A has no real solution. See the picture below for details.
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
Well id have to know how many packs of crackers there was and possibly i might need to know the # of crackers in a pack
