Question

Supposed a sample of 1453 tankers is drawn. Of these ships 989 did not have spills. Using the data construct the 99% confidence interval for the population proportion of all tankers that have spills each month. Ranch your answers to three decimal places

Answer 1

Sample: 1453

989= Without spills

99% Confidenece interval is given by:

`ConfidenceInterval=Z_c*\sqrt[\placeholder{⬚}]{(p(1-p))/(n)}`

For 99% confidence, the Z_c is:

`Z_c=2.576`

And p is given by:

`\begin{gathered} p=(989)/(1453)=0.68 \\ 1-p=0.319 \end{gathered}`

Substituing:

`ConfidenceInterval=2.576*\sqrt[\placeholder{⬚}]{(0.68*(0.319))/(1453)}=0.0315`

Finally, the way to find the intervals is given by:

`p\pm ConfidenceInterval`

ANSWER:

Upper endpoint:

`0.68+0.0315=0.7115\approx0.712`

Lower endpoint:

`0.68-0.0315=0.6485\approx0.648`