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
Get Answers For Free
Most questions answered within 1 hours.