Question

In: Math

An analysis is conducted to compare mean time to pain relief (measured in minutes) under four...

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)

           

Solutions

Expert Solution

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.


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