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The police department in your city was asked by the mayor’s office to estimate the cost...

The police department in your city was asked by the mayor’s office to estimate the cost of crime. The police began their study with burglary records, taking a random sample of 500 files since there were too many crime records to calculate statistics for all of the crimes committed. If the average dollar loss in a burglary, for this sample size 500 is $678, with a standard deviation of $560, construct the 95% confidence interval for the true average dollar loss in burglaries rounded to the nearest dollar amount. We can be 95% confident that the upper limit of our interval of dollar loss in crimes committed is $________.

Math   Statistics-And-Probability   
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Answer #1

n = 500     

x-bar = 678     

s = 560     

% = 95     

Standard Error, SE = σ/√n =    560 /√500 = 25.04396135

z- score = 1.959963985     

Width of the confidence interval = z * SE =     1.95996398454005 * 25.0439613479976 = 49.08526227

Lower Limit of the confidence interval = x-bar - width =      678 - 49.0852622722886 = 628.9147377

Upper Limit of the confidence interval = x-bar + width =      678 + 49.0852622722886 = 727.0852623

The 95% confidence interval is [$629, $727]

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