Question

In a survey of 300 T.V. viewers, 120 said they watch network nightly news programs. Construct a 95% confidence interval for the proportion of viewers that watch network nightly news.

(0.372, 0.428)

(0.396, 0.403)

(0.30, 0.50)

(0.345, 0.455)

Write a summary sentence for the confidence interval you calculated in the prior problem.

Answer 1

Solution :

Given that,

n = 300

x = 120

Point estimate = sample proportion = = x / n = 120/300=0.4

1 -   = 1- 0.4 =0.6

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.4*0.6) /300 )

= 0.0554

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.4 - 0.0554< p <0.4+ 0.0554

0.345< p < 0.455

The 95% confidence interval for the population proportion p is : 0.345, 0.455