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 )

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