In: Statistics and Probability
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A)
Ho: µ1=µ2=µ3=µ4
H1: not all means are equal
.....
B)
F stat will be used
treatment | G1 | G2 | G3 | G4 | ||||
count, ni = | 7 | 7 | 7 | 7 | ||||
mean , x̅ i = | 118.286 | 122.00 | 121.71 | 129.86 | ||||
std. dev., si = | 3.1 | 5.6 | 4.2 | 3.7 | ||||
sample variances, si^2 = | 9.571 | 31.667 | 17.905 | 13.476 | ||||
total sum | 828 | 854 | 852 | 909 | 3443 | (grand sum) | ||
grand mean , x̅̅ = | Σni*x̅i/Σni = | 122.96 | ||||||
( x̅ - x̅̅ )² | 21.889 | 0.930 | 1.563 | 47.511 | ||||
TOTAL | ||||||||
SS(between)= SSB = Σn( x̅ - x̅̅)² = | 153.223 | 6.509 | 10.938 | 332.580 | 503.25 | |||
SS(within ) = SSW = Σ(n-1)s² = | 57.429 | 190.000 | 107.429 | 80.857 | 435.7143 |
no. of treatment , k = 4
df between = k-1 = 3
N = Σn = 28
df within = N-k = 24
mean square between groups , MSB = SSB/k-1 =
503.25/3= 167.7500
mean square within groups , MSW = SSW/N-k =
435.7143/24= 18.1548
F-stat = MSB/MSW = 167.75/18.1548=
9.24
P value = 0.0003
anova table | ||||||
SS | df | MS | F | p-value | ||
Between: | 503.3 | 3 | 167.8 | 9.24 | 0.000 | |
Within: | 435.7 | 24 | 18.2 | |||
Total: | 939.0 | 27 |
F =9.24
................
c)
α = 0.01
F-critical = 4.718
so, f stat should be greater than 4.718
d)
Decision: f stat > critical , reject null
hypothesis
so, there is difference between drying times
....
e)
tukey kramer rule:
critical value = q*√(MSE/2*(1/ni+1/nj))
if absolute difference of means > critical
value,means are significnantly different ,otherwise
not
........................
Please let me know in case of any doubt.
Thanks in advance!
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