Answer:
The 99% confidence interval for the population proportion of oil tankers that have spills each month.
(0.6489 , 0.7311)
Step-by-step 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
[tex]p^{-} = \frac{x}{n} = \frac{577}{836} = 0.690[/tex]
The 99% confidence interval for the population proportion is determined by
[tex](p^{-} - Z_{0.01} \sqrt{\frac{p(1-p)}{n} } , p^{-} + Z_{0.01} \sqrt{\frac{p(1-p)}{n} })[/tex]
Step(ii):-
The Z- value = 2.576
[tex](0.690 - 2.576 \sqrt{\frac{0.690(1-0.690)}{836} } , 0.690 + 2.576 \sqrt{\frac{0.690(1-0.690)}{836} })[/tex]
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)
