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
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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