Question

The following data are from a completely randomized design. Treatment A B C 163 145 123 142 157 121 166 129 137 145 145 141 149 132 154 189 144 128 Sample mean 159 142 134 Sample variance 310 103.2 156.8 a. Compute the sum of squares between treatments. b. Compute the mean square between treatments. c. Compute the sum of squares due to error. d. Compute the mean square due to error (to 1 decimal). e. Set up the ANOVA table for this problem. Round all Sum of Squares to nearest whole numbers. Round all Mean Squares to one decimal place. Round F to two decimal places. Round your p-value to 4 decimal places. Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments Error Total f. At the  level of significance, test whether the means for the three treatments are equal. Calculate the value of the test statistic (to 2 decimals). The -value is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 16 What is your conclusion? - Select your answer -Conclude that not all treatment means are equalDo not reject the assumption that the means for all three treatments are equalItem 17

Answer 1

Ans:

	Group 1	Group 2	Group 3	Total
Sum	954	852	804	2610
Count	6	6	6	18
Mean, Sum/n	159	142	134
Sum of square, Ʃ(xᵢ-x̅)²	1550	516	784
Standard deviation	17.6068	10.1587	12.5220

Number of treatment, k =	3
Total sample Size, N =	18
df(between) = k-1 =	2
df(within) = N-k =	15
df(total) = N-1 =	17
SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N =	1956
SS(within) = SS1 + SS2 + SS3 =	2850
SS(total) = SS(between) + SS(within) =	4806
MS(between) = SS(between)/df(between) =	978
MS(within) = SS(within)/df(within) =	190
F = MS(between)/MS(within) =	5.1474
p-value = F.DIST.RT(5.1474, 2, 15) =	0.0199

Null and Alternative Hypothesis:
Ho: µ1 = µ2 = µ3
H1: At least one mean is different.
Test statistic:
F =	5.1474
Critical value:
Critical value Fc = F.INV.RT(0.05, 2, 15) =	3.6823
p-value:
p-value = F.DIST.RT(5.1474, 2, 15) =	0.0199
Decision:
P-value < α, Reject the null hypothesis.

a)SS(between)=1956

b)MS(between)=1956/2=978

c)SS(error)=2850

d)MS(error)=2850/15=190

e)

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	1956	2	978	5.15	0.0199
Within Groups	2850	15	190
Total	4806	17

f)Test statistic,F=978/190=5.15

alpha=0.05

The p-value is between .01 and .025
Conclude that not all treatment means are equal

(as p-value<0.05)