Question

In a simple random sample of 20 residents of the city of Dallas, the mean paper products recycled per person per day was 0.95 pounds with a standard deviation of 0.32 pounds. Determine the 99% confidence interval for the mean paper products recycled per person per day for the population of Dallas, assuming the population is approximately normal

Answer 1

The 99% confidence interval for the mean paper products recycled per person per day for the population of Dallas is

`0.7454 <  \mu <  1.1546`

Explanation:

From the question we are told that

The sample size is n = 20

The sample mean is
`\= x = 0.95 \ pounds`

The standard deviation is
`\sigma = 0.32 \ pounds`

Generally given that the sample size is small , n< 30 we will be making use of t distribution table

Generally the degree of freedom is mathematically represented as

`df = 20 - 1`

=>
`df = 19`

From the question we are told the confidence level is 99% , hence the level of significance is

`\alpha = (100 - 99 ) \%`

=>
`\alpha = 0.01`

Generally from the t distribution table the critical value of
`(\alpha )/(2)` at a degree of freedom of
`df = 19` is

`t_{(\alpha )/(2), 19 } =  2.86`

Generally the margin of error is mathematically represented as

`E =t_{(\alpha )/(2), 19 } *  (\sigma )/(√(n) )`

=>
`E = 2.86 *  (0.32)/(√(20) )`

=>
`E = 0.2046`

Generally 99% confidence interval is mathematically represented as

`\= x -E <  \mu <  \=x  +E`

=>
`0.95  -0.2046 <  \mu <  0.95  + 0.2046`

=>
`0.7454 <  \mu <  1.1546`