Question

Problem 1 – Election poll The incumbent governor of a province in Indonesia believes that he will succeed in the upcoming regional election in winning more than 50% of the votes amongst the 13,324,525 registered voters. A random sample of 1,000 voters were selected. 510 people stated that they would vote for the governor. Estimate with 99% confidence the true proportion of voters who would vote for the governor. ​

a) You are recently hired as a junior data analyst working for the governor’s office. Specify the appropriate formula you would use to solve the problem. Provide a brief reason why you chose the formula.

​ ​ b) Obtain the 99% confidence interval estimate of the true proportion of voters who would vote for the governor. Display working.

​ ​ c) Provide an interpretation of the answers you obtained in part b) in the context of the problem. ​

d) Present brief statements of two or three sentences to the governor’s office with regard to the possibility of him winning the election on the basis of the interval estimation conducted. Is it very much conclusive that he will win the election? Why?

e) Assuming all of the other variables to do the calculation of confidence interval held constant, what would happen to the width of the interval when a lower sample size is used?

Answer 1

1.

a.
TRADITIONAL METHOD
given that,
possible chances (x)=510
sample size(n)=1000
success rate ( p )= x/n = 0.51
I.
sample proportion = 0.51
standard error = Sqrt ( (0.51*0.49) /1000) )
= 0.016
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
margin of error = 2.576 * 0.016
= 0.041
III.
CI = [ p ± margin of error ]
confidence interval = [0.51 ± 0.041]
= [ 0.469 , 0.551]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=510
sample size(n)=1000
success rate ( p )= x/n = 0.51
CI = confidence interval
confidence interval = [ 0.51 ± 2.576 * Sqrt ( (0.51*0.49) /1000) ) ]
= [0.51 - 2.576 * Sqrt ( (0.51*0.49) /1000) , 0.51 + 2.576 * Sqrt ( (0.51*0.49) /1000) ]
= [0.469 , 0.551]
-----------------------------------------------------------------------------------------------
b.
99% confidence interval estimate of the true proportion of voters who would vote for the governor [ 0.469 , 0.551]
c.
interpretations:
1. We are 99% sure that the interval [ 0.469 , 0.551] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion
d.
Given that,
possibile chances (x)=510
sample size(n)=1000
success rate ( p )= x/n = 0.51
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p>0.5
level of significance, α = 0.01
from standard normal table,right tailed z α/2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.51-0.5/(sqrt(0.25)/1000)
zo =0.632
| zo | =0.632
critical value
the value of |z α| at los 0.01% is 2.326
we got |zo| =0.632 & | z α | =2.326
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: right tail - Ha : ( p > 0.63246 ) = 0.26354
hence value of p0.01 < 0.26354,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
test statistic: 0.632
critical value: 2.326
decision: do not reject Ho
p-value: 0.26354
we do not have enough evidence to support the claim that he will succeed in the upcoming regional election in winning more than 50% of the votes amongst the 13,324,525 registered voters.
e.
Increasing the sample size decreases the width of confidence intervals, because it decreases the standard error.
Decreasing the sample size causes the error bound to increase, making the confidence interval wider.