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?
(A)Solution :
Given that,
n = 1000
Point estimate = sample proportion = = 0.40
1 - = 1 - 0.40 = 0.60
At 95% confidence level
= 1 - 95%
=1 - 0.95 =0.05
/2
= 0.025
Z/2
= Z_{0.025 = 1.96}
Margin of error = E = Z_{ / 2} * (( * (1 - )) / n)
= 1.96 (((0.40*0.60) /1000 )
E = 0.0304
(B)
Solution :
Given that,
= 0.40
1 - = 1 - 0.40= 0.60
margin of error = E = 1% = 0.01
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z_{/2} = Z_{0.025} = 1.96
Sample size = n = (Z_{/2} / E)^{2} * * (1 - )
= (1.96 / 0.01)^{2} * 0.40* 0.60
= 9220
Sample size = 9220
2.) You are interested in calculating the sample error and size of the upcoming local elections....
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