Question

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 $________.

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]