In a gallop poll of 1000 adults 930 said that restaurants an

In a gallop poll of 1000 adults, 930 said that restaurants and bars should refuse service to patrons who had too much to drink. Find a 90% confience interval for the true percentage of patrons with this view. Identify the point estimate for the percentage and the margin of error.

Solution

point estimate=930/1000 = 0.93

Given a=0.1, Z(0.05)=1.645 (from standard normal table)

the margin of error = Z*sqrt(p*(1-p)/n)

=1.645*sqrt(0.93*0.07/1000)

=0.01327261

So 90% confidence interval is

p +/- Z*sqrt(p*(1-p)/n)

--> 0.93 +/- 0.01327261

--> (0.9167274, 0.9432726)