Question

A highway safety institution conducts experiments in which cars are crashed into a fixed barrier at 40 mph. In the institute's 40-mph offset test, 40% of the total width of each vehicle strikes a barrier on the driver's side. The barrier's deformable face is made of aluminum honeycomb, which makes the forces in the test similar to those involved in a frontal offset crash between two vehicles of the same weight, each going just less than 40 mph. You are in the market to buy a family car and you want to know if the mean head injury resulting from this offset crash is the same for large family cars, passenger vans, and midsize utility vehicles (SUVs). The data in the accompanying table were collected from the institute's study. Complete parts (a) through (d) below. Click the icon to view the data table. (a) State the null and alternative hypotheses. O A. Ho: HCars = H Vans = HSUvs and Hy: at least one mean is different O B. Ho: Cars = Hvans = Hsuvs and Hy: all means are different OC. Ho: 'Cars = Hvans = Hsuvs and Hy: HCars <Hvans <HSUVs

P value

State your conclusion. Would we perform post-hoc procedures for this data?

Answer 1

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

	Large Car	Passenger	SUV
Total	2277	2754	2246
n	7	7	7
Mean	325.29	393.43	320.86
Sum Of Squares	233903.429	212849.7143	99076.8572
Variance	38983.9048	35474.9524	16512.8095
SD	197.4434	188.3480	128.5022

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(a) The Hypothesis: Option B

H0:

Mcars = Hvans = MSUV

H1: At least one mean is different.

(b) Option C: yes the requirements are met.

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The ANOVA table is as below

Source	SS	DF	Mean Square	F	p
Between	23169.49	2	11584.74	0.382	0.6879
Within/Error	545830.00	18	30323.89
Total	568999.49	20

The p Value: The p value is calculated for F = 0.38 for df1 = 2 and df2 = 18 and is = 0.6879 (P value > 0.10)

The Decision Rule:

If p-value is < $\alpha$ , Then reject H0.

The Decision:

Also since p-value (0.6879) is > $\alpha$ (0.01), We Fail to Reject H0.

The Conclusion: There isn't sufficient evidence at the 99% level of significance to warrant rejection of the claim that the head injury for each vehicle type is the same.

No, we would not conduct post hoc procedures for this data.

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Calculations For the ANOVA Table:

Overall Mean = [325.29 + 393.43 + 320.86] / 3 = 346.52

SS treatment = SUM [n* ( $\bar{x_i}$ - overall mean)²] = 7 * (325.29 - 346.52)² + 7 * (393.43 - 346.52)² + 7 * (320.86 - 346.52)² = 23169.49

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment/df1 = 23169.49 / 2 = 11584.74

SSerror = SUM (Sum of Squares) = 233903.429 + 212849.7143 + 99076.8572 = 545830

df2 = N - k = 21 - 3 = 18

Therefore MS error = SSerror/df2 = 545830 / 18 = 30323.89

F = MSTR/MSE = 11584.74 / 30323.89 = 0.382

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Mcars = Hvans = MSUV

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