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The national average cost of cable TV bills was reported as $99. You collect a random sample of 75 people in your town a...

The national average cost of cable TV bills was reported as $99. You collect a random sample of 75 people in your town and find the average cable bill is $103 with a standard deviation of $13.50. You want to compute a 99% confidence interval estimate for the mean difference between the national average and your town average. What is the margin of error?

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ca 99% -0.99 ,on - 75, 5 = $13.50 a=1-0.99 = 0.01 dfs 75-1=74 cl2 = 0.005 toleraf at 0.006,74 = 2.644 Margin of exta E = tan

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