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SSTR = 6,750 H0: μ1=μ2=μ3=μ4 SSE = 8,000 Ha: at least one mean is different nT...

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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Homework Answers

Answer #1

Given that

SSTR = 6,750 H0: μ1=μ2=μ3=μ4
SSE = 8,000 Ha: at least one mean is different
nT = 20

F statistic = [SSTR/df(R)]/[SSE/df(E)]

setting SSTR = 6750, df(R) = number of categories - 1 = 4-1 = 3, SSE = 8000 and df(E) = nT - number of categories = 20-4 = 16

setting the values, we get

F statistic = [6750/3]/[8000/16]

= 2250/500

= 4.5

p value = F.DIST.RT(x,df(R),df(E))

= F.DIST.RT(4.5,3,16)

= 0.018

So, p value is less than 0.05 significance level

Therefore, null hypothesis should be rejected

option A

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