Question

Treatment A Treatment B Treatment C

32 44 34

30 43 37

30 44 36

26 46 37

32 48 41

Sample mean 30 45 37

Sample variance 6 4 6.5

a. At the X= 0.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal?

Compute the values below (to 1 decimal, if necessary).

Sum of Squares, Treatment

Sum of Squares, Error

Mean Squares, Treatment

Mean Squares, Error

Calculate the value of the test statistic (to 2 decimals).

The p-value is

What is your conclusion?

b. Calculate the value of Fisher's LSD (to 2 decimals). Use Fisher's LSD procedure to test whether there is a significant difference between the means for treatments A and B, treatments A and C, and treatments B and C. Use x=0.05.

Difference Absolute Value Conclusion

xA-xB

xA-xC

XB-xC

c. Use Fisher's LSD procedure to develop a 95% confidence interval estimate of the difference between the means of treatments A and B (to 2 decimals). Since treatment B has the larger mean, the confidence interval for the difference between the means of treatments A and B (xA-xB ) should be reported with negative values.

( , )

Answer 1

a)

Applying one way ANOVA:

Source of variation		SS	df	MS	F	p vlaue
treatments		563.333	2	281.6667	51.212	0.000
error		66.000	12	5.5000
total		629.333	14

sum of square; treatment=	563.33
sum of square; error=	66.00
mean square; treatment=	281.67
mean square; error=	5.50

test statistic =

51.21

pvalue =0.000

b)

Fisher's (LSD) for group i and j =                  (tN-k)*(sp*√(1/ni+1/nj)           =

3.23

Difference	Absolute Value	Conclusion
x1-x2	15.00	significant difference
x1-x3	7.00	significant difference
x2-x3	8.00	significant difference

c)

95% confidence interval -18.23 , -11.77