Question

A polling organization announces that the proportion of American voters who favor congressional term limits is 64%, with a 95% confidence margin of error of 3%. If the opinion poll had announced the margin of error for 80% confidence rather than 95% confidence, this margin of error would be

Answer 1

Answer:

Hence when C.I. is 80% then Margin of error will be 1.9591%

Step-by-step explanation:

Given:

1st Margin of error(MOE)=3%

1st C.I.=95%

2nd C.I.=80%

To Find:

MOE at 80%

Solution:

The proportion is 64 % i.e. is constant for both MOE

Proportion is given by (P),

P=x/n

Where x= favors congressional terms

n= total voters or sampled voters.

There values are same or constant.

i.e standard deviation is also constant

Now Formula for MOE is given by

$MOE=Z*[Standard devation/Sqrt(n)]$

here Z is value for confidence interval

MOE is directly proportional to the Z-value

So

$MOE=K*Z-value$         ... where k is proportionality constant.

$MOE/Z-value$              ....ratio is constant

So , for 95 % Z=1.96 and for 80% Z=1.28

$MOE(1st)/(Z-value 1st)=MOE(2nd)/(Z-value 2nd)$

$3/1.96=MOE(2nd)/1.28$

$MOE(2nd)=(3/1.96)*1.28$

$MOE(2nd)=1.5306*1.28$

$=1.9591$ %