Sampling Proportion

Question

Suppose that 50% of the subscribers of a cable television company watch the shopping channel at least once a week. The cable company is trying to decide whether to replace this channel with a new local station. A survey of 100 subscribers will be undertaken. The cable company has decided to keep the shopping channel if the sample proportion is greater than .56. What is the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only .50? (Round the answer to four decimal places.)

P(p > .56) =  

Answer 1

1 - P(p-hat <= .56) = 1 - P(N <= .56*100) = 1 - P(N <= 56)

We can use a normal approximation, or get the exact probability from Excel.

The exact probability is 1 - binom.dist(56,100,.5,TRUE) = , which we can round to .0967.

Note: the normal approximation of P(N <= 56), using the continuity correction, is

P(z >= (56.5-50)/sqrt(.5*.5*100)) = P(z <= 6.5/5) = P(z <= 1.3) = 1 - normsdist(1.3) = .0968. The normal approximation is very close.