Statistics
Essay by subhashsharma • November 2, 2016 • Research Paper • 1,756 Words (8 Pages) • 1,223 Views
Chapter 13
1. a) From the problem, the given data is
The sample mean treatments for A,B and C are
= 156[pic 1]
= 142[pic 2]
= 134[pic 3]
Number of treatments, k = 3
== = 6[pic 4][pic 5][pic 6]
= [pic 7][pic 8]
= 6 + 6 + 6
= 18
The formula for the overall sample mean is
= [pic 9][pic 10]
= = 14[pic 11]
The sum of squares between treatments is
SSTR = [pic 12]
= 6(156 – 144)2 + 6(142 – 144)2 + 6(134 – 144)2
= 1488
b) Mean square treatment is
MSTR = [pic 13]
= = 744[pic 14]
c) The given sample variances are
= 164.4[pic 15]
= 131.2 [pic 16]
= 110.4[pic 17]
Sum of the squares after the error is followed by
SSE =[pic 18]
= (6 – 1)164.4 + (6 – 1)131.2 + (6 – 1)110.4
= 2030
d) Mean square due to the error is given by: MSE = [pic 19]
= [pic 20]
= 135.3
e) ANOVA table for the entire randomized design is
Variation source  Sum of squares  Degrees of freedom
 Mean square  F  Pvalue

Treatment  SSTR  k  1  MSTR= [pic 21]  [pic 22]  
Error  SSE   k[pic 23]  MSE= [pic 24]  
Total  SST   1[pic 25] 
The ANOVA table for the given experiment is
Variation Source  Sum of Squares  Degrees of Freedom  Mean Square  F  p  value 
Treatment  1488  2  744  5.50  0.0162 
Error  2030  15  135.3  
Total  3518  17 
f) In order to test the equality of three means for the test statistic is given by F = [pic 26]
= = 5.50[pic 27]
degree of freedom for the numerator is k – 1 = 2
degree of freedom for the denominator is 18 – 3[pic 28]
= 15.
Here we reject null hypothesis, because of the greater values of the test statistics.
Here the pvalue is the higher tail area of F distribution to the right side of the test statistic.
F=5.50
Area in the Upper Tail  0.1  0.05  0.025  0.01 
F value (=2,=15)[pic 29][pic 30]  2.7  3.64  4.77  6.36 
F=5.50 which is between 4.77 and 6.36
The area in the upper tail at F=5.50 is between 0.025 and 0.01.
So the pvalue is between 0.01 and 0.025.
pvalue = 0.0162.
will be rejected With pvalue ≤ α = 0.05[pic 31]
It is proved from the test that the three treatments are not equal
11) Assume the population 1 as drying times for paint 1, population 2 for the paint 2 and population 3 for the paint 3 and population 4 for the paint 4.
Here
= population 1 mean drying time [pic 32]
= population 2 mean drying time [pic 33]
= population 3 mean drying time [pic 34]
= population 4 mean drying time[pic 35]
Total treatments, k = 4
and = ====5[pic 36][pic 37][pic 38][pic 39][pic 40]
= + +++[pic 41][pic 42][pic 43][pic 44][pic 45][pic 46]
= 5+5+5+5+5
= 25
Treatment j sample mean = = [pic 47][pic 48]
given data for the sample mean yields at the three different temperature levels are
= 133[pic 49]
= 139[pic 50]
= 136[pic 51]
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