Question

The following data are from a completely randomized design.

	Treatment
	A	B	C
	163	142	126
	142	158	121
	168	129	138
	145	142	143
	147	133	153
	189	148	123
Sample mean	159	142	134
Sample variance	325.2	108.4	162.4

1. Compute the sum of squares between treatments. Round the intermediate calculations to whole number.
     
   
2. Compute the mean square between treatments.
     
   
3. Compute the sum of squares due to error.
     
   
4. Compute the mean square due to error (to 1 decimal).
     
   
5. Set up the ANOVA table for this problem. Round all Sum of Squares to the nearest whole number. Round all Mean Squares to one decimal place. Round F to two decimal places. Round p-value to four decimal places.
   

   Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
   Treatments
   Error
   Total

Answer 1

a)

	A	B	C
count, ni =	6	6	6
mean , x̅ i =	159.000	142.00	134.00
std. dev., si =	18.033	10.412	12.744
sample variances, si^2 =	325.200	108.400	162.400
total sum	954	852	804		2610	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =				145.00
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	196.000	9.000	121.000
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	1176.000	54.000	726.000		1956
SS(within ) = SSW = Σ(n-1)s² =	1626.000	542.000	812.000		2980.0000

sum of squares between treatments = 1956

b)

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   18
df within = N-k =   15

mean square between treatments , = SSB/k-1 = 1956/2 = 978

c)

sum of squares due to error=2980

d)

mean square due to error MSE = SSE/N-k = 2980/15 = 198.7

e)

	SS	df	MS	F	p-value
Treatments	1956	2	978.0	4.92	0.0227
Error	2980	15	198.7
Total:	4936	17