Question

In a random sample of 200 people, 154 said that they watched educational television. Find the 90% confidence interval of the true proportion of people who watched educational television. If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 154 / 200 = 0.77

Z/2 = 1.645

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

= 1.645 * (((0.77 * 0.23) / 200)

E = 0.049

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.77 - 0.049 < p < 0.77 + 0.049

0.721 < p < 0.819

(0.721 , 0.819)

Yes

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