Question

A researcher examines 39 seawater samples for lead concentration. The mean lead concentration for the sample data is 0.771 cc/cubic meter with a standard deviation of 0.0589. Determine the 90% confidence interval for the population mean lead concentration. Assume the population is normally distributed. 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

Given that,

=  0.771

s =0.0589

n = 39

Degrees of freedom = df = n - 1 = 39- 1 = 38

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,38 =1.686 ( using student t table)

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

= 1.686* ( 0.0589/ 39) = 0.0159

The 90% confidence interval estimate of the population mean is,

- E < < + E

0.771 - 0.0159< < 0.771+ 0.0159

0.7551 < < 0.7869

( 0.7551 , 0.7869)