Question

A poll is taken in which 359 out of 600 randomly selected voters indicated their preference for a certain candidate.

Find a 90% confidence interval for p.

Find the margin of error for this 90% confidence interval for p.

Without doing any calculations, indicate whether the margin of error is larger or smaller or the same for an 80% confidence interval.

Answer 1

sOL:

Sample proportion=p^=x/n=359/600= 0.5983333

zcrit for 90%=1.645

90% confidence interval for p is

p^-z*sqrt(p^*(1-p^)/n,p^-z*sqrt(p^*(1-p^)/n

0.5983333-1.645*sqrt(0.5983333*(1-0.5983333)/600),0.5983333+1.645*sqrt(0.5983333*(1-0.5983333)/600)

  0.5654107, 0.6312559

90% confidence interval for p lies in between  0.5653946 and 0.631272

  0.5654107, 0.6312559

Find the margin of error for this 90% confidence interval for p.

z*sqrt(p^*(1-p^)/n

1.645*sqrt(0.5983333*(1-0.5983333)/600)

= 0.03292265

0.03292265

margin of error is smaller for an 80% confifence interval comapred to 90% as z crit for 80% is less compared to 90%

Smaller