SOLUTION: A market gardener is preparing a report for her seed supplier. She wants to include information about the mean weight of pumpkins she grows. A 95% confidence interval for the mea

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Question 835614: A market gardener is preparing a report for her seed supplier. She wants to include
information about the mean weight of pumpkins she grows. A 95% confidence interval for
the mean weight when 40 pumpkins are sampled gives 4.71kg   5.93kg .
i) What is the mean weight of the 40 pumpkins?
ii) What is the margin of error in this confidence interval?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A 95% confidence interval for the mean weight when 40 pumpkins are sampled gives 4.71kg 5.93kg .
i) What is the mean weight of the 40 pumpkins?
mean = (5.93+4.71)/2 = 5.32
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ii) What is the margin of error in this confidence interval?
ME = (5.93-4.71)/2 = 0.61
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Cheers,
Stan H.
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