Homework Answers
| A | B | C | |||
| count, ni = | 5 | 5 | 5 | ||
| mean , x̅ i = | 201.000 | 199.40 | 210.60 | ||
| std. dev., si = | 3.082 | 2.302 | 2.702 | ||
| sample variances, si^2 = | 9.500 | 5.300 | 7.300 | ||
| total sum | 1005 | 997 | 1053 | 3055 | |
| grand mean , x̅̅ = | Σni*x̅i/Σni = | 203.67 | |||
| square of deviation of sample mean from grand mean,( x̅ - x̅̅)² | 7.111 | 18.204 | 48.071 | ||
| TOTAL | |||||
| SS(between)= SSB = Σn( x̅ - x̅̅)² = | 35.556 | 91.022 | 240.356 | 366.9333 | |
| SS(within ) = SSW = Σ(n-1)s² = | 38.000 | 21.200 | 29.200 | 88.4000 |
no. of treatment , k = 3
df between = k-1 = 2
N = Σn = 15
df within = N-k = 12
mean square between groups , MSB = SSB/k-1 =
183.4667
mean square within groups , MSW = SSW/N-k =
7.3667
F-stat = MSB/MSW = 24.9050
| SS | df | MS | F | p-value | F-critical | |
| Between: | 366.93 | 2 | 183.47 | 24.90 | 0.0001 | 3.89 |
| Within: | 88.40 | 12 | 7.37 | |||
| Total: | 455.33 | 14 | ||||
| α = | 0.05 | |||||
| conclusion : | p-value<α , reject null hypothesis |
F = 24.9050
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