Question

In a survey of 600 adults, 450 reported that e-mails are easy to misinterpret.

A. Construct a 99% confidence interval estimate for the population proportion of adults who report that e-mails are easy to misinterpret.

a. [0.601 , 0.816]
b. [0.704 , 0.796]
c. [0.729 , 0.739]
d. [0.684 , 0.895]
e. [0.736 , 0.785]

Answer 1

Solution :

Given that,

n = 600

x = 450

$\hat p$ = x / n = 450 / 600 = 0.750

1 - $\hat p$ = 1 - 0.750 = 0.250

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 * ($\sqrt$((0.750 * 0.250) / 600)

= 0.046

A 99% confidence interval for population proportion p is ,

$\hat p$ - E < P <$\hat p$ + E

0.750 - 0.046 < p < 0.750 + 0.046

0.704 < p < 0.796

(0.704 , 0.796)

Option b) is correct .