Clearly state the null and alternative hypothesis and your assumptions. The perc
ID: 3233187 • Letter: C
Question
Clearly state the null and alternative hypothesis and your assumptions.
The percent reduction in blood sugar a certain time after injection of insulin into rabbits is given below. A group of 24 rabbits was divided at random into six groups of four rabbits each, and each rabbit received an injection of insulin. Two factors were involved, the dose at three levels and the preparation of insulin in two ways, A and B. Analyze this using a two-way analysis of variance for a factorial design (repeated measurements). (10 points).
Dose Level
Preparation I ii iii
A 17 64 62
21 49 72
49 34 61
54 63 91
B 33 41 56
37 64 62
40 34 57
16 64 72
Explanation / Answer
Two way anova with m observations per cell:
H01 : There is no significant difference in blood sugar after injecting insulin due to doses.
H02 : There is no significant difference in blood sugar after injecting insulin due to preparation.
H03: There is no significant difference in blood sugar after injecting insulin due to interaction affect between doses and preparation.
Alternate hypotheses:
H11 : There is a significant difference in blood sugar after injecting insulin due to doses.
H12 : There is a significant difference in blood sugar after injecting insulin due to preparation.
H13: There is a significant difference in blood sugar after injecting insulin due to interaction affect between doses and preparation.
1
2
3
A
17
64
62
21
49
72
49
34
61
54
63
91
B
34
41
56
37
64
62
40
34
57
16
64
72
1
2
3
A
141
210
286
B
127
203
247
Analysis:
Grand Total = 1214
CF = G2/N = 12142/24 =61408.17
Total sum of squares = Raw S.S – CF = 172 + 642 + … + 342+412 +…+642+722 - 61408.17
= 8113.833
M = 4; p=3 ; q=2
1
2
3
T.j.
T.j.2
A
141
210
286
637
405769
B
127
203
247
577
332929
Ti..
268
413
533
1214
738698
Ti..2
71824
170569
284089
526482
Sum of squares due to treatments (dose) = Ti..2/mq – CF
= 526482/4*2 - 61408.17 = 4402.083
Sum of squares due to variety (preparation) = T.j.2/mp – CF
=738698/4*3 - 61408.17 = 150
Sum of squares due to interaction = ( Tij.2/m – CF)- SST-SSV
Tij.2 = (141)2+(210)2+(286)2+(127)2+(203)2+(247)2 = 264124
S.S due to interaction = (264124/4)- 61408.17-4402.083-150 = 70.75
SS due to error = 8113.833-4402.083-150-70.75 = 3491
Anova table
Source of variation
Degrees of freedom
Sum of squares
M.S.S
Variance ratio
F at 5% level
Dose
p-1 =2
4402.083
=4402.083/2
=2201.0415
F1 = 2201.0415/193.944
=11.3489
F(2,18) =3.5546
Preparation
q-1=1
150
=150/1
=150
F2=150/193.944
=0.7734
F(1,18)=4.4139
Interaction
(p-1)(q-1)=2
70.75
=70.75/2
=35.375
F3 = 35.375/193.944
=0.1824
F(2,18)=3.5546
Error
18
3491
=3491/18
=193.944
Total
mpq-1=23
Inference:
1)Since Fcal =11.3489 > Ftab = 3.5546, we reject the null hypothesis and conclude that there is significant difference in blood sugar after injecting insulin due to doses.
2) Since Fcal =0.7734< Ftab = 4.4139, we accept the null hypothesis and conclude that there is no significant difference in blood sugar after injecting insulin due to preparations.
3) Since Fcal =0.1824 < Ftab = 3.5546, we accept the null hypothesis and conclude that there is no significant difference in blood sugar after injecting insulin due to interaction effect between dose and preparation.
1
2
3
A
17
64
62
21
49
72
49
34
61
54
63
91
B
34
41
56
37
64
62
40
34
57
16
64
72
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