Question

A consultant wants to know how American cities are providing facilities for the less wealthy. He takes a random sample of 37 metropolitan areas and finds they have an average of 19.3 public pools per city with a standard error of 1.9. Help him calculate a 88% confidence interval for the average number of public pools per American metropolitan area

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 19.3

standard error=SE=1.9

At 88% confidence level the z is ,

Z/2 = Z0.06 = 1.55 ( Using z table )

Margin of error = E =   Z/2    * SE
= 1.55* 1.9

E= 2.9450
At 88% confidence interval estimate of the population mean
is,

- E < < + E

19.3 - 2.9450 <   < 19.3 + 2.9450

16.3550 <   <22.2450
( 16.3550 ,22.2450 )