Question

1. Braking distance was evaluated for 4 different brake materials. Run an analysis of variance to determine if there is a difference in treatments?   Use the six steps of hypothesis testing at a 0.05 level of significance. Put the below data into Excel to perform the analysis.

Braking Distance
Disk Material 1	Disk Material 2	Disk Material 3	Disk Material 4
69.8	78.1	77.7	90.6
65.6	81.0	80.9	91.0
71.7	81.9	80.8	89.4
70.5	81.8	79.8	88.1
68.0	78.8	81.1	87.5
69.3	78.9	82.0	92.0
71.0	81.4	80.0	90.7
68.8	82.2	80.8	88.4
66.7	79.8	80.9	89.7
72.0	76.7	79.8	88.1
72.8	80.1	80.1	89.4
67.2	78.7	80.4	89.2

2. Is the average braking distance for material 2 different than the average braking distance for material 3? Prove your conclusion by using the confidence interval formula in chapter 12

Answer 1

a)

Analysis of variance:

Excle > Data > Data Analysis > Anova: Single Factor

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Disk Material 1	12	833.4000	69.4500	5.1282
Disk Material 2	12	959.4000	79.9500	3.0464
Disk Material 3	12	964.3000	80.3583	1.1008
Disk Material 4	12	1074.1000	89.5083	1.8572
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	2420.488	3	806.8294	289.8986	5.48E-29	2.8165
Within Groups	122.4583	44	2.7831
Total	2542.947	47

Hypothesis:

H0: μ1 = μ2 = μ3 = μ4

Ha: Not all means are equal

Test:

F stat = 289.8986

P value = 0

Decision:

P value < 0.05, Reject H0

Conclusion:

There is enough evidence to conclude that there is a difference in treatments at 5% significance level

b)

	Disk Material 2	Disk Material 3
	78.1	77.7
	81	80.9
	81.9	80.8
	81.8	79.8
	78.8	81.1
	78.9	82
	81.4	80
	82.2	80.8
	79.8	80.9
	76.7	79.8
	80.1	80.1
	78.7	80.4
Mean	79.9500	80.3583
SD	1.7454	1.0492
n	12	12

Hypothesis:

H0: μ2 = μ3

Ha: μ2 not = μ3

Equal varaince test:

Fstat	2.767409194	S1^2/S2^2
Fc	2.81793047	From t table

F stat < Fc, Do not reject the null hypothesis of equal variance

95% Confidence interval for the difference between the average braking distance for material 2 and the average braking distance for material 3

CI	Equal variance
tc	2.073873068	T.INV.2T(alpha,Df)
Upper	0.810889538	(X1 bar-X2 bar )+tc*Sp*SQRT(1/n1 + 1/n2)
Lower	-1.627489538	(X1 bar-X2 bar )-tc*Sp*SQRT(1/n1 + 1/n2)

CI = (-1.6275, 0.8109)

We do not reject null hypothesis why because confidence interval contains null value

There is not enough evidence to conclude that the average braking distance for material 2 different than the average braking distance for material 3