Question

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.

Answer 1

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