Question

1) A media analyst is trying to determine the proportion of people living in St. Louis that watch a particular television show. The analyst collects a random sample of 204 people from St. Louis. Of the 204 people in the sample, 56 people said that they did watch the show.

a) Find a 95% confidence interval for the true population proportion of people in St. Louis who watch the show.

b) Provide the right endpoint of the interval as your answer.

Round your answer to 4 decimal places.

Answer 1

Solution :

n =204

x = 56

= x / n = 56 / 204 = 0.275

1 - = 1 - 0.274 = 0.725

a ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

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

= 1.960 * (((0.275 * 0.725) / 900)

= 0.061

A 95 % confidence interval for population proportion p is ,

b ) + E

0.275 +0.061

Right endpoint = 0.336