Question

Opinion polls are a popular method for assessing product preference, political preference, and more. As a simple example, consider that a poll was taken ten days prior to a civic election to try to predict what proportion of the electorate would vote for the incumbent mayor. You are required to analyze the results of the poll and predict what proportion of voters will vote for the incumbent.

5. What is the margin of error of the estimated proportion?

6. What is the lower 90% confidence limit on the proportion of voters who will vote for the incumbent?

5 and 6 or NOT the answers of; 0.0304, 0.4928.

DATA:

0
1
1
1
1
1
1
1
1
0
1
0
1
1
0
1
1
1
1
0
1
1
1
1
0
0
0
0
1
0
1
1
1
1
1
0
1
1
0
0
1
1
0
0
1
1
0
0
1
0
0
0
1
0
0
0
0
0
0
1
0
0
1
1
0
1
0
1
0
0
1
1
0
0
0
0
1
1
1
0
1
1
1
1
0
1
1
1
1
0
1
1
1
1
1
1
1
0
1
0
1
1
1
0
1
0
0
1
0
1
0
1
1
1
1
1
1
0
1
1
1
0
1
0
1
0
0
1
1
1
1
0
0
1
0
0
0
1
0
1
0
0
0
1
1
0
1
1
0
1
0
0
0
1
0
0
1
0
0
1
1
1
1
0
1
0
1
0
1
1
1
0
1
0
0
1
0
0
0
1
1
0
1
0
1
1
0
1
0
0
0
1
0
0
0
0
1
1
0
0
1
0
1
0
0
1
1
0
1
0
1
1
0
1
1
0
1
0
1
1
0
0
0
1
0
1
1
1
1
1
0
1
1
1
0
0
1
0
1
1
0
0
1
0
0
1
1
1
1
0
0
0
0
0
1
1
0
0
0
1
0
1
1
0
0
1
0
0
1

Answer 1

Sol:

	Count
0	123
1	146
Total	269

From the given data:

5. What is the margin of error of the estimated proportion?

1-vote for incumbent

0-do not vote for incumbent

we are interested in margin of error who vote

=Z crit*sqrt(p^(1-p^)/n

n=269

p^=x/n=146/269=0.542751

z alpha/2for 90%=1.645

margin of error=1.645*sqrt(0.542751*(1-0.542751)/269)

margin of error=0.04996

What is the lower 90% confidence limit on the proportion of voters who will vote for the incumbent?

lower limit=sample prop-margin of error

=0.542751-0.04996

=0.492791

but answers we are getting that only

is rounding needed?

to how many decimals