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n = n1 = n2 = 5
d-bar = 1.68
s (of d) = 4.61
Ho: d-bar ≥ 0
Ha: d-bar < 0
α = 0.05
Degrees of freedom = 5 - 1 = 4
Critical t- score = -2.131846782
Reject Ho if t < -2.131846782
SE = s/√n = 4.61/√5 = 2.061654675
t = d-bar/SE = 1.68/2.06165467525481 = 0.814879436
p- value = 0.230438158
Decision (in terms of the hypotheses):
Since 0.814879436 > -2.132 we fail to reject Ho
Conclusion (in terms of the problem):
There is no sufficient evidence that variety A had a higher yield than variety B.
(a) Ho: mu1 > mu2 and Ha: mu1 < mu2
(b) t = 0.815
webwork / math243fall-masden /week 8b - ch20 inference about a population mean / 3 Week 8b...
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