Question

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?

Answer 1

a)

population mean (since it is the estimate of average loss suffered by all casino patrons )

b)

point estimate =sample mean =300

c)

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

d)

margin of error for this confidence interval estimate =19.93