P value
State your conclusion. Would we perform post-hoc procedures for this data?
The summary statistics obtained from the given data are as below.
Large Car | Passenger | SUV | |
Total | 2277 | 2754 | 2246 |
n | 7 | 7 | 7 |
Mean | 325.29 | 393.43 | 320.86 |
Sum Of Squares | 233903.429 | 212849.7143 | 99076.8572 |
Variance | 38983.9048 | 35474.9524 | 16512.8095 |
SD | 197.4434 | 188.3480 | 128.5022 |
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(a) The Hypothesis: Option B
H0:
H1: At least one mean is different.
(b) Option C: yes the requirements are met.
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The ANOVA table is as below
Source | SS | DF | Mean Square | F | p |
Between | 23169.49 | 2 | 11584.74 | 0.382 | 0.6879 |
Within/Error | 545830.00 | 18 | 30323.89 | ||
Total | 568999.49 | 20 |
The p Value: The p value is calculated for F = 0.38 for df1 = 2 and df2 = 18 and is = 0.6879 (P value > 0.10)
The Decision Rule:
If p-value is < , Then reject H0.
The Decision:
Also since p-value (0.6879) is > (0.01), We Fail to Reject H0.
The Conclusion: There isn't sufficient evidence at the 99% level of significance to warrant rejection of the claim that the head injury for each vehicle type is the same.
No, we would not conduct post hoc procedures for this data.
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Calculations For the ANOVA Table:
Overall Mean = [325.29 + 393.43 + 320.86] / 3 = 346.52
SS treatment = SUM [n* ( - overall mean)2] = 7 * (325.29 - 346.52)2 + 7 * (393.43 - 346.52)2 + 7 * (320.86 - 346.52)2 = 23169.49
df1 = k - 1 = 3 - 1 = 2
MSTR = SS treatment/df1 = 23169.49 / 2 = 11584.74
SSerror = SUM (Sum of Squares) = 233903.429 + 212849.7143 + 99076.8572 = 545830
df2 = N - k = 21 - 3 = 18
Therefore MS error = SSerror/df2 = 545830 / 18 = 30323.89
F = MSTR/MSE = 11584.74 / 30323.89 = 0.382
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Mcars = Hvans = MSUV
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