Question

1. It was found that in a sample of 90 teenage boys, 70% of them have received speeding tickets.

a) What is the 90% confidence interval of the true proportion of teenage boys who have received speeding tickets?

b) Given the above information, how many boys must be sampled to be within 2% of true proportion in a 90% confidence interval?

c) If no information is given for the proportion, how many boys must be sampled?

Answer 1

a)

Given that,

n =90

= 0.70

1 - = 1 - 0.70= 0.30

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90= 0.1

/ 2 = 0.1 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * [( * (1 - )] / n)

= 1.645 * {[(0.70*0.30)] / 90}

= 0.0795

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.70 - 0.0795 < p < 0.70 + 0.0795

0.6205 < p < 0.7795

b)

E= 0.02

= 0.70

1 - = 1 - 0.70= 0.30

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90= 0.1

/ 2 = 0.1 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

sample size = ( Z_/2 / E)² * * (1 - )     

   = ( 1.645/ 0.02 )² *0.30 *0.70

                   = 1420.7

                   = 1421

c)

E= 0.02

= 0.50

1 - = 1 - 0.50= 0.50

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90= 0.1

/ 2 = 0.1 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

sample size = ( Z_/2 / E)² * * (1 - )     

   = ( 1.645/ 0.02 )² *0.50 *0.50

                   = 1691.3

                   = 1691