Question

Given the following information obtained from four normally distributed populations, construct an ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "SS" to 2 decimal places, "MS" to 4 decimal places, and "F" to 3 decimal places.)

SST = 65.21; SSTR = 15.95; c = 4; n₁ = n₂ = n₃ = n₄ = 15

b. At the 1% significance level, what is the conclusion to the ANOVA test of mean differences?

• Reject H₀; we can conclude that some means differ.

• Do not reject H₀; we cannot conclude that some means differ.

• Do not reject H₀; we can conclude that some means differ.

• Reject H₀; we cannot conclude that some means differ.

Answer 1

SST = 65.21

SSTR = 15.95

c = 4

n₁ = n₂ = n₃ = n₄ = 15

df_total = 15*4 - 1 = 60 - 1 = 59

df_TR = c - 1 = 4 - 1 = 3

df_error = df_total - df_TR

df_error = 59 - 3 = 56

SSE = SST - SSTR

SSE = 65.21 - 15.95

SSE = 49.26

MSE = SSE/df_error = 49.26/55 = 0.8956

MSTR = SSTR/ df_TR = 15.95/3 = 5.3168

F = MSTR/MSE = 5.3168/0.8956

F = 5.94

The corresponding critical value at alpha = 1% = 0.01 at df (3, 56 )

F_c = 4.152

As we reject Ho if F > F_c

Hence here 5.94 > 4.152

null hypothesis is rejecetd.

Mean are not equal

Reject H₀; we can conclude that some means differ.