Question

Allied Corporation wants to increase the productivity of its line workers. Four different programs have been suggested to help increase productivity. Twenty employees, making up a sample, have been randomly assigned to one of the four programs and their output for a day's work has been recorded. The results are shown below. The company wants to test if there us significance different I in productivity across programs.

	Program A	Program B	Program C	Program D
	150	150	185	175
	130	120	220	150
	120	135	190	120
	180	160	180	130
	145	110	175	175
Sample Mean	145	135	190	150

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	Source of Variation	Sum of Squares	Degrees of Freedom
	Treatment	8,750	3
	Error	7,600	16
	Total	16,350	19

a.	State the null and alternative hypotheses.
b.	Calculate the Mean Square due to Treatment (MSTR) and Mean Square Error (MSE).
c.	As the statistical consultant to Allied, what would you advise? Use a .05 level of significance.
d.	Use Fisher's LSD procedure and determine which population mean (if any) is different from the others. Let α = .05.

Answer 1

Here, Allied corporation wants to increase the productivity of its line workers. So, for that 20 employees making up a sample have been randomly assigned to one of 4 programs & their output has been recorded. We have to test that if there is significant difference in productivity across programs. Hence, Hypothesis can be stated as-

Null hypothesis ( H0 ) : there is any significant difference in productivity across 4 programs. i.e. α1 = α2 = α3 = α4

VS Alternative hypothesis ( H1 ) : there is no any significant difference between productivity across 4 programs. i.e. α1 ≠ α2 ≠ α3 ≠ α4

To calculate mean square due to treatment and mean square error, we have ANOVA table as follows -

source of variation	degrees of freedom	sum of square	mean sum of square	F- ratio
Treatment	t-1	SST	MSST=SST/(t-1)	F=MSST/MSSE
Error	n-t	SSE	MSSE=SSE/(n-t)	-
Total	n-1	TSS	-	-

So, from given table values and above ANOVA table -

WE have, MST = SST/(t-1)

Here, SST = 8.750, t-1 = 3

  MST = 8.750/3 = 2.9166

SSE = 7.600, n-t = 16

MSE = 7.6/16 = 0.475

Hence, F-ratio = MSST/MSSE = 2.9166/0.475 = 6.1402

Hence, ANOVA table is as follows-

source of variation	degrees of freedom	sum of square	mean sum of square	F- ratio
Treatment	t-1=3	SST=8.75	MSST=2.9166	F=6.1402
Error	n-t=16	SSE	MSSE=0.475	-
Total	n-1	TSS	-	-

Test statistic -  

Tab F = Ft-1,n-t,α = F3,16,0.05 = 2.4618

Cal F = 6.1402

So, Cal F > Tab F ,Hence we can reject Ho at 5% of level of significance.

Conclusion - There is an evidence that there is no significant difference between productivity across 4 programs.

As a statistical consultant, I would advise that any of 4 programs can be implement to increase the productivity of line workers.

Also, we can see that there is no significant difference between productivity across 4 programs, so there is no need to determine which procedure is differnt from other.