In: Statistics and Probability
Imagine you are testing for the effects of two experimental drugs (data sets B and C)relative to a control group (Data set A) on a physiological variable.
Data set A: 105.51,111.46,104.35,115.27,113.98,119.18,116.45,119.98
Data set B: 93.63,88.90,95.03,86.08,85.86,98.22,105.46,96.50
Data set C: 95.12,97.12,94.08,99.78,101.44,101.27,104.70,103.70
a) Conduct an ANOVA on the data set. Show all calculations, table of results, and state your conclusion.
Please show the calculations.
One Way Analysis of Variance (ANOVA)
treatment | A | B | C | |||
count, ni = | 8 | 8 | 8 | |||
mean , x̅ i = | 113.273 | 93.71 | 99.651 | |||
std. dev., si = | 5.826 | 6.665 | 3.886 | |||
sample variances, si^2 = | 33.947 | 44.416 | 15.100 | |||
grand mean , x̅̅ = | Σni*x̅i/Σni = | 102.21 | ||||
square of deviation of sample mean from grand mean,( x̅ - x̅̅)² | 122.351 | 72.271 | 6.554 | |||
TOTAL | ||||||
SS(between)= SSB = Σn( x̅ - x̅̅)² = | 978.810 | 578.170 | 52.429 | 1609.408825 | ||
SS(within ) = SSW = Σ(n-1)s² = | 237.629 | 310.911 | 105.697 | 654.2368 |
no. of treatment , k = 3
df between = k-1 = 2
N = Σn = 24
df within = N-k = 21
mean square between groups , MSB = SSB/k-1 =
804.7044
mean square within groups , MSW = SSW/N-k =
31.1541
F-stat = MSB/MSW = 25.8298
anova table | ||||||
SS | df | MS | F | p-value | F-critical | |
Between: | 1609.409 | 2 | 804.704 | 25.830 | 0.0000 | 3.467 |
Within: | 654.237 | 21 | 31.154 | |||
Total: | 2263.646 | 23 | ||||
α = | 0.05 |
value of test stat= 25.83
p value is 0.0000
Decision: p-value<α , reject
null hypothesis
conclusion : there is enough evidence of significant
mean difference among three
treatments