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 $________.

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raphealnwobi raphealnwobi · Answer 1 · 2022-04-01T15:09:40+02:00

The 95% confident upper limit of dollar loss in crimes committed is $628.90

Margin of error (E) is used to show the deviation from the population.

sample = 500, mean = $678, standard deviation of $560, z score of 95% confidence = 1.96, hence:

[tex]E=z_\frac{\alpha }{2} *\frac{standard\ deviation}{\sqrt{sample\ size} } \\\\E=1.96*\frac{560}{\sqrt{500} } =49.1[/tex]

Confidence interval = mean ± E = 678 ± 49.1 = (628.9, 727.1)

The 95% confident upper limit of dollar loss in crimes committed is $628.90

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