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You measure 41 watermelons' weights, and find they have a mean weight of 61 ounces. Assume...

You measure 41 watermelons' weights, and find they have a mean weight of 61 ounces. Assume the population standard deviation is 11.7 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight.

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Answer #1

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This is given by:

Mean + / - Z*stdev/sqrt(n)

Z for 90% Confidence interval is 1.28. So:

= 61 +/- 1.28*11.7/sqrt(41)

= 58.6611 to 68.3389

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