Question

Suppose a sample of 1069 tankers is drawn. Of these ships, 856 did not have spills. Using the data, construct the 98% confidence interval for the population proportion of oil tankers that have spills each month. Round your answers to three decimal places.

Answer 1

Solution :

Given that,

856 did not have spills .

So, 1069 - 856 = 213 have spills .

n = 1069

x = 213

= x / n = 213 / 1069 = 0.199

1 - = 1 - 0.199 = 0.801

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.326 * (((0.199 * 0.801) / 1069)

= 0.028

A 98% confidence interval for population proportion p is ,

- E < P < + E

0.199 - 0.028 < p < 0.199 + 0.028

0.171 < p < 0.228

(0.171 , 0.228)