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 =0.9, n=160,  α =0.2 p̂ =0.9, 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 used

b.  p̂ =0.45, n=140,  α =0.2 p̂ =0.45, 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

Please provide correct answer. thanks.

Answer 1

Solution :

Given that,

a) Point estimate = sample proportion = = 0.9

1 - = 1 - 0.9 = 0.1

Z/2 = Z0.01 = 2.33

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.33 (((0.9 * 0.1) / 160)

= 0.0553

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.9 - 0.0553 < p < 0.9 + 0.0553

( 0.8447 < p < 0.9553 )

b) Point estimate = sample proportion = = 0.45

1 - = 1 - 0.45 = 0.55

Z/2 = Z0.01 = 2.33

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.33 (((0.45 * 0.55) / 140)

= 0.0980

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.45 - 0.0980 < p < 0.45 + 0.0980

( 0.3520 < p < 0.5480 )