The accompanying data table contains chest deceleration measurements (in g, wher

Question

The accompanying data table contains chest deceleration measurements (in g, where g is the force of gravity) from samples of small, midsize, and large cars. Shown are the technology results for analysis of variance of this data table. Assume that a researcher plans to use a 0.05 significance level to test the claim that the different size categories have the same mean chest deceleration in the standard crash test. Complete parts (a) and (b) below.

a. What characteristic of the data specifically indicates that one-way analysis of variance should be used?

A. There are three samples of measurements

B. The population means are approximately normal.

C. The measurements are categorized according to the one characteristic of size.

D. Nothing specifically indicates that one-way analysis of variance should be used.

b. If the objective is to test the claim that the three size categories have the same mean chest deceleration, why is the method referred to as analysis of variance?

A. The method is based on showing that the population variances are similar.

B. The method is based on showing that the population variances are different.

C. The method is based on estimates of multiple varying population variances.

D. The method is based on estimates of a common population variance.

Chest Develeration Measurements (g) from a Standard Crash Test small 44 39 37 54 44 42 midsize 36 53 43 42 49 41 large 32 45 41 38 38 33

Explanation / Answer

Solution-

(a) Assumtions for one way anova are-

Thus the correct option is - B: The population means are approximately normal.

P-value for the test is 0.061

This is greater than 0.05 hence null hypothesis is not rejected and thus means of population are equal.

Looking at the option correct option is D: The method is based on estimates of a common population variance.

Answers

TY!

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