Question

The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day. Assume that the population standard deviation is 2.4 gallons. The mean water usage per family was found to be 19.5 gallons per day for a sample of 3034 families. Construct the 98% confidence interval for the mean usage of water. Round your answers to one decimal place.

Answer 1

Solution :

Given that,

= 19.5

= 2.4

n = 3034

At 98% confidence level the z is ,

  = 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

Margin of error = E = Z_/2* (/n)

= 2.326 * (2.4/ 3034 )

= 0.1

At 98% confidence interval estimate of the population mean is,

- E < < + E

19.5 - 0.1 < < 19.5 + 0.1

19.4 < < 19.6

(19.4, 19.6)