Question

SSTR = 6,750

SSE = 8,000

n_{T} = 20

H_{0}: μ_{1} = μ_{2} = μ_{3} =
μ_{4}

H_{a}: 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

Answer #1

H_{0}: μ_{1} = μ_{2} = μ_{3} =
μ_{4}

H_{a}: Not all population means are equal

SSTR = 6,750

SSE = 8,000

n_{T} = 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

SSTR = 6,750
H0:
μ1=μ2=μ3=μ4
SSE = 8,000
Ha: at least one mean is different
nT = 20
Refer to Exhibit 10-11. The null hypothesis
Question 15 options:
should be rejected
should not be rejected
was designed incorrectly
None of these alternatives is correct.

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Ha: at least one mean is different
At Alpha=.05 the null hypothesis:
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