In: Statistics and Probability
The head injury data are shown in the following table.
Subcompact: 681 428 917 898 420
Compact: 643 655 442 514 525
Midsize: 469 727 525 454 259
Full-size: 384 656 602 687 360
Table 5: Head Injury & Car Crash
At α = 0.05 level of significance, test the claim that different weight categories have the same mean by using the data in table 5.
(a) Clearly state H0 and H1.
H0 :
H1 :
(b) Find all related critical values, draw the distribution, clearly mark and shade the critical region(s). Give the name of the program you used for this step as well as any degrees of freedom. Drawing & Shading Required.
(c) Find the computed test statistic, and the P-value.
C.T.S. :
P-Value :
(d) Use non-statistical terminology to state your final conclusion.
Subcompact | Compact | Midsize | Fullsize |
681 | 643 | 469 | 384 |
428 | 655 | 727 | 656 |
917 | 442 | 525 | 602 |
898 | 514 | 454 | 687 |
420 | 525 | 259 | 360 |
(a)
(a) H0: There is no difference between the treatment means; that is, μ1 = μ2 = μ3 = μ4 | ||||||
H1: At least one among μ1, μ2, μ3 and μ4 is different from the other three |
(b) Df (Treatment) = 3, Df (Error) = 16, α = 0.05, Critical F score = 3.2389
(c) ANOVA
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Subcompact | 5 | 3344 | 668.8 | 58542.7 | ||
Compact | 5 | 2779 | 555.8 | 8272.7 | ||
Midsize | 5 | 2434 | 486.8 | 28110.2 | ||
Fullsize | 5 | 2689 | 537.8 | 23905.2 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 88425 | 3 | 29475 | 0.992167 | 0.42157 | 3.238872 |
Within Groups | 475323.2 | 16 | 29707.7 | |||
Total | 563748.2 | 19 |
Test F- score = 0.9922, p- value = 0.4216
Since the p- value > 0.05, we fail to reject Ho
There is no sufficient evidence that the four groups have significantly different means.
[I have used Excel for the ANOVA table and STATDISK to get the plot]