Question

suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. Five hundred randomly selected adukt residents in this city are surveyed ti determine whether they have cell phones. Of the 500 hundred people sampled, 421 responded yes- they own cell phones.

using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.

Answer 1

solution:

Let X = No.of people having cell phones in the given sample

Total adults sampled (n) = 500

No.of people having cell phones (x) = 421

Sample proportion (p') = x/n = 421/500 = 0.842

For 95% confidence level , = 1-CL = 1- 0.95 = 0.05

Here, Z() = Z(0.025) = 1.96 [ use standard normal distribution table ]

The confidence interval for population proportion is given by

CI : P' ± Z() *

: 0.842   ± 1.96*

: 0.842 ± 0.032

: (0.810 , 0.874 )

This interval contains true proportion of adult residents of this city who have cell phones

we estimate that 95% confidence that between 81% and 87.4% of all adult residents of this city have cell phones

answered by: anonymous