Question

You measure 39 backpacks weights, and find they have a mean weight of 33 ounces. Assume the population standard deviation is 6.6 ounces. Based on this, what is the maximal margin of error associated with a 95% confidence interval for the true population mean backpack weight. Give your answer as a decimal, to two places 30.93 35.07 x ounces
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Answer #1

as we know that margin of error =z*std deviation/sqrt(n)

for 95% CI ; critical value of z =1.96

therefore margin of error =1.96*6.6/sqrt(39)=2.07

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