Question

(2 pts) 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 2.3 gallons. The mean water usage per family was found to be 18.5 gallons per day for a sample of 717 families. Construct the 80% confidence interval for the mean usage of water. Round your answers to one decimal place. Answer: Lower endpoint: Upper endpoint:

Answer 1

we have given :

sample size = n = 717

xbar = sample mean = 18.5 gallon

assume known population standard deviation = 2.3 gallon

## construct 80 % confidence interval for the mean usage of water . ( μ )

here we have sample size is large ( n > 30 ) and known population standard deviation hence we can use

z test to find out confidence interval :

  μ =   ( xbar ± Z critical value * σ / sqrt(n) )

Z critical value = Z ( α /2 ) = ± 1.28 ( from table)

  μ =   ( 18.5 ± 1.28 * (2.3  / sqrt(717) )

  μ = ( 18.5  ± 0.1099)

μ = ( 18 . 3900 , 18. 6099)

lower endpoint = 18.3 gallons

and upper endpoint = 18.6 gallons

we can say 80% confident that population mean for the mean usage of water is lies within 18.3 gallons and

18.6 gallons .