Question

You want to estimate the proportion of students at OSU who have ever used an online dating
app. You select a random sample of 140 OSU students and find that 43% have used an online
dating app. If you want to construct a 99% confidence interval, what will the margin of error
be? Try not to do a lot of rounding along the way until you get to the end of your calculations.

Answer 1

Solution :

Given that,

n = 140

Point estimate = sample proportion = $\hat p$ =0.43

1 - $\hat p$ = 1-0.43=0.57

At 99% confidence level the z is ,

$\alpha$  = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z_$\alpha$/2 = Z_0.005 = 2.576

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 2.576 ($\sqrt$((0.43*0.57) / 140)

= 0.10778