## Description

1132 tankers is drawn 861 did not have spills p=0.239. Construct the 98% confidence interval for the population proportion of oil tankers that have spills each month lower bound and upper bound round answers to 3 decimal places

## Explanation & Answer

To find the 98% confidence interval we first find Z_crit, which is given by:

N.inv ( p = 0.99 , mean = 0, sd = 1 ) = 2.326

We then find the standard error, which so long as your population is large can be estimated by:

SE = sqrt[ p*( 1 - p ) / n ] where p = 0.239 and n = 1132:

SE = sqrt(0.239*(0.761)/1132) = 0.0127

Multiply SE by Z_crit:

0.0127 * 2.326 = 0.0295

So the interval is given by 0.239-0.0295 , 0.239+0.0295 = (0.210, 0.269) (to 3 decimal places)

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