Consider the following Oneway ANOVA computer output below Ho

Consider the following One-way ANOVA computer output below How many levels of the treatment were used in this experiment How many replicates were used for each treatment Estimate the variance of the error term Find the value for (1) through (6) Find the critical value for Fat a=0.05 -venation (5 If) Make your conclusion on the effect of treatments at a=0.05 w.th an explanation

Solution

a) No of levels of treatment = DF treatment + 1

DF treatment = DF Total - DF Error = 19 - 15 = 4

No of levels of treatment = 4 + 1 = 5

b) Total df = 19

Total df = No of levels * No of replicates - 1

19 = 5 * No of replicates - 1

No of replicates = 4

c) Variance of Error term MS Error = SS Error/DF Error = 167.5/15 = 11.17

d)

(1) DF treatment = DF Total - DF Error = 19 - 15 = 4

(2) SS Treatment = SS Total - SS Error = 326.2 - 167.15 = 159.05

(3) MS Treatment = SS Treatment/DF Treatment = 159.05/4 = 39.76

(4) MS Error = SS Error/DF Error = 167.5/15 = 11.17

(5) F = MS Treatment / MS Error = 39.76 / 11.17 = 3.56

(6) P value = 0.0312

e) Critical value for F at aplha 0.05 is 3.056 (Use the F table for DF numerator 4 and DF denominator 15)

f) Since the test statistic is larger than the critical value, we reject the null hypothesis