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What is the margin of error for a 95% confidence interval of a simple random sample of...

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surfer3 | Valedictorian

Posted June 1, 2012 at 1:34 AM via web

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What is the margin of error for a 95% confidence interval of a simple random sample of applicants for credit cards with n=50 and σ is known to be 68

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 2, 2012 at 2:54 PM (Answer #1)

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The margin of error is given by: ME = critical value*standard error. Standard error is `sigma/sqrt n` .

Here, sigma is known to be 68. The number of samples is 50. This gives standard error as `68/sqrt 50 = 9.6166` . For the 95% confidence interval the critical value is 1.96. The margin of error is 1.96*9.6166 = 18.84

The required margin of error is 18.84

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