5 In a random sample of 50 households in a certain city 38 o

#5

In a random sample of 50 households in a certain city, 38 owned a computer. Give a 90% C.I. for the true fraction of all household that own computers A comparison of the mean life of two brands of light bulbs produced the data summary:

Solution

p=38/50 =0.76

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) = 0.76 -1.645*sqrt(0.76*(1-0.76)/50) =0.6606442

So the upper bound is

p + Z*sqrt(p*(1-p)/n) = 0.76 +1.645*sqrt(0.76*(1-0.76)/50) =0.8593558