Question

To test whether the mean time needed to mix a batch of material is the same for machines produced by three manufacturers, the Jacobs Chemical Company obtained the following data on the time (in minutes) needed to mix the material.

Manufacturer
	1	2	3
	18	27	24
	24	24	21
	24	30	24
	21	27	21

1. Use these data to test whether the population mean times for mixing a batch of material differ for the three manufacturers. Use  = .05.  
   
   Compute the values below (to 2 decimals, 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 the mean time needed to mix a batch of material is not the same for all manufacturersDo not reject the assumption that mean time needed to mix a batch of material is the same for all manufacturersItem 7
   
2. At the   = .05 level of significance, use Fisher's LSD procedure to test for the equality of the means for manufacturers 1 and 3.  
   
   Calculate Fisher's LSD Value (to 2 decimals).  
   
   
   
   What is your conclusion about the mean time for manufacturer 1 and the mean time for manufacturer 3?  
   
   Select Cannot conclude there is a difference in the mean time for these manufacturers  These manufacturers have different mean times Item 9

Answer 1

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

Source	SS	df	MS	F	P value
Between	64.500	2	32.25	5.6087	0.026
Within	51.750	9	5.75
Total	116.250	11

a)

sum of sq;treatment=	64.50
sum of sq; error=	51.75
mean sq;treatment=	32.25
mean square; error=	5.75

value of the test statistic =5.61

The  p-value is between .025 and .05

Conclude the mean time needed to mix a batch of material is not the same for all manufacturers

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)   =					3.84

Cannot conclude there is a difference in the mean time for these manufacturers