Question

2.) You are interested in calculating the sample error and size of the upcoming local elections. You asked 1000 people if they are planning on voting for candidate A. 40% said “Yes”. Calculate the margin of sample error at 95%. What sample size will you need to have a sample error of only 1% at the 95% confidence level?

Answer 1

(A)Solution :

Given that,

n = 1000

Point estimate = sample proportion = = 0.40

р

1 - = 1 - 0.40 = 0.60

р

At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z_{0.025 = 1.96}

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ ((
р
* (1 -
р
)) / n)

= 1.96 ($\sqrt$((0.40*0.60) /1000 )

E = 0.0304

(B)

Solution :

Given that,

$\hat p$ = 0.40

1 - $\hat p$ = 1 - 0.40= 0.60

margin of error = E = 1% = 0.01

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.96

Sample size = n = (Z_$\alpha$/2 / E)² * $\hat p$ * (1 - $\hat p$ )

= (1.96 / 0.01)² * 0.40* 0.60

= 9220

Sample size = 9220