Question

The following data are from a completely randomized design.

	Treatment	Treatment	Treatment
	A	B	C
	32	46	33
	30	45	36
	30	46	35
	26	48	36
	32	50	40
Sample mean	30	47	36
Sample variance	6	4	6.5

1. At the  = .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 Selectless than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 6  
   
   What is your conclusion?
   SelectConclude that not all treatment means are equalDo not reject the assumption that the treatment means are equalItem 7  
   
2. 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  = .05.
   

   Difference	Absolute Value	Conclusion
   A -  B		SelectSignificant differenceNo significant differenceItem 10
   A -  C		SelectSignificant differenceNo significant differenceItem 12
   B -  C		SelectSignificant differenceNo significant differenceItem 14

   
   
3. 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 ( _A -  _B) should be reported with negative values.
   (  ,   )

Answer 1

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

Source	SS	df	MS	F	P value
Between	743.333	2	371.67	67.5758	0.000
Within	66.000	12	5.50
Total	809.333	14

a)

sum of sq;treatment=	743.33
sum of sq; error=	66.00
mean sq;treatment=	371.67
mean square; error=	5.50

test statistic =

67.58

The p-value is  less than .01

Conclude that not all treatment means are equal

b)

critical value of t with 0.05 level and N-k=12 degree of freedom=				tN-k=	2.179
Fisher's (LSD) for group i and j =(tN-k)*(sp*√(1/ni+1/nj)   =					3.23

Difference	Absolute Value	Conclusion
x1-x2	17.00	significant difference
x1-x3	6.00	significant difference
x2-x3	11.00	significant difference

95% confidence interval =-20.23 to -13.77