Question

Question: A recent poll of 285 New Jersey residents indicated that 203 plan to vote in the next election.  

  a) Find a 99 % confidence interval for the population proportion of New Jersey resident who plan to vote in the next election. Show all calculations.

  b) Find a 90% confidence interval for the population proportion of New Jersey residents who plan to vote in the next election. Show all calculations.

Answer 1

Solution :

Given that,

n = 285

x = 203

= x / n = 203 / 285 = 0.712

1 - = 1 - 0.712 = 0.288

(a)

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.712 * 0.288) / 285)

= 0.069

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.712 - 0.069 < p < 0.712 + 0.069

0.643 < p < 0.781

(0.643 , 0.781)

(b)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645* (((0.712 * 0.288) / 285)

= 0.044

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.712 - 0.044 < p < 0.712 + 0.044

0.668 < p < 0.756

(0.668 , 0.756)