Question

An environmentalist wants to find out the fraction of oil tankers that have spills each month.

Step 2 of 2:

Suppose a sample of 669 tankers is drawn. Of these ships, 173had 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,

n = 669

x = 173

Point estimate = sample proportion = = x / n = 173/669=0.259

1 -   = 1- 0.259 =0.741

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

= 2.326 (((0.259*0.741) /669 )

E = 0.039

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.259-0.039 < p < 0.259+0.039

0.220< p < 0.298