page 16: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/16.jpg)
Independent samples are drawn from k populations (treatments).
X11
x21
.
.
.Xn1,1
1
1x
n
X12
x22
.
.
.Xn2,2
2
2x
n
X1k
x2k
.
.
.Xnk,k
k
kx
n
Sample sizeSample mean
X is the “response variable”.The variables’ value are called “responses”.
Notation
![Page 17: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/17.jpg)
Terminology• In the context of this problem…
Response variable – distance
Experimental unit – golf club when we record distance figures.
Factor or treatment – the criterion by which we classify the populations (the treatments). In this problems the factor is the type of golf clubs.
![Page 18: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/18.jpg)
The rationale of the name of Analysis of Variance (ANOVA)
• We are testing the different between means but why ANOVA?
• Two types of variability are employed when testing for the equality of the population means: Within Samples and Between Samples
![Page 19: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/19.jpg)
Graphical demonstration:Employing two types of variability: Within Samples and Between Samples
One Way Analysis of Variance
![Page 20: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/20.jpg)
20
25
30
1
7
Treatment 1 Treatment 2 Treatment 3
10
12
19
9
Treatment 1Treatment 2Treatment 3
20
161514
1110
9
10x1
15x2
20x3
10x1
15x2
20x3
The sample means are the same as before,but the larger within-sample variability makes it harder to draw a conclusionabout the population means.
A small variability withinthe samples makes it easierto draw a conclusion about the population means.
![Page 21: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/21.jpg)
••••
•
One-Factor ANOVA Example: Scatter Diagram
270
260
250
240
230
220
210
200
190
•••••
•••••
Distance
227.0 x
205.8 x 226.0x 249.2x 321
Club 1 Club 2 Club 3254 234 200263 218 222241 235 197237 227 206251 216 204
Club1 2 3
3x
1x
2x x
From scatter diagram, we can clearly see sample means
difference because of small within-sample variability
![Page 22: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/22.jpg)
Test Statistics (F), Critical Value & Rejection Criterion
• Test statistic:
where MSB is mean squares between varianceswhere MSW is mean squares within variances
• Rejection Region: F > F, k-1,n-k
• Degrees of freedom– df1 = k – 1 (k = levels or treatments)
– df2 = n – k (n = sum of sample sizes from all populations)
MSW
MSBF
H0: μ1= μ2 = … = μ k
HA: At least two population means are different
The hypothesis test:
![Page 23: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/23.jpg)
One-Factor ANOVA Example Computations
Club 1 Club 2 Club 3254 234 200263 218 222241 235 197237 227 206251 216 204
x1 = 249.2
x2 = 226.0
x3 = 205.8
x = 227.0
n1 = 5
n2 = 5
n3 = 5
n = 15
k = 3
MSB = 4716.4 / (3-1) = 2358.2
MSW = 1119.6 / (15-3) = 93.325.275
93.3
2358.2F
kn
SSWMSW
T
1
k
SSBMSB
MSW
MSBF
SSB = 4716.4
SSW = 1119.6
n
x
n
T
n
T
n
T2
3
23
2
22
1
21SSB
3
23
2
22
1
212 )(SSW
n
T
n
T
n
Tx
![Page 24: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/24.jpg)
F = 25.275
One-Factor ANOVA Example Solution
H0: μ1 = μ2 = μ3
HA: μi not all equal = .05df1= k-1 =3-1 =2 df2 = n-k =15-3 =12
Test Statistic:
Decision: Test statistic F is greater than critical value
Conclusion:Reject H0 at = 0.05
There is evidence that at least one μi differs from the rest
0
= .05
F.05 = 3.885Reject H0Do not
reject H0
25.27593.3
2358.2
MSW
MSBF
Critical Value: F, k-1,n-k = F, 2,12 = 3.885
![Page 25: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/25.jpg)
SUMMARY
Groups Count Sum Average Variance
Club 1 5 1246 249.2 108.2
Club 2 5 1130 226 77.5
Club 3 5 1029 205.8 94.2
ANOVA
Source of Variation
SS df MS F P-value F crit
Between Groups
4716.4 2 2358.2 25.275 4.99E-05 3.885
Within Groups
1119.6 12 93.3
Total 5836.0 14
ANOVA Single Factor: Excel OutputEXCEL: tools | data analysis | ANOVA: single factor
25.27593.3
2358.2
MSW
MSBF F, k-1,n-k = F, 2,12 = 3.885
![Page 26: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/26.jpg)
Rationale 1: Variability Between Sample
• If H0: μ1= μ2 = … = μk is true, we would expect all the sample means to be close to one another.
• If the alternative hypothesis is true, at least some of the sample means would differ.
• Thus, we measure variability between sample means (and hence MSB or MSTr).
![Page 27: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/27.jpg)
• Large variability within the samples weakens the “ability” of the sample means to represent their corresponding population means.
• Therefore, even though sample means may markedly differ from one another, we have to consider the “within samples variability” (and hence MSW or MSE).
Rationale II: Variability Within
![Page 28: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/28.jpg)
Interpreting One-Factor ANOVA F Statistic
• The F statistic is the ratio of the between estimate of variance and the within estimate of variance– The ratio must always be positive– df1 = k -1 will typically be small– df2 = n - k will typically be large
The test statistic F ratio should be close to 1 (SSB small due to small sample means difference) if
H0: μ1= μ2 = … = μk is true
The ratio will be larger than 1 (SSB large due to large sample means difference) if
H0: μ1= μ2 = … = μk is false
MSW
MSBF
![Page 29: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/29.jpg)
Example 2
A study was conducted to determine if the drying time for a certain paint is affected by the type of applicator used. The data in the table on the next screen represents the drying time (in minutes) for 3 different applicators when the paint was applied to standard wallboard.
Is there any evidence to suggest the type of applicator has a significant effect on the paint drying time at the 0.05 level?
Notation: The type of applicator is the treatment, factor or level .
hence k = 3
![Page 30: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/30.jpg)
Notation Used in ANOVAFactor Levels
Sample from Sample from Sample from Sample from
Replication Level 1 Level 2 Level 3 Level k
n = 1 x 1,1 x 2,1 x 3,1 x k ,1
n = 2 x 1,2 x 2,2 x 3,2 x k ,2
n = 3 x 1,3 x 2,3 x 3,3 x k ,3
Column T 1 T 2 T 3 T k T
Totals T = grand total = sum of all x 's = x = T i
. . .
. . .
. . .
![Page 31: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/31.jpg)
Applicator (Level)Brush Roller Pad
(i = 1 ) (i = 2) (i = 3)39.1 31.6 32.739.4 33.4 33.231.1 30.2 28.733.7 41.8 29.230.5 33.9 25.834.6 31.4
26.729.5
Sum 208.4 170.9 237.2
Mean 34.73 34.18 29.65
1T 2T 3T
1x 2x 3x
Sample Results
![Page 32: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/32.jpg)
Solution• Assumptions:
– The data (samples) was randomly collected and all observations are independent.
– The populations are (approximately) normally distributed.
– Populations have equal variances.
• The null and the alternative hypothesis:
Ho: 1 = 2 = 3
The mean drying time is the same for each applicator
Ha: At least one mean is different Not all drying time means are equal
![Page 33: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/33.jpg)
Partition of Total Variation
Variation Due to Factor/Treatment (SSB)
Variation Due to Random Sampling (SSW)
Sum of Squares Total (SST)
Commonly referred to as: Sum of Squares Within (SSW) Sum of Squares Error (SSE) Sum of Squares Unexplained Within Groups Variation
Commonly referred to as: Sum of Squares Between (SSB) Sum of Squares Treatment (SSTr) Sum of Squares Factor Sum of Squares Among Sum of Squares Explained Among Groups Variation
= +
Total variation SST can be split into two parts: SST = SSB + SSW
![Page 34: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/34.jpg)
n
x
n
T
n
T
n
T2
3
23
2
22
1
21SSB
3
23
2
22
1
212 )(SSW
n
T
n
T
n
Tx
![Page 35: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/35.jpg)
x and x2 Calculator: Enter xi data, retrieve x and x2
• Enter Statictics SD: Mode Mode 1• Clear old data: Shift Clr 1 =
• Enter xi data: 39.1 DT 39.4 DT 31.1 DT 33.7 DT 30.5 DT 34.6 DT …29.5 DT
• Find (x): Shift S-SUM 2 = 616.5• Find (x2): Shift S-SUM 1 = 20,316.69
35
![Page 36: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/36.jpg)
89.31280.2000369.2031619
5.61669.20316)(SST
22
2
n
xx
97.1088.2000377.20112
19
)5.616(
8
2.237
5
9.170
6
4.208
SSB
2222
2
3
23
2
22
1
21
n
x
n
T
n
T
n
T
92.20397.10889.312
SSBSSTSSW
Variation Sums of Squares
93.203
20112.77-20316.69
)(SSW3
23
2
22
1
212
n
T
n
T
n
Tx
![Page 37: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/37.jpg)
Mean SquareThe mean square for the factor being tested and for the error is obtained by dividing the sum-of-square value by the corresponding number of degrees of freedom
75.1216
92.203df(error)SS(error)
MS(error)
49.542
97.108df(factor)SS(factor)
MS(factor)
Calculations:
Numerator degrees of freedom = df(factor) = k 1 = 3 1 = 2
df(total) = n 1 = 19 1 = 18
Denominator degrees of freedom = df(error) = n k = 19 3 = 16
![Page 38: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/38.jpg)
One-Way ANOVA Table
Source of Variation
dfSS MS
Between Samples
SSB MSB =
Within Samples n - kSSW MSW =
Total n - 1SST =SSB+SSW
k - 1 MSBMSW
F ratio
SSBk - 1SSWn - k
F =
The sums of squares and the degrees of freedom must check SS(factor) + SS(error) = SS(total) or SSB + SSW = SST df(factor) + df(error) = df(total) or df(between) + df(within) = df(total)
An ANOVA table is often used to record the sums of squares and to organize the rest of the calculations. Format for the ANOVA Table:
![Page 39: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/39.jpg)
Source df SS MS
Factor 2 108.97 54.59
Error 16 203.92 12.75
Total 18 312.89
The Completed ANOVA Table
The Complete ANOVA Table:
27.475.1249.54
MS(error)MS(factor)
* FThe Test Statistic:
![Page 40: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/40.jpg)
Solution Continued
The Results
a. Decision: Reject Ho at = 0.05b. Conclusion: There is evidence to suggest the three population means are not all the same. The type of applicator has a significant effect on the paint drying time at the 0.05 level of significance.
Critical Value: F, k-1,n-k = F, 2,16 = 3.63
The Test Statistic F = 4.27 is in the rejection region.
Reject H0
F.05 = 3.63
Do not reject H0
= .05
![Page 41: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/41.jpg)
One-Way ANOVA F-Test: Exercise 1•You’re a trainer for Microsoft Corp. Is there any evidence to suggest the type of training method has a significant effect on the learning time at the 0.05 level?•The data in the table represents the learning times (in hours) of 12 people using 4 different training methods.
M1 M2 M3 M410 11 13 18
9 16 8 235 9 9 25
© 1984-1994 T/Maker Co.
Answer: Critical Value = 4.07. Test statistic = 11.6
![Page 42: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/42.jpg)
Hey! Lets get our hand dirty… Using SPSS….
![Page 43: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/43.jpg)
One Way Analysis of Variance Using SPSS
• Suppose we want to know whether students who have to work many hours outside school to support themselves find their grade suffering.
• We examine this question by comparing the GPAs of students who work various hours outside school.
• Let’s examine this question using data in Student file. File>Open> Student
![Page 44: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/44.jpg)
One Way Analysis of Variance Using SPSS
• First examine the average GPA for each of the three work categories (0 hrs, 1-19hrs, >20hrs) - WorkCat
• Graph>Boxplot then choose Simple and click Define. Select GPA as the variable and WorkCat for the Category Axis. Click
• Option
![Page 45: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/45.jpg)
After Clicking Options…, click off Display groups defined by missing value, and click
Continue then OK.
• You’ll get this
![Page 46: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/46.jpg)
What is the Box-plot telling us?
• Some variation across the groups• See median GPAs (dark line in the middle
of box) differ slightly between groups. • So, should we attribute the observed
difference to sampling error or they genuinely differ?
• Neither box-plot nor the median offer decisive evidence. Hence we need ANOVA.
![Page 47: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/47.jpg)
One Way Analysis of Variance Using SPSS
• We are testing: H0 : 1 = 2 = 3
H1: At least two means differ
• Before attempting ANOVA, need to review the ANOVA assumptions. (i)Independent samples (ii) Normality (iii) Variances equality. We can test both (ii) & (iii).
• Analyze>Descriptive Statistics>Explore
![Page 48: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/48.jpg)
Analyze>Descriptive Statistics>Explore
• In the Explore dialog box, select GPAs as the dependent List variable, WorkCat as the Factor List variable and Plot as the Display. Next, click Plot…
• We are interested in a normality test, select
Select this & deselect this only. Click Continue
and OK. See next slide…
![Page 49: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/49.jpg)
The Output has several parts, let focus on the tests of normality
• The Kolmogorov-Smirnov test assesses whether there is significant departure from normality in the population distribution of the 3 groups. H0: Distributions are normal.
• Look at the p-values, all are > 0.05. Do not reject H0. Hence no evidence of non-normality.
![Page 50: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/50.jpg)
One Way Analysis of Variance Using SPSS
• We still need to validate the homogeneity of variance assumption. We do this within ANOVA.
• Analyze>Compare Means>One-Way ANOVA• Dependent List variable is GPA and Factor variable is WorkCat.Click Option,
![Page 51: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/51.jpg)
One Way Analysis of Variance Using SPSS
• under Statistics, select Descriptive and Homogeneity of variance test. Click Continue & OK
• H0: Variances are equal. One-Way ANOVA output consists many parts. Look at the p-value > 0.05.
• Hence do not reject H0.
![Page 52: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/52.jpg)
Normality & Homogeneity of variances assumptions met … hence
• Let find out whether students who work various hours outside school differ in their GPAs.
• The P-value of .000 is very small, hence we reject Ho and conclude that
the means GPAs are not all the same. Where are the differences? Hence Post-Hoc test…
![Page 53: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/53.jpg)
End of ANOVA
See U Later…
![Page 54: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/54.jpg)
One-Way ANOVA F-Test: Exercise 1 Solution
•You’re a trainer for Microsoft Corp. Is there any evidence to suggest the type of training method has a significant effect on the learning time at the 0.05 level?•The data in the table represents the learning times (in hours) of 12 people using 4 different training methods.
M1 M2 M3 M410 11 13 18
9 16 8 235 9 9 25
© 1984-1994 T/Maker Co.
![Page 55: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/55.jpg)
Summary Table Solution*
Source ofSource ofVariationVariation
DegreesDegrees ofofFreedomFreedom
Sum ofSum ofSquaresSquares
MeanMeanSquareSquare
(Variance)(Variance)
FF
TreatmentTreatment((MethodsMethods))
4 - 1 = 34 - 1 = 3 348348 116116 11.611.6
ErrorError 12 - 4 = 812 - 4 = 8 8080 1010
TotalTotal 12 - 1 = 1112 - 1 = 11 428428
![Page 56: Anova by Hazilah Mohd Amin](https://reader034.vdocuments.site/reader034/viewer/2022052301/55847ce1d8b42a15768b50ff/html5/thumbnails/56.jpg)
FF00 4.074.07
One-Way ANOVA F-Test Solution*
•H0: 1 = 2 = 3 = 4
•Ha: Not All Equal• = .05•1 = 3 2 = 8 •Critical Value(s):
Test Statistic: Test Statistic:
Decision:Decision:
Conclusion:Conclusion:Reject at Reject at = .05 = .05
There Is Evidence Pop. There Is Evidence Pop. Means Are DifferentMeans Are Different
= .05= .05
FFMSBMSB
MSEMSE
116116
10101111 66..