1. A consumer agency tested weight loss from 3 drugs over 3 months. They recorded weight loss from 15 people for each drug. They want to determine if the mean weight loss is the same for each drug using a one-way ANOVA test at a 10% significance level. 2. They analyzed IQ test results from students in different departments to see if there are differences in IQ levels between departments. 3. They surveyed time spent online each day by people in 3 cities to determine if there are differences between the cities at a 10% significance level using a one-way ANOVA test.

2. ONE WAY ANOVA:
1.A consumer agency wanted to find out if the mean weight
loss for each of the 3 types of drugs for loosing weight is
same. The following table records the weight loss per Kg
by 15 people after eating these drugs for 3 months.
• Using 10% significant level, will you conclude that mean
weight loss is the same for each of the 3 drugs.
DRUG 1 DRUG 2 DRUG 3
7.5 9.5 8.5
10 9 10
8 7.5 6.5
6 10 11
6.5 6 8

3. DRUG DRUG 1
(x1)
x1^2 DRUG 2
(x2)
x2^2 DRUG 3
(x3)
x3^2
7.5 56.25 9.5 90.25 8.5 72.25
10 100 9 81 10 100
8 64 7.5 56.25 6.5 42.25
6 36 10 100 11 121
6.5 42.25 6 36 8 64
TOTAL 38 298.5 42 363.5 44 399.5

4. H0=the weight loss in different drugs are same.
H1=the weight loss in different drugs are different.
T=Σx1+Σx2+ Σx3
T=38+42+44=124
CF=T^2/n
CF=124^2/15=1025
SST= Σx1^2+Σx2^2+ Σx3^2 – CF
SST=298.5+363.5+399.5-1025=36.5
SSTR= Σxj^2/nj-CF
SSTR=288.5+352.8+387.2-1025=3.5

5. SSE=SST-SSTR
SSE=36.5-3.5=33
MSTR=SSTR/r-1
MSTR=3.5/3-1=1.75
MSE=SSE/n-r
MSE=33/15-3=2.75
F=MSE/MSTR
F=2.75/1.75=1.57

6. • F(.10,12,2)=9.39
• F(1-α,12,2)=0.106
• (0.106,9.39)
As the value of test statistic F=1.57 is smaller than the
critical value F=9.39, it falls in the acceptation region .
Consequently, we accept the null hypothesis and
conclude that the weight loss in different drugs are
same.

7. ONE WAY ANOVA:
2.A research is concerned about the IQ level of the student
of different department in a school. Data of a IQ test
result in term of percentage is recorded. Is there any
significant difference between the different department of
studies?
SCIENCE ARTS COMMERCE
44 58 65
56 25 45
78 65 58
85 73 38

8. LET THE ARBITRARY VALUE BE 55.
DEPT SCIENC
E(x1)
x1^2 ARTS
(x2)
x2^2 COMME
RCE(x3)
x3^2
SEC 1 -11 121 3 9 10 100
SEC 2 1 1 -30 900 -10 100
SEC 3 23 529 10 100 3 9
SEC 4 30 900 18 324 -17 289
TOTAL 43 1551 1 1333 -14 498

9. H0=there is no difference in IQ level in different department.
H1=there is difference in IQ level in different department.
T=Σx1+Σx2+ Σx3
T=43+1-14=30
CF=T^2/n
CF=30^2/12=75
SST= Σx1^2+Σx2^2+ Σx3^2 – CF
SST=1551+1333+498-75=3307
SSTR= Σxj^2/nj-CF
SSTR=462.25+.25+49-75 = 436.5

10. SSE=SST-SSTR
SSE=3307-436.5= 287.5
MSTR=SSTR/r-1
MSTR=436.5/3-1=218.25
MSE=SSE/n-r
MSE=287.5/12-3 = 31.9
F=MSTR/MSE
F=218.25/31.9=6.833

11. • F(.05,2,9)=4.26
• F(1-α,2,9)=.2347
• (0.2347,4026)
Because the value of test statistic F=6.833 is greater than the
critical value F=4.26, it falls in the rejection region .
Consequently, we reject the null hypothesis and conclude that
the IQ level in 3 departments are different.

12. ONE WAY ANOVA
3. A survey has conducted to check how much time person
sit on the internet per day. Data is collected from a
sample of 18 people were chosen randomly from 3
cities. Is there any significant difference at 10%
significant level
DELHI MUMBAI CHENNAI
10 12 9
6 9 4
5 10 5
8 5 8
6 2 6
1 2
7

13. DELHI(x
1)
x1^2 MUMBA
I(x2)
x2^2 CHENN
AI(x3)
x3^2
10 100 12 144 9 81
6 36 9 81 4 16
5 25 10 100 5 25
8 64 5 25 8 64
6 36 2 4 6 36
1 1 2 4
7 49
TOTAL 35 261 39 355 41 275

14. H0=time persons spend in each city is same
H1=time persons spend in each city is different.
T=Σx1+Σx2+ Σx3
T=35+39+41=115
CF=T^2/n
CF=115^2/18= 734.72
SST= Σx1^2+Σx2^2+ Σx3^2 – CF
SST=261+355+275-734.72=156.3
SSTR= Σxj^2/nj-CF
SSTR=245+253.5+240-734.72=3.92

15. SSE=SST-SSTR
SSE=156.3+3.92=160.22
MSTR=SSTR/r-1
MSTR=3.92/3-1=1.96
MSE=SSE/n-r
MSE=160.22/18-3=10.68
F=MSE/MSTR
F=10.68/1.96=5.45

16. • F(.10,15,2)=9.42
• F(1-α,15,2)=0.106
• (0.106,9.42)
As the value of test statistic F=5.45 is smaller than the
critical value F=9.42, it falls in the acceptation region .
Consequently, we accept the null hypothesis and
conclude that the time persons spend in each city is
same

17. TWO WAY ANOVA
4.For arranging a Christmas party Raj, ram, ravi were
collecting funds on Monday, Tuesday and Wednesday
are tabulated below (in thousand). Is there any
significant different in their collection and in different
days at 5% significant level?
RAJ RAM RAVI
MONDAY 10 14 8
TUESDAY 12 9 15
WEDNESDAY 11 12 16

18. RAJ
(x1)
x1^2 RAM
(x2)
x2^2 RAVI
(x3)
x3^2 TOTAL
MONDAY 10 100 14 196 8 64 32
TUESDAY 12 144 9 81 15 225 36
WEDNES
DAY
11 121 12 144 16 256 39
TOTAL 33 365 35 421 39 545 107

19. ROW WISE:
H0=The funds of Raj, Ram, Ravi are same.
H1=The funds of Raj, Ram, Ravi are different
COLUMN WISE:
H0=The funds collected in all the 3 days are same.
H1=The funds collected in all the 3 days are different
T=Σx1+Σx2+ Σx3
T=33+35+39=107
CF=T^2/n
CF=107^2/9=1272.1
SST= Σx1^2+Σx2^2+ Σx3^2 – CF
SST=365+421+545-1272=59
SSTR= Σxj^2/nj-CF
SSTR=363+408.33+507-1272.11=6.22

20. SSR= Σrow^2/n-CF
SSR=32^2/3+36^2/3+39^2/3-1272 = 8.333
SSE=SST-(SSTR+SSR)
SSE=59-(6.22+8.33) = 44.44
MSTR=SSTR/c-1
MSTR=6.22/3-1 = 3.11
MSR=SSR/r-1
MSR=8.33/3-1 = 4.165
MSE=SSE/(r-1)(c-1)
MSE=44.44/(3-1)(3-1) = 11.11

21. ROW
F=MSE/MSTR
F=11.11/3.1 = 3.57
F(0.05,4,2) = 19.25
F(1-α,4,2) = 0.051
COLUMN
F=MSE/MSR
F=11.11/4.16 = 2.67
F(0.05,4,2)=3.57
F(1-α,4,2) = 0.051
As the value of test statistic F=3.57 and F=2.67is smaller than the critical
value F=19.25 it falls in the acceptation region . Consequently, we
accept the null hypothesis and conclude that the funds of Raj, Ram,
Ravi collected in all 3 days are same.

22. Two way ANOVA
5.The manufacturer of the batteries cell wants to check
whether the 3 batteries produced in his factory has the
same life span. He used the cell in remote, clock, toys
and radio. Is their any significant difference in batteries
and also the effect of their uses in different items at 5%
significant level.
REMOTE CLOCK TOY RADIO
BATTERY1 60 40 35 25
BATTERY2 69 45 39 30
BATTERY3 57 35 30 21

23. Let the arbitrary value be 45
REM
OTE
(x1)
x1^2 CLOC
K
(x2)
x2^2 TOY
(x3)
x3^2 RADIO
(x4)
x4^2 TOTAL
BATT
ERY-
1
15 225 -5 25 -10 100 20 400 20
BATT
ERY-
2
24 576 0 0 -6 36 -15 225 3
BATT
ERY-
3
12 144 -10 100 -15 225 -24 576 -22
TOTAL 51 945 -15 125 -31 361 -19 1201 1

24. ROW WISE:
H0=The performance of batteries in different items are same.
H1=The performance of batteries in different items are different
COLUMN WISE:
H0=The life span of each batteries are same.
H1=The life span of each batteries are different
T=Σx1+Σx2+ Σx3
T=51-15-31-19=-14
CF=T^2/n
CF=(-14)^2/12=16.33
SST= Σx1^2+Σx2^2+ Σx3^2 – CF
SST=945+125+361+1201-16.33=2615.7
SSTR= Σxj^2/nj-CF
SSTR=867+75+320.33+120.33-16.33=1366.33

25. SSR= Σrow^2/n-CF
SSR=20^2/4+3^2/4+-22^2/4-16.33 = 206.9
SSE=SST-(SSTR+SSR)
SSE=2615.7-(1366.33+206.9) = 1042.47
MSTR=SSTR/c-1
MSTR=1366.33/4-1= 455.44
MSR=SSR/r-1
MSR=206.9/3-1 = 103.45
MSE=SSE/(r-1)(c-1)
MSE=1042.47/(3-1)(4-1) = 173.74

26. ROW
F=MSTR/MSE
F=455.44/103.45 = 4.4
F(0.05,6,3) = 8.94
F(1-α,6,3) = 0.111
COLUMN
F=MSE/MSR
F=173.74/03.45 = 1.68
F(0.05,2,6)=3.57
F(1-α,4,2) = 0.28
As both the value of test statistic F=4.4 and F=1.68 is smaller than the critical
value it falls in the acceptation region . Consequently, we accept the null
hypothesis and conclude that the performance of batteries in different items
are same.