Question

A political candidate is anxious about the outcome of the election. Complete parts​ (a) and​ (b) below.

a) To have his next survey result produce a 95% confidence interval with a margin of error of no more than 0.034 how many eligible voters are needed in the​ sample?

(b) If the candidate fears that​ he's way behind and needs to save​ money, how many voters are needed if he expects that his percentage among voters p is near 0.25?

Answer 1

Solution :

Given that,

= 0.5

1 - = 1 - 0.5= 0.5

margin of error = E = 0.034

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )

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

= (1.96 / 0.034)2 * 0.5* 0.5

= 831

Sample size = 831

b.

Solution :

Given that,

= 0.25

1 - = 1 - 0.25 = 0.75

margin of error = E = % = 0.034

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )

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

= (1.96 / 0.034)2 * 0.25 * 0.75

= 623

Sample size = 623