Question

a research study in an area that is of interest to you we are going to analyze the study using a one-way ANOVA by doing the following:

• State the null and alternative hypotheses
• Make a decision about the null hypothesis
• Draw a conclusion from the analysis
• Relate the results of the analysis to the research hypothesis

Answer 1

In a group of 15 people, the data shows us how much is the life of tyres when fixed on different cars The data is as follows and is given in months

#	BMW	AUDI	Jaguar
1	34	32	44
2	36	31	37
3	40	47	40
4	33	40	31
5	37	35	28

A car dealer says that the mean tyrage is the same. Are his claims valid at the 0.05 level of significance?

The summary statistics obtained from the given data are as below.

Total	180	185	180
n	5	5	5
Mean	36.00	37.00	36.00
Sum Of Squares	30	174	170
Variance	7.5	43.5	42.5

The Hypothesis:

H0: There is no difference between the means of the of the 3 cars: .

Ha: The mean of at least one car is different from the others..

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The ANOVA table is as below. the p value is calculated for F = 0.05 for df1 = 2 and df2 = 12

The Fcritical is calculated at = 0.05 for df1 = 2 and df2 = 12

Source	SS	DF	Mean Square	F	Fcv	p
Between	3.33	2	1.67	0.05	3.8853	0.9495
Within/Error	383.47	12	31.96
Total	386.80	14

The Decision Rule:

If Ftest is > F critical, Then Reject H0.

Also if p-value is < Then reject H0.

The Decision:

Since Ftest (0.05) is < F critical (3.8853), We Fail to Reject H0.

Also since p-value (0.9495) is > (0.05), We Fail to Reject H0.

The Conclusion: There isn't sufficient evidence at the 95% level of significance to warrant rejection of the claim that the means of tire life of the three cars are different.

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We use Tukeys HSD to relate to what we have proved

Tukeys Post Hoc Table

Where Mi and Mj are the 2 means being compared and their positive (absolute) differences is taken.

n = number of replicates in each sample. Here n = 5, MSerror = 31.17

The Rule is that if Tukeys observed is > Tukeys critical, then there is a significant difference between groups.

Tukeys critical is found from the critical value tables for = 0.05, for k (# of columns) = 3 on the horizontal and df error = 12 on the vertical.

The Critical Value = 3.773

M1 = 36, M2 = 37, M3 = 36,

The Tabular Format is as below, taking 2 groups at a time.

	Absolute Difference	Observed	Critical	Obs > Crit
M1 - M2	1	0.4005	3.44	No
M1 - M3	0	0.0000	3.44	No
M2 - M3	1	0.4005	3.44	No

Therefore there is no significant difference between any 2 groups taken at a time.

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Calculations For the ANOVA Table:

Overall Mean = [(5*36 + (5*37) + (5*36) / 15 = 545/15 = 36.33

SS treatment = SUM [n* ( - overall mean)²] = 5 * (36.33 - 36)² + 5 * (36.33 - 37)² + 5 * (36.33 - 36)² = 3.33

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment/df1 = 3.33 / 2 = 1.67

SSerror = SUM (Sum of Squares) = 30 + 174 + 170 = 374

df2 = N - k = 15 - 3 = 12

Therefore MS error = SSerror/df2 = 374 / 12 = 31.17

F = MSTR/MSE = 1.67 / 31.17 = 0.05

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