Question

An environmentalist wants to find out the fraction of oil tankers that have spills each month. Suppose a sample of 836 tankers is drawn. Of these ships, 577 did not have spills. Using the data, construct the 99% confidence interval for the population proportion of oil tankers that have spills each month.

Answer 1

The 99% confidence interval for the population proportion of oil tankers that have spills each month.

(0.6489 , 0.7311)

Explanation:

Step(i):-

Given sample size 'n' = 836

Suppose a sample of 836 tankers is drawn. Of these ships, 577 did not have spills.

sample proportion

`p^(-) = (x)/(n) = (577)/(836) = 0.690`

The 99% confidence interval for the population proportion is determined by

`(p^(-) - Z_(0.01) \sqrt{(p(1-p))/(n) } , p^(-) + Z_(0.01) \sqrt{(p(1-p))/(n) })`

Step(ii):-

The Z- value = 2.576

`(0.690 - 2.576 \sqrt{(0.690(1-0.690))/(836) } , 0.690 + 2.576 \sqrt{(0.690(1-0.690))/(836) })`

On calculation , we get

(0.690 - 0.0411 , 0.690 + 0.0411)

(0.6489 , 0.7311)

Final answer:-

The 99% confidence interval for the population proportion of oil tankers that have spills each month.

(0.6489 , 0.7311)