Question

A 90% confidence interval for a population mean is given as (17.68, 22.02). This confidence interval is based on a simple random sample of 36 observations. Calculate the sample mean and standard deviation. Use the t-distribution in any calculations.

Answer 1

The correct solution is "7.704". The further explanation is given below.

Explanation:

According to the question,

`t_(\alpha/2),df = 1.690`

Sample mean,

`\bar{x} = ((17.68 + 22.02))/(2)`

`=19.85`

The Margin of error (E),

=
`Upper \ confidence \ interval - \bar{x}`

=
`22.02 - 19.85`

=
`2.17`

Now,

⇒
`s = ( E* \sqrt n )/(t_(\alpha/2),df)`

`=( 2.17* \sqrt 36 )/( 1.690)`

`= 7.704`