SOLUTION: Suppose the average watermelon contains 144.0 brown seeds. Farmer Brown just picked 25 ripe melons from the south side of his big garden and found that the average was 141.0 seeds

Question 138368: Suppose the average watermelon contains 144.0 brown seeds. Farmer Brown just picked 25 ripe melons from the south side of his big garden and found that the average was 141.0 seeds and the standard deviation was 2.4 seeds. The same test statistic was obtained from another sample of 36 melons picked from the north side of the garden, but this time the standard deviation was 3.6 seeds. What was the average number of seeds in each watermelon picked from the north side?

a. 140.25
b. 140.40
c. 140.54
d. 141.75

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Suppose the average watermelon contains 144.0 brown seeds. Farmer Brown just picked 25 ripe melons from the south side of his big garden and found that the average was 141.0 seeds and the standard deviation was 2.4 seeds. The same test statistic was obtained from another sample of 36 melons picked from the north side of the garden, but this time the standard deviation was 3.6 seeds. What was the average number of seeds in each watermelon picked from the north side?
a. 140.25
b. 140.40
c. 140.54
d. 141.75
--------------------
Test statistic for the south side:
t(141) = (141-144)/[2.4/sqrt(25)] = -6.25
----------------------------------------
Test statistic for the north side:
Let "x" be the average number of seeds in the 36 watermelons on north side.
t(x) = (x-144)/[3.6/sqrt(36)] = -6.25
x-144 = (3.6/6][-6.25]
x-144 = 0.6*-6.25
x = 144-3.75
x = 140.25
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Cheers,
Stan H.

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