Question

1. A researcher wishes to see whether there is any difference in the weight gains of athletes following one of three special diets. Athletes are randomly assigned to three groups and placed on the diet for 6 weeks. The weight gains (in pounds) are shown here. At α = 0.05, can the researcher conclude that there is a difference in the diets?

Diet A             Diet B             Diet C             Total

                        3                      10                   8

6                      12                   3

7                      11                   2

4                      14                   5

                                    8

                                    6

the critical value = 3.98

Conclusion:

Analysis of Variance Summary Table

P-value

Source                             Sum of Squares       D.F.    Mean Square F ratio p-value

Between groups                      

Within                            

SSB                                                           SSW   

MSB = SSBk-1                                               MSW=SSWN-k

F = MSBMSW

ANOVA 1 TUKEY B

                                    ID         N

                                    ____________________________

                                    3           4       4.50

                                        ____________________________

                                    1          4          5.00             

                                    ____________________________

                                    2          6                                  10.17

                                    _____________________________

                                    Sig.                 .954               1.00

Answer 1

	A	B	C
count, ni =	4	6	4
mean , x̅ i =	5.000	10.17	4.50
std. dev., si =	1.826	2.858	2.646
sample variances, si^2 =	3.333	8.167	7.000
total sum	20	61	18		99	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	7.07
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	4.291	9.580	6.612
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	17.163	57.483	26.449		101.0952
SS(within ) = SSW = Σ(n-1)s² =	10.000	40.833	21.000		71.8333

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   14
df within = N-k =   11
  
mean square between groups , MSB = SSB/k-1 =    50.5476
  
mean square within groups , MSW = SSW/N-k =    6.5303
  
F-stat = MSB/MSW =    7.7405

anova table
	SS	df	MS	F	p-value	F-critical
Between:	101.10	2	50.55	7.7405	0.0080	3.98
Within:	71.83	11	6.53
Total:	172.93	13
					α =	0.05
			conclusion :	p-value<α , reject null hypothesis

......................

Level of significance	0.05
no. of treatments,k	3
DF error =N-k=	11
MSE	6.530
q-statistic value(α,k,N-k)	3.8200

critical value = q*√(MSE/n)

			confidence interval
population mean difference	critical value	lower limit	upper limit	result
µ1-µ2	-5.17	4.46	-9.62	-0.71	means are different
µ1-µ3	0.50	4.88	-4.38	5.38	means are not different
µ2-µ3	5.67	4.46	1.21	10.12	means are different

...................................

Please revert back in case of any doubt.

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