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 2.2 gallons. The mean water usage per family was found to be 15.8 gallons per day for a sample of 669 families. Construct the 90% confidence interval for the mean usage of water. Round your answers to one decimal place.

Answer 1

The 90% confidence interval is
`15.66< \mu < 15.94`

Explanation:

From the question we are told that

The population standard deviation is
`\sigma = 2.2`

The sample mean is
`\= x = 15.8 \ gallon`

The sample size is n = 669

From the question we are told the confidence level is 95% , hence the level of significance is

`\alpha = (100 - 90 ) \%`

=>
`\alpha = 0.10`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.645`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } *  (\sigma )/(√(n) )`

=>
`E = 1.645  *  ( 2.2)/(√(669) )`

=>
`E = 0.1399`

Generally 90% confidence interval is mathematically represented as

`\= x -E <  \mu <  \=x  +E`

=>
`15.8  - 0.1399 <\mu<  15.8  + 0.1399`

=>
`15.66< \mu < 15.94`