Question

A researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic meter with a standard deviation of 0.093. Determine the 90% confidence interval for the population mean iron concentration. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

Explanation:

We are given that a researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic meter with a standard deviation of 0.093.

Firstly, the pivotal quantity for 90% confidence interval for the population mean iron concentration is given by;

P.Q. =
`(\bar X - \mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample mean iron concentration = 0.802 cc/cubic meter

s = sample standard deviation = 0.093

n = number of water samples = 27

`\mu` = population mean

Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean,
`\mu` is ;

P(-1.706 <
`t_2_6` < 1.706) = 0.90 {As the critical value of t at 26 degree of

freedom are -1.706 & 1.706 with P = 5%}

P(-1.706 <
`(\bar X - \mu)/((s)/(√(n) ) )` < 1.706) = 0.90

P(
`-1.706 * {(s)/(√(n) ) }` <
`{\bar X - \mu}` <
`1.706 * {(s)/(√(n) ) }` ) = 0.90

P(
`\bar X -1.706 * {(s)/(√(n) )` <
`\mu` <
`\bar X +1.706 * {(s)/(√(n) )` ) = 0.90

90% confidence interval for
`\mu` = [
`\bar X -1.706 * {(s)/(√(n) )` ,
`\bar X +1.706 * {(s)/(√(n) )` ]

= [
`0.802 -1.706 * {(0.093)/(√(27) )` ,
`0.802 +1.706 * {(0.093)/(√(27) )` ]

= [0.771 , 0.832]

Therefore, 90% confidence interval for the population mean iron concentration is [0.771 , 0.832].