A newspaper reported that 50 of people sy coffee shops are o

A newspaper reported that 50% of people sy coffee shops are overpriced. The source of this information was a telephone survey of 40 adults.

Find and interpret a 90% confidence interval for the parameter of interest.

P=(__,__) Round to two decimal places for each.

Solution

Given a=1-0.9=0.1, Z(0.05) = 1.645 (from standard normal table)

So the lower bound is

p - Z*sqrt(p*(1-p)/n) = 0.5 -1.645*sqrt(0.5*0.5/40 ) =0.37

So the upper bound is

p + Z*sqrt(p*(1-p)/n) = 0.5 +1.645*sqrt(0.5*0.5/40 ) =0.63