urgent;llyyyy a. What characteristic of the data specifically indicates that one-way analysis of variance should be used?O A. There are three samples of measurements.O B. The measurements are categorized according to the one characteristic of size.O C. The population means are approximately normal.O 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?O A. The method is based on estimates of a common population variance.O B. The method is based on estimates of multiple varying population variances.O C. The method is based on showing that the population variances are similar.OD. The method is based on showing that the population variances are different. 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 foranalysis 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 thestandard crash test. Complete parts (a) and (b) below.More InfoSPSS Sum ofResults SquaresBetween 200.857GroupsWithinGroupsTotal549.714750.571df21820PrintMean Square100.42930.540DoneF3.288Sig.0.061XDustMore InfoChest Deceleration Measurements (g) from a Standard Crash Test37543944434252413837SmallMidsizeLarge443632395345PrintDone4938424133X

PART A) Here we can see that in data measurement is taken Chest Measurements So the correct option…

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. More Info SPSS Sum of Results Squares Between 200.857 Groups Within Groups Total 549.714 750.571 df 2 18 20 Print Mean Square 100.429 30.540 Done F 3.288 sig. 0.061 X bu More Info Chest Deceleration Measurements (g) from a Standard Crash Test 37 54 44 43 49 41 38 Small Midsize Large 44 36 32 39 53 45 Print 42 38 Done 39 52 37 42 41 33 X

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. More Info SPSS Sum of Results Squares Between 200.857 Groups Within Groups Total 549.714 750.571 df 2 18 20 Print Mean Square 100.429 30.540 Done F 3.288 sig. 0.061 X bu More Info Chest Deceleration Measurements (g) from a Standard Crash Test 37 54 44 43 49 41 38 Small Midsize Large 44 36 32 39 53 45 Print 42 38 Done 39 52 37 42 41 33 X

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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