Question

. A poll found that 36% of U.S. teens aged from 13 to 17 years old have a computer with Internet access in their rooms. The poll was based on a random sample of 1028 teens.

A) Give the 98% confidence interval for the population proportion, p.

B) What does the interval in part (A) mean in this context. Be as specific as possible

C) How big of a sample must we take in order to ensure that we are within 1% of the true proportion with 98% confidence?

Answer 1

Solution :

a ) Given that

n = 1028

x = 225

= 0.360

1 - = 1 - 0.360 = 0.740

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z/2 = Z0.01 = 2.326

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.326 * (((0.360 * 0.740) / 1028)

= 0.035

b ) A 98 % confidence interval for population proportion p is ,

- E < P < + E

0.360 - 0.035 < p < 0.360 + 0.035

0.325 < p < 0.2395  

c ) Given that,

= 0.36

1 - = 1 - 0.36 = 0.74

margin of error = E = 1% = 0.01

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z/2 = Z0.01 = 2.326

Sample size = n = ((Z / 2) / E)2 * * (1 - )

= (2.326 / 0.01)2 * 0.36 * 0.47

= 12469

n = sample size = 12469