SSTR= 6750 SSE = 8000 nt = 20Ho: u1=u2=u3=u4 Ha: at least one mean is differnt...
SSTR= 6750
SSE = 8000
nt = 20Ho: u1=u2=u3=u4
Ha: at least one mean is differnt
At alpha=.05 the null hypothesis
A. was designed incorrectly
B. None of these
C. Should be rejected
D. Should not be rejected
Homework Answers
Solution:
Number of treatments = k = 4
d.f.TR = k - 1 = 3
nt = 20
d.f.E = (nt) - k = 20 - 4 = 16
| d.f. | SS | MSS | F | ||
| Treatment | 3 | 6750 | 2250 | 4.5 | |
| Error | 16 | 8000 | 500 | ||
| Total | 79 |
Using F table , at alpha=.05 , critical value is
F0.05,3,16 = 3.239
4.5 > 3.239
So ,
Reject the null hypothesis
Answer is
C. Should be rejected
Add Answer to:
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