Question

An analysis is conducted to compare mean time to pain relief (measured in minutes) under four competing treatment regimens. Summary statistics on the four treatments are shown below. The ANOVA table presented below is not completed.

Treatment	Sample Size	Mean Time to Relief	Sample Variance
A	5	33.8	17.7
B	5	27.0	15.5
C	5	50.8	9.7
D	5	39.6	16.8

Source of Variation	SS	df	MS	F
Between Groups			508.13
Within Groups	3719.48
Total

           

a. What is the within group (error) degrees of freedom value (df₂)?

b. Based on the data and ANOVA table provided in Q44, compute the MSE. (round to 2 decimal places)

c.  Based on the data and ANOVA table provided in Q44, compute the F test statistic. (round to 2 decimal places)

d.  What is the critical value for the hypothesis test you performed in Q44-Q46? (2 decimal places)

           

Answer 1

Question 1

within group (error) degrees of freedom value = (n₁ -1) + (n₂ -1) + (n₃ -1) + (n₄ -1) = (5 -1) + (5 - 1) + (5 - 1) + (5 - 1) = 16

Question 2  Based on the data and ANOVA table provided in Q44, compute the MSE.

MSE = SS (Withing groups)/ dF(error) = 3719.48/16 = 232.4675

Question 3 Here first we will complete the table

dF(treatments) = Number of treatments - 1 = 4 - 1= 3

dF (Between Groups) = MS (Between Groups) * dF(treatments) = 3 * 508.13 = 1524.4

MS(Withing groups) = SS(Within groups)/dF(errror) = 3719.5/16 = 232.468

Source of Variation	SS	df	MS	F
Between Groups	1524.4	3	508.13	2.1858109
Within Groups	3719.5	16	232.468
Total	5243.9	19

Based on the data and ANOVA table provided in Q44, compute the F test statistic.

F = 508.13/232.468 = 2.186

d.Here dF1 = 3, dF2 = 16 and alpha = 0.05

F(critical) = FINV(3,16, 0.05) = 3.239

Here as F < F(critical) so we failed to reject the null hypothesis.