To test for any significant difference in the number of hours between breakdowns for four machines,
the following data were obtained:
At the 0.05 level of significance, is there any difference in the population mean among the four machines?
Machine No. of hours between breakdowns
M1 6.4 7.8
M2 8.7 7.4
M3 11.1 10.3
M4 9.9 12.8
m 4
r 5
n 20
source of variability sum of squares degrees of freedom
variability between groups 0.786666666666665 5
variability within groups 28.7933333333333 12
total variability 29.58 17
r machines,
e four machines?
y. (y.-y..)^2 (yij-yi.)^2
5.3 7.4 8.4 7.3 7.1 5.175625
9.4 10.1 9.2 9.8 9.1 0.075625
9.7 10.3 9.2 8.8 9.9 0.275625
12.1 10.8 11.3 11.5 11.4 4.100625
y.. 9.375
S1 29.58
variance test statistics F critical
0.157333 0.06557073397 3.105875
2.399444
(yij-y..)^2
Sr 28.79
Method using totals Method using means
m
S1 r yi y
2 2
1 m 2 Y
S1 Yi
r i 1 n i 1
m r
Sr yij yi
m r m 2
1
Sr y Yi2 2
ij
i 1 j 1 r i 1 i 1 j 1
m r 2
Sc yij y
m r 2
Y
Sc yij2
i 1 j 1 n i 1 j 1
Sc 0.78
y y
2
i
i 1
m r
y yi
2
ij
1 j 1
m r 2
y
1 j 1
ij y
example2
Compare rent for bussines areas in different cities A,B,C,D,E
Use alpha=0,05 to identify if there is a significant difference.
City rent €/m^2
A 30 40 28 44 46 38
B 21 21 26 33 23 18
C 14 13 14 19 26 21
D 14 10 11 9 17 13
E 11 14 9 7 19 12
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
29.874526986656 4 59.417115 14.854279 16.720365
39.8327026488747 4 58.919206 14.729802 21.547785
28.2148310429529 4 59.417115 14.854279 56.570109
43.8159729137622 4 68.047534 17.011883 141.57678
46.4714864236872 4 84.478524 21.119631 16.846617
38.1730067051716 4 63.566355 15.891589 17.599539
32.3640709022107 4 69.043351 17.260838 168.06676
0 4 37.841068 9.4602669 143.12853
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 299.40248 7 42.771783 0.5878712 0.7591494 2.4226285
Within Groups 1746.1694 24 72.75706
Total 2045.5719 31
Page 5
example2
32 0
28 25
27 13
15 0
0 0
Page 6
example2
Page 7
Test at the 0.05 level of significance to see if there are significant differences in the mean corn yields per acre
due to different types of fertilizers and irrigation. Use the following data:
Type of irrigation
fertilizer I. II. III.
A 4.3 4.2 5.1
B 3.7 5.4 7.5
C 4.8 6.3 8.1
D 5.7 6.2 5.8
E 4.1 5.8 7.2
Anova: Two-Factor Without Replication
SUMMARY Count Sum Average Variance
4.3 2 9.3 4.65 0.405
3.7 2 12.9 6.45 2.205
4.8 2 14.4 7.2 1.62
5.7 2 12 6 0.08
4.1 2 13 6.5 0.98
II. 5 27.9 5.58 0.722
III. 5 33.7 6.74 1.553
ANOVA
Source of VariationSS df MS F P-value F crit
Rows 7.174 4 1.7935 3.724818 0.115423 6.388233
Columns 3.364 1 3.364 6.986501 0.057388 7.708647
Error 1.926 4 0.4815
Total 12.464 9
orn yields per acre
A market study on the beer preference was conducted:
Beer preference
Gender
light regular dark
male 20 40 20
female 30 30 10
Use a 0.05 level of significance and test for independence between gender and
beer preference. What is your conclusion?
Observed (Male Light) Observed (Male Regular) Observed (Male Dark) Observed (Female Light)
20 40 20 30
observed value
Beer preference
Gender
light regular dark
male 20 40 20
female 30 30 10
expected value
Beer preference
Gender
light regular dark
male 26.66667 37.33333 16
female 23.33333 32.6667 14
If there is a relationship between gender and preference we may be interested in stregth of this relationship
Cramer's V Pearson's C
m
k
Cramer's V weak relationship
Pearson's C
Chuprov's T2
Observed (Female Regular) Observed (Female Dark) Expected (Male Light)
30 10 26.6666666666667
RESULTS
Chi-Square Statistic Degrees of Freedom
6.12244897959184 2
s relationship
Chuprov's T^2
0-0,3->>weak relationship
0,3-0,5->>medium relationship
0,5-0,7(0,8)->>strong relationship
more than 0,8(0,9)->>very strong relationship
Expected (Male Regular) Expected (Male Dark) Expected (FemaleExpected
Light) (Female
Expected
Regular)
(Female Dark)
37.3333333333333 16 23.3333333333333 32.66667 14
Critical Value (0.05) P-Value
5.99146454710798 0.0468303168524196
cted (Female Dark)
