Question

Samples of 100 8-hour shifts were randomly selected from the police records for each of two districts in a large city. The number of police emergency calls was recorded for each shift. The sample statistics are listed below.

		Region
		1	2
Sample Size		100	100
Sample Mean		2.6	3.4
Sample Variance		1.24	2.54

Find a 90% confidence interval for the difference in the mean numbers of police emergency calls per shift between the two districts of the city. (Use μ₁ − μ₂. Round your answers to two decimal places.)

________ calls per shift to ________ calls per shift

Answer 1

since sample size is large ; we can use standard normal distribution (z distribution)

		region 1	region 2
sample mean x =		2.60	3.40
std deviation σ=		1.114	1.594
sample size n=		100	100
std error σx1-x2=√(σ21/n1+σ22/n2)    =			0.194

Point estimate of difference '=x1-x2 =			-0.800
for 90 % CI value of z=			1.645	from excel:normsinv((1+0.9)/2)
margin of error E=z*std error =			0.320
lower bound=(x1-x2)-E =			-1.120
Upper bound=(x1-x2)+E =			-0.480
from above 90% confidence interval for population mean =(-1.12,-0.48)

-1.12 calls per shift to -0.48 calls per shift