The CBS' television show 60 Minutes has been successful for many years. That show recently had a share of 20, which means that among the TV sets in use at the time the show aired, 20% were tuned to 60 Minutes. Assume that this is based on a sample size of 5000 - which is a typical sample size for this kind of experiments. Construct a 95% confidence interval for the true proportion of TV sets that are tuned to 60 Minutes.

We are asked for the confidence interval of the proportion, therefore we make use of the proportion distribution formulas.

Let us say that:

p = probability of success = 20% = 0.20

q = probability of failure = 1 - p = 0.80

n = number of samples = 5000

Now we use the confidence interval formula for proportion:

Confidence interval = p ± z sqrt (p q / n)

We can find for the value of z at the specified confidence level of 95% using the standard probability tables:

z = 1.96

Substituting the values:

Confidence interval = 0.20 ± (1.96) sqrt (0.20 * 0.80 / 5000)

Confidence interval = 0.20 ± 0.0111

Confidence interval = 0.189, 0.211

Answer:

0.189 < p < 0.211

or

0.19 < p < 0.21