A simple random sample of 81 is selected from a population with a standard deviation of...

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A simple random sample of 81 is selected from a population witha standard deviation of 17. The degree of confidence is 90%. Whatis the margin of error for the mean?

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Z score for 90% confidence interval = Z0.05 = 1.645

Margin of error = Z0.05 * sd / sqrt(n) = 1.645 * 17 / sqrt(81) = 3.107

                                                                                                                                                                                                                


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