MODULE OUTLINE
UNIVERSITY OF ZULULAND
FACULTY OF SCIENCE AND AGRICULTURE
TEST 1
LECTURER:
MISS J. M. BATIDZIRAI
1. Define the following:
a. Analysis of variance- Is a statistical procedure use to test the null hypothesis that the means
of more than 2population are equal
b. A factor - -Is a controlled independent variable; a variable whose levels are set by the
experimenter.
c. Randomization-random allocation of experimental units to the treatments of the study. 
d. Experimentwise type I error rate, (  E ).When we use the Fisher’s LSD procedure to
sequentially test two sets of hypothesis, the type I error rate associated with this is called the
experiment wise type 1 error rate. 
e. Blocking -This is the procedure by which experimental units are grouped into homogeneous
clusters in an attempt to improve the comparison of treatments by randomly allocating the
treatments within each cluster or 'block'.
Hence blocks are the homogeneous groups into which the experimental units are grouped
2. State the 4 assumptions for Analysis of variance of a Completely Randomized design.
a. The population from which the samples are drawn are (approximately) normal
b. The population from which the samples are drawn have the same variance (or standard
deviation) i.e. σ1 = σ2 = ………….. = σk 
c. The samples drawn from different populations are independent. 
d. The experimental units are allocated into the treatments in a completely random
manner
3. Given the model for a Randomized Block Design as 𝑦𝑖𝑗 = 𝜇 + 𝛼𝑖 + 𝛽𝑗 + 𝜀𝑖𝑗 . State the
assumptions of the model.
The assumptions of the model are that yij , the response on treatment i in block j is
2
normally distributed with mean    i   j and variance  e .
yij ~ N (  i   j ; e2 )
i.e.
4. Three brands of batteries are under study. It is s suspected that the lives (in weeks) of the
three brands are different. Five randomly allocated batteries of each brand are tested with
the following results:
Weeks
of life
Brand Brand Brand
1
2
3
100
76
108
96
80
100
92
75
96
96
84
98
92
82
100
Ti
476
397
502
a. Perform a statistical test to check whether or not the lives of these brands of batteries
are different.
If necessary do the multiple comparison tests, using the Turkey’s procedure. (Use 𝛼
=5%)
STEP 1
𝐻0 : 𝜇1 = 𝜇2 = 𝜇3 (The lives of these brands of batteries are NOT different).
𝐻1 : 𝜇1 = 𝜇2 = 𝜇3 (The lives of these brands of batteries are different). 
STEP II
Since we are testing for means from more than 2 populations, we use an Fdistribution
Use 0.05
STEP III
2
 4 ni

  yij 
2
100  96  .....  98  100  13752

i 1 j 1


CM 


=126041.7
n
15
15
4
ni
TSS   yij2  CM 
i 1 j 1
= (Sum of all squares of all values of y) – CM
= (1002  962  ............ 100 2) 126041.7
= 127425 – 137601.3
=1383.33
Ti 2
 CM 
i 1 ni
4
SST  
 4762 397 2 502 2 
=


  CM
5
5 
 5
= 636189=126041.7
=1196.133
SSE = TSS - SST
=1383.33-1196.133
=187.2
ANOVA TABLE
source
Treat
Error
Total
DF
2
12
14
SS
1196.133
187.2
1383.33
MS
598.7
15.6
F
38.34
STEP IV
Fcritic  F(2.12) 0.05 =3.89
•
STEP V
Since Fcalc  Fcritic , we reject H 0 and conclude that the lives of these brands of batteries
are different. 
MULTIPLE COMPARISON:
𝐻0 : 𝜇1 = 𝜇2
𝐻1 : 𝜇1 = 𝜇2
𝐻0 : 𝜇1 = 𝜇3
𝐻1 : 𝜇1 = 𝜇3

𝐻0 : 𝜇2 = 𝜇3
𝐻1 : 𝜇2 = 𝜇3
𝑞(𝑘;𝑑𝑓) ∝ = 𝑞(3;12) ∝ =3.77
𝑴𝑺𝑬
TSD=(𝑞(𝑘;𝑑𝑓) ∝ )√
𝒏
𝟏𝟓.𝟔
= 𝟑. 𝟕𝟕√
𝟓
= 𝟑. 𝟕𝟕 ∗ 𝟏. 𝟕𝟔𝟔 = 𝟔. 𝟔𝟔
Samples
XmaxXmin
Comparison
with TSD
1&2
15.8
>TSD
1&3
5.2
<TSD
2&3
21
>TSD
Decision
Reject
H0
Do Not
Reject
H0
Reject
H0
1&2 and 2&3 of the brands are different, 
[20 Marks]
b. Construct a 95% interval estimate on the mean life of battery brand 2, using a Fisher’s
Least Significance Difference.
𝜶
𝑴𝑺𝑬
𝒏

CI= ̅̅̅
𝒚𝒊. ± 𝒕𝑵−𝒑 (𝟐 ) √
𝟏𝟓.𝟔
𝟓
=79.4 ± 2.179√
=79.4 ± 3.849
= (75.551 ; 83.249)
c.
Construct a 99% interval estimate on the mean difference between the lives of
battery brands 2 and 3, using a Fisher’s Least Significance Difference.
𝜶
𝑪𝑰 = ̅̅̅
𝒚𝒊. − ̅̅̅
𝒚𝒋. ± 𝒕𝑵−𝒑 (𝟐 ) √𝟐
𝑴𝑺𝑬
𝒏

=𝟕𝟗. 𝟒 − 𝟏𝟎𝟎. 𝟒 ± 𝟑. 𝟎𝟓𝟓√
=(-28.631;-13.369)
𝟐(𝟏𝟓.𝟔)
𝟓


d. Using results in Question 4c) above, test if brand 2 and brand 3 have different lives.
Since the confidence interval do not include a 0, we reject 𝑯𝟎 and conclude that they
are different
Where
𝐻0 : 𝜇2 = 𝜇3 (The lives of brands of batteries 2 and 3 are NOT different).
𝐻1 : 𝜇2 = 𝜇3 (The lives of these brands of batteries 2 and 3 are different).
5. An aluminum master alloy manufacturer produces grain refiners in ingot form. This company
produces the product in four furnaces. Each furnace is known to have its own unique operating
characteristics, so any experiment run in the foundry that involves more than one furnace will
consider furnace a nuisance variable. The process engineers suspect that stirring rate impacts the
grain size of the product. Each furnace can be run at four different stirring rates. A randomized
block design is run for a particular refiner and the resulting grain size data is shown below.
Stirring
Rate
a.
b.
Furnace
1
2
5
8
4
10
14
5
15
14
6
20
17
9
From the information above, state which one is the
i.
Experimental unit.- grain refiners
ii.
Response variable.- grain size
iii.
Treatment.- Stirring Rate
iv.
Blocks.- The 4 furnaces
v.
Levels of the Stirring rate-5, 10, 15 and 20
3
5
6
9
3
4
6
9
2
6
Is there any evidence that stirring rate impacts grain size?
𝐻0 : 𝑇1 = 𝑇2 = 𝑇3 = 𝑇4
𝐻1 : 𝑇1 ≠ 𝑇2 ≠ 𝑇3 ≠ 𝑇4 
𝐻0 : 𝐵1 = 𝐵2 = 𝐵3 = 𝐵4
𝐻1 : 𝐵1 ≠ 𝐵2 ≠ 𝐵3 ≠ 𝐵4 
AND
We use an F- distribution and 0.05 level of significance.
Stirring
Rate

Furnace
1
8
14
14
17
2
4
5
6
9
𝐵1 =53
𝐵2 =24
5
10
15
20
3
5
6
9
3
4
6
9
2
6
𝑇1 =23
𝑇2 =34
𝑇3 =31
𝑇4 =35
𝐵3 =23 𝐵3 =23 ∑ 𝑦𝑖𝑗 =123
2
𝐶𝑀 =
(𝑦𝑖𝑗 )
𝑁
=
1232
16
=945.5625
𝑇𝑆𝑆 = ∑ 𝑦𝑖𝑗 2 − 𝐶𝑀 = (82 + 142 + ⋯ … 62 ) − 𝐶𝑀 = 1211 − 945.5625 =265.4375
𝑆𝑆𝐹𝑢𝑟𝑛𝑎𝑐𝑒 =
∑ 𝑇𝑖 2
𝑏
− 𝐶𝑀 = (
532 +242 +232 +232
)−
4
𝐶𝑀 =
4443
−
4
𝐶𝑀 =1110.75−945.5625 =165.1875
𝑆𝑆𝑆𝑡𝑖𝑟𝑖𝑛𝑔 𝑅𝑎𝑡𝑒 =
∑ 𝐵𝑖 2
𝑝
232 +342 +312 +352
)−
4
− 𝐶𝑀 = (
𝐶𝑀 =
3871
4
945.5625 = 22.1875
𝑆𝑆𝐸 = 𝑇𝑆𝑆 − 𝑆𝑆𝑇𝑟𝑒𝑎𝑡 − 𝑆𝑆𝐵𝑙𝑜𝑐𝑘𝑠 =78.0625
ANOVA TABLE
Source
DF
Furnace
3
S. Rate
3
Error
9
Total
15
SS
MS
𝑭𝒐𝒃𝒔
165.1875 55.0625 6.348279
22.1875 7.395833 0.852682
78.0625 8.673611
265.4375
𝑭𝒄𝒓𝒊𝒕𝒊𝒄
5.08
5.08
𝐹𝑐𝑟𝑖𝑡𝑖𝑐 = 𝐹(3.9) 0.05 = 3.86 for both Treatments and Blocks
a) Since 𝑭𝒐𝒃𝒔 > 𝑭𝒄𝒓𝒊𝒕𝒊𝒄 ,we reject H 0 and conclude that stirring rate impacts the grain size.

b) Since 𝑭𝒐𝒃𝒔 < 𝑭𝒄𝒓𝒊𝒕𝒊𝒄,we do not reject H 0 and conclude that the Furnace does not impact the
grain size. 