Question

A biologist examines 29 seawater samples for iron concentration. The mean iron concentration for the sample data is 0.334 cc/cubic meter with a standard deviation of 0.0139. Determine the 99% 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

solution

Given that,

= 0.334

s =0.0139

n = 29

Degrees of freedom = df = n - 1 =29 - 1 = 28

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

critical value  = t /2  df = t0.005,28 =   2.763    ( using student t table)

Margin of error = E = t/2,df * (s /n)

=  2.763 * (0.0139 / 29) = 0.007

The 99% confidence interval of the population mean is,

- E < < + E

0.334 -0.007 < <0.334 + 0.007

0.327 < <0.341

( 0.327 , 0.341)