Question

A student examines 88 seawater samples for nitrate concentration. The mean nitrate concentration for the sample data is 0.424 cc/cubic meter with a standard deviation of 0.0827. Determine the 90% confidence interval for the population mean nitrate 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

Solution :

Given that,

Point estimate = sample mean = = 0.424

Population standard deviation =    = 0.0827
Sample size = n =88

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2    * ( /n)

=1.645 * (0.0827 /  88 )

= 0.0145
At 90% confidence interval estimate of the population mean
is,

- E < < + E

0.424 - 0.0145 <   <0.424 + 0.0145

0.4095 <   < 0.4385

( 0.4095 , 0.4385 )