1.) A randomly selected sample of 70 casino patrons has an average loss of $300 with a sample standard deviation of $100. Use this information to complete the following:
a) Does the information above provide an estimate of a population mean or a population proportion?
b) Provide a point estimate for the population mean loss of all casino patrons.
c) Construct and interpret a 90% confidence interval for the parameter you selected in part (a).
d) What is the margin of error for this confidence interval estimate?
population mean (since it is the estimate of average loss suffered by all casino patrons )
point estimate =sample mean =300
|sample mean 'x̄=||300.000|
|sample size n=||70|
|std deviation s=||100.000|
|std error ='sx=s/√n=100/√70=||11.9523|
|for 90% CI; and 69 df, value of t=||1.667||from excel: t.inv(0.95,69)|
|margin of error E=t*std error =||19.93|
|lower bound=sample mean-E =||280.073|
|Upper bound=sample mean+E =||319.927|
|from above 90% confidence interval for population mean =(280.07<μ<319.93)|
|above interval gives 90% confidence to contain true value of population mean|
margin of error for this confidence interval estimate =19.93
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