Question

Random sample of coin weights
Quarter Nickel Penny
7.9 9.5 6.7
7.2 7.0 7.1
7.8 8.7 6.5
8.1 7.6 7.1
7.9 N/A 5.5
Test at a 1% level of significance whether the population mean weights of any two watchband
types are different? State the required assumptions for the test. Hint: Use sum of x^2=794.02

Answer 1

H0: μ1 = μ2 = μ3

H1: There is at least one mean difference among the populations.

Applying one way ANOVA":

	Quarter	Nickel	Penny
	7.9	9.5	6.7
	7.2	7	7.1
	7.8	8.7	6.5
	8.1	7.6	7.1
	7.9		5.5
count ni=	5	4	5
Average==ΣXi/ni=	7.7800	8.2000	6.5800
Total=ΣX2i   =	303.110	272.7000	218.2100
SS=ΣX2i -(Σxi)2/ni=	0.468	3.7400	1.7280
Grand average Xgrand =∑xi/N =		104.6/9=	7.4714

i	ni	x̅i	ni*(Xi-Xgrand)2	SS=(ni-1)*s2
Quarter	5	7.7800	0.476	0.468
Nickel	4	8.2000	2.123	3.74
Penny	5	6.5800	3.973	1.728
Total	14		6.573	5.9360
			SSTr	SSE
df treatments =	number of treatments-1=			2
df error =	N-k=	14-3=		11
df total=	N-1=	14-1=		13
MS(treatment) =SSTr/df(Tr)=6.57/2=			3.28629
MS(error) =SSE/df(error)=5.94/11=			0.53964
test statistic F =MSTr/MSE =3.29/0.54=			6.090

for 1% level and (2,11) degree of freedom, critical value =7.206

Since test statistic (6.090 ) < crtiical value we fail to reject null hypothesis

we do not have sufficient evidence at 1% level to conclude that   population mean weights of any two watchband
types are different