Question

The following data are from a completely randomized design. In the following calculations, use

α = 0.05.

	Treatment 1	Treatment 2	Treatment 3
	63	82	69
	46	73	55
	54	87	62
	37	70	46
xj	50	78	58
sj2	123.33	62.00	96.67

a) Find the value of the test statistic. (Round your answer to two decimal places.)

Find the p-value. (Round your answer to three decimal places.)

p-value =

b)

Use Fisher's LSD procedure to determine which means are different.

Find the value of LSD. (Round your answer to two decimal places.)

LSD =

Answer 1

Applying one way ANOVA: (use excel: data: data analysis: one way ANOVA: select Array):

Source	SS	df	MS	F	P value
treatment	1664.00	2	832.00	8.85	0.0075
error	846.00	9	94.00
Total	2510.00	11

a)

value of the test statistic F =8.85

p value =0.0075

b)

critical value of t with 0.05 level and N-k=9 degree of freedom=				tN-k=	2.262
Fisher's (LSD) for group i and j =(tN-k)*(sp*√(1/ni+1/nj)   =					15.51

Difference	Absolute Value	Conclusion
x1-x2	28.00	significant difference
x1-x3	8.00	not significant difference
x2-x3	20.00	significant difference

treatment 1 and 2 as well treatment 2 and 3 are different