Question

A poll was conducted by a home mortgage company regarding home ownership in the United States. The company polled 1,488 Americans and found that 69% of those polled own a home. What is the approximate margin of error, assuming a 99% confidence level?

Answer 1

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*$\sqrt{p(1-p)/n}$

Z score when confidence level is 99%=2.576.

Margin of error=2.576*$\sqrt{0.69(1-0.69)/1488}$

=2.576*$\sqrt{(0.69*0.31)/1488}$

=2.576*$\sqrt{0.2139/1488}$

=2.576*$\sqrt{0.0001437}$

=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.

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