Question

Calculate the margin of error and construct a confidence interval for the population proportion using the normal approximation to the  p̂  p̂ -distribution (if it is appropriate to do so).

Standard Normal Distribution Table

a.  p̂ =0.85, n=140,  α =0.2 p̂ =0.85, n=140,  α =0.2

E=E=

Round to four decimal places

Enter 0 if normal approximation cannot be used

  < p <  < p <  

Round to four decimal places

Enter 0 if normal approximation cannot be used

b.  p̂ =0.3, n=160,  α =0.2 p̂ =0.3, n=160,  α =0.2

E=E=

Round to four decimal places

Enter 0 if normal approximation cannot be used

  < p <  < p <  

Round to four decimal places

Enter 0 if normal approximation cannot be use

Please provide correct answers. thanks

Answer 1

Solution :

Given that,

a) Point estimate = sample proportion = $\hat p$ = 0.85

1 - $\hat p$ = 1 - 0.85 = 0.15

Z$\alpha$/2 = Z0.1 = 2.33  

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

= 2.33 ($\sqrt$((0.85 * 0.15) / 140)

= 0.0703

A 98% confidence interval for population proportion p is ,

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

0.85 - 0.0703 < p < 0.85+ 0.0703

( 0.7797 < p < 0.9203 )

b) Point estimate = sample proportion = $\hat p$ = 0.3

1 - $\hat p$ = 1 - 0.3 = 0.7

Z$\alpha$/2 = Z0.1 = 2.33  

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

= 2.33 ($\sqrt$((0.3 * 0.7) / 160)

= 0.0844

A 98% confidence interval for population proportion p is ,

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

0.3 - 0.0844 < p < 0.3 + 0.0844

( 0.2156 < p < 0.3844 )