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Sagot :
If the sample size is 1488 and confidence interval of 99% then the margin of error is 0.03088.
Given sample size of 1488, percentage of those polled own a home be 69% and confidence level be 99%.
We are required to find the approximate margin of error.
Margin of error is the difference between calculated values and real values.
n=1488
p=0.69
Margin of error=z*[tex]\sqrt{p(1-p)/n}[/tex]
Z score when confidence level is 99%=2.576.
Margin of error=2.576*[tex]\sqrt{0.69(1-0.69)/1488}[/tex]
=2.576*[tex]\sqrt{(0.69*0.31)/1488}[/tex]
=2.576*[tex]\sqrt{0.2139/1488}[/tex]
=2.576*[tex]\sqrt{0.0001437}[/tex]
=2.576*0.01198
=0.03088
Hence if the sample size is 1488 and confidence interval of 99% then the margin of error is 0.03088.
Learn more about margin of error at https://brainly.com/question/10218601
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