Question

# SSTR = 6,750 SSE = 8,000 nT = 20 H0: μ1 = μ2 = μ3 =...

SSTR = 6,750

SSE = 8,000

nT = 20

H0: μ1 = μ2 = μ3 = μ4

Ha: Not all population means are equal

a. The mean square between treatments (MSTR) equals

b. The mean square within treatments (MSE) equals

c. The test statistic to test the null hypothesis equals

d. The critical F value at alpha = 0.05 is

e. The null hypothesis is

H0: μ1 = μ2 = μ3 = μ4

Ha: Not all population means are equal

SSTR = 6,750

SSE = 8,000

nT = 20

df_TR = 4 - 1 = 3

.a. The mean square between treatments (MSTR) equals

MSTR = SSTR/df_TR

MSTR = 6750/3

MSTR = 2250

b. The mean square within treatments (MSE) equals

MSE = SSE/df_E

df_E = df_total - df_TR

df_total = nT - 1 = 20 - 1 = 19

df_E = 19 - 3 = 16

MSE = 8000/16

MSE = 500

c. The test statistic to test the null hypothesis equals

F = MSTR/MSE

F = 2250/500

F = 4.5

d. The critical F value at alpha = 0.05 is

critical F = 3.2389 at alpha = 0.05 and df = (3,16)

e. The null hypothesis is

As F = 4.5 > F_c = 3.2389

Hence, reject the null hypothesis

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