Question

F р Questions: Use the source table shown below to answer questions 1 to 4. Source SS df MS Subjects 323 19 Treatment 72 4 18 Error 228 76 3 Total 623 99 6 <0.05 1. Identify which type of ANOVA (Repeated-Measures or Factorial) was used to generate this source table, and explain how you determined your response. [1 mark] 2. Would the results of this test significant at a=0.05? How do you know this? [1 mark] 3. Would the results of this test significant at a=0.01? How do you know this? [1 mark] 4. [Let's play pretend!] Oops! The professor made a mistake! He was supposed to run a between- subjects ANOVA instead of the one conducted above. How could you correct this mistake using the data given in the source table? [1 mark]

P.S. Please do not post that "hand-written" response that contains many errors. Thanks for answering this for me!

Answer 1

Solution

Part (1)

Repeated-Measures ANOVA Answer 1

Explanation

If it were a factorial experiment, there must have been at least two treatments. Further, F is calculated only for ‘Treatment’ so ‘subjects’ is treated as repeat measurement.

Part (2)

Yes Answer 2

Explanation

Given p-value (p) is less than the given significance level of 0.05

Part (3)

Yes. Answer 3

Explanation

Under the null hypothesis of no treatment effect, F ~ F_4,76 and so given α = 0.01, the critical value is the upper 1% point of F_4,76 which is 3.57 [using Excel Function: Statistical FINV]

Since F value > critical value, null hypothesis stands rejected implying that the results are significant.

Note:

Depending solely on the given information only, no definite conclusion can be made, because p < 0.05 does not necessarily imply it is less than 0.01 – it could as well be lying between 0.01 and 0.05. However using Excel Function: Statistical FDIST,

p-value [i.e., P(F_4,76 > 6)] can be found to be 0.0003. This being less than 0.01, null hypothesis stands rejected]    

Part (4)

Here, ‘subjects’ would become treatment. This would lead to:

MS = 323/19 = 17; F = 17/3 = 5.67; p-value = P(F_19,76 > 5.67) < 0.01;

Hence, the null hypothesis stands rejected at α = 0.01 and 0.05 implying the results are significant. Answer 4

DONE