Question

A commercial for a breakfast cereal is shown during a certain television program. The manufacturer of the cereal wants to estimate the percent of television viewers who watch the program. The manufacturer wants the estimate to have a margin of error of at most 0.10 at a level of 80 percent confidence. Of the following, which is the smallest sample size that will satisfy the manufacturer’s requirements? please show your work

A. 40

B. 50

C. 100

D. 1,700

E. 2,500

Answer 1

Solution,

Given that,

=  1 - = 0.5

margin of error = E = 0.10

At 80% confidence level

= 1 - 80%

= 1 - 0.80 = 0.20

/2 = 0.10

Z/2 = Z0.10  = 1.28

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

= (1.28 / 0.10)2 * 0.5 *0.5

= 40.96

sample size = n = 40