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

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      

GroupsCountSumAverageVariance  

Subcompact53444688.848502.7  

compact52879575.84582.7  

Midsize52534506.818720.2  

Full size52689537.823905.2  

ANOVA      

Source of VariationSSdfMS            F             P-valueF crit

Between Groups94825331608.331.321    0.302           3.24

Within Groups382843.21623927.7    

Total477668.219    

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

p value = 0.302

c

step-by-step explanation:

where is abdul's work? don't see that anywhere posted : /

step-by-step explanation: