Question

A company is considering the purchase of advertising time on a local radio show and will do so if it believes that the percent of all radio listeners that listen to the show is greater than 30% = 0.30. To make this decision, the company randomly selects 120 regular radio listeners and asks each whether he or she listens to the show. Of these 120 listeners, 42 report that they listen to the show.

Estimate with 95 percent confidence the proportion of all local radio listeners that listen to the show.

From this interval, can it be believed that more than 30 percent of all local radio show listeners listen to the show

How many listeners should be selected to estimate the proportion of all local radio listeners that listen to the show with 95% confidence and within a margin of error of 3%? Use the proportion from the initial sample as your initial guess.

Answer 1

Solution :

1) Given that,

n = 120

x = 42

Point estimate = sample proportion = = x / n = 42 / 120 = 0.35

1 - = 1 - 0.35 = 0.65

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.960

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.96 (((0.35 * 0.65) / 120)

= 0.085

A 95% confidence interval for population proportion p is ,

± E  

= 0.35 ± 0.085

= ( 0.265, 0.435 )

Yes, can it be believed that more than 30 percent of all local radio show listeners listen to the show, because 30% less than upper limit.

2) Given that,

= 1 - = 0.5

margin of error = E = 0.03

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025  = 1.96

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.96 / 0.03)2 * 0.5 * 0.5

= 1067.11

sample size = n = 1068