a)
Ho: µ1=µ2=µ3
H1: not all means are equal
excel data analysis tool for one factor anova,steps are:
write data>menu>data>data analysis>anova :one
factor>enter required labels>ok, and following o/p Is
obtained,
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
0 | 8 | 13 | 1.625 | 1.98 | ||
24 | 8 | 25 | 3.125 | 2.4 | ||
48 | 8 | 40 | 5.00 | 4.00 | ||
72 | 8 | 60 | 7.50 | 10.57 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 154 | 3 | 51.38 | 10.84 | 0.000 | 2.95 |
Within Groups | 133 | 28 | 4.74 | |||
Total | 287 | 31 |
p value < α=0.05, reject Ho
conclusion :there is significant difference in four means
b)
Level of significance | 0.0500 |
no. of treatments,k | 4 |
DF error =N-k= | 28 |
MSE | 4.738 |
t-critical value,t(α/2,df) | 2.048 |
Fishers LSD critical value=tα/2,df √(MSE(1/ni+1/nj))
if absolute difference of means > critical value,means are
significnantly different ,otherwise not
absolute mean difference | critical value | result | |||||
µ1-µ3 | 3.375 | 2.2293 | means are different | ||||
µ2-µ4 | 4.375 | 2.2293 | means are different |
Note: for all problems, assume a.05 1. A common assumption is that sleep deprivation influences aggression. To test this assumption, volunteer subjects are randomly assigned by a researcher to sl...