Survey in regards of speeding tickets with teenage boys.... How do you solve?

It was found that in a sample of 80 teenage boys, 70% had gotten speeding tickets. What is the 99% confidence interval of the true proportion, p, of teenage boys who have gotten speeding tickets?





possible solutions....








a. 0.544 %26lt; p %26lt; 0.856





b. 0.616 %26lt; p %26lt; 0.784





c. 0.600 %26lt; p %26lt; 0.800





d. 0.581 %26lt; p %26lt; 0.819





e. 0.568 %26lt; p %26lt; 0.832





f. 0.556 %26lt; p %26lt; 0.844





g. none of these

Survey in regards of speeding tickets with teenage boys.... How do you solve?
Since we do not know the Standard Deviation, we have to use the following formula:





n= 80, p=.70





s = sqrt ( [p * (1-p) ] / n)





(sqrt is the square root)





so we get





p +/- Z * s





or for this data:





s = sqrt ((.70 * .30)/80) = 0.051





z= 2.575 for 99% see Z table for .4950





.7 +/- 2.575*.051


.7 +/- .132


.568 %26lt; p %26lt; .832