Question

Research suggests that nowadays people read newspapers only occasionally. In a small study, only 30 of the 230 respondents said that they read at least one newspaper everyday. Estimate the true proportion of the people who read at least one newspaper everyday using a 90% confidence interval.

(a) The limits of the 90% confidence interval are calculated by p̂ ±zp̂ (1−p̂ )n‾‾‾‾‾‾‾‾‾√p^±zp^(1−p^)n, where

p̂ = (Round to 3 decimal places.) Please explain how you got it!

z= (Round to 3 decimal places.) Please explain

n=

(b) The 90% confidence interval for the true proportion is:

(  ,  ) (Round to 3 decimal places.)

Answer 1

Solution :

n = 230

x = 30

a ) = x / n = 30 / 230 = 0.130

1 - = 1 - 0.130 = 0.870

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (((0.130 * 0.870) / 230)

= 0.036

b ) A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.130 - 0.036 < p < 0.130 + 0.036

0.094 < p < 0.166

The 90% confidence interval for the population proportion p is : ( 0.094 , 0.166 )