Question

A random sample of 1025 adults in a certain large country was asked "do you pretty much think televisions are a necessity or a luxury you could do without?" of the 1025 adults surveyed, 524 indicated that televisions are a luxury they could do without.

Construct and interpret a 95% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without.

Answer 1

Solution:

    We are given that:

n = sample size = 1025

x = of adults in the country who believe that televisions are a luxury they could do without = 524

We have to find a 95% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without.

Formula:

where

and

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

Thus

Thus limits of confidence interval are:

Thus we are 95% confident that the true population proportion of adults in the country who believe that televisions are a luxury they could do without is in between ( 0.4806 , 0.5418 ).