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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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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