Answer:
see explaination
Explanation:
The statistical procedure for comparing the 3 groups was the F test with 2 degrees of freedom in the numerator (groups - 1) and 23 df in the denominator (N total - groups)
It is calculated as SSB/2/(SSW/23) = 10.22754
As Fcrit =
from Excel, finv(.05,2,23) = 3.4221, we can reject the null hypothesis of no difference between groups.
SSW = the sum of std i ^ 2 * (n i - 1)
SSB = the sum of ni (mean i - mean)^2
2. From Excel, we get the pvalue from fdist(10.22754,2,23) = .000664
3. For LSD, we calculate (mean 1 - mean 2)/(s * sqrt(1/n1+1/n2))
This is the pooled s = sqrt(SSW/23)
Then, we found t crit from tinv(.05,23) = 2.068
Making use of the chart i made, We found significant differences between std and lac as well as std and veg, but no significant difference between lac and veg.