Question

uppose 300 randomly selected people are surveyed to determine whether or not they rent their home. Of the 300 surveyed, 111 reported renting their home. Find the margin of error for the confidence interval for the population proportion with a 90% confidence level.

Answer 1

Answer:

$M.E=0.046$

Step-by-step explanation:

From the question we are told that:

Sample size $n=300$

No. of rental $x=111$

Confidence Interval $CI=0.90\ \alpha=0.10$

Probability of rental

$P(x)=\frac{111}{300}$

$P(x)=0.370$

Generally the equation for Margin of Error is mathematically given by

$M.E=Z_c*\sqrt{\frac{P(1-P)}{n}}$

$M.E=1.645*\sqrt{\frac{0.370(1-0.370)}{300}}$

$M.E=0.046$