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Sloan [31]
3 years ago
14

All cars can be classified into one of four​ groups: the​ subcompact, the​ compact, the​ midsize, and the​ full-size. There are

five cars in each group. Head injury data​ (in hic) for the dummies in the​ driver's seat are listed below. Use a 0.05 significance level. Find the​ P-value to test the null hypothesis that the different weight categories have the same mean.
Subcompact: 681 528 917 898 420

​Compact: 643 655 542 514 525

​Midsize: 469 727 525 454 359

​Full-size: 384 656 602 687 360
Mathematics
1 answer:
Olegator [25]3 years ago
6 0

Answer:

p value = 0.302

Step-by-step explanation:

Given that all cars can be classified into one of four​ groups: the​ subcompact, the​ compact, the​ midsize, and the​ full-size. There are five cars in each group. Head injury data​ (in hic) for the dummies in the​ driver's seat are listed below.

H_0: All cars have same mean values

H_a: atleast two cars have different mean values

(Two tailed anova test)

Anova: Single Factor      

     

SUMMARY      

Groups Count Sum Average Variance  

Subcompact 5 3444 688.8 48502.7  

compact 5 2879 575.8 4582.7  

Midsize 5 2534 506.8 18720.2  

Full size 5 2689 537.8 23905.2  

     

     

ANOVA      

Source of Variation SS df MS            F             P-value F crit

Between Groups 94825 3 31608.33 1.321    0.302           3.24

Within Groups 382843.2 16 23927.7    

     

Total 477668.2 19    

Since p value of 0.302 is greater than 0.05 significance level we accept null hypothesis.

p value = 0.302

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