Question

An article a newspaper reported on the topics that teenagers most want to discuss with their parents. The​ findings, the results of a​ poll, showed that​ 46% would like more discussion about the​ family's financial​ situation, 37% would like to talk about​ school, and​ 30% would like to talk about religion. These and other percentages were based on a national sampling of 513 teenagers. Estimate the proportion of all teenagers who want more family discussions about school. Use a 90​% confidence level. Express the answer in the form ModifyingAbove p with caretplus or minusE and round to the nearest thousandth.

Answer 1

Solution:

Given:

Sample Size = n = 513

Sample proportion of teenagers who want more family discussions about school =

We have to construct a 90​% confidence interval for  the proportion of all teenagers who want more family discussions about school.

Formula:

where

Z_c is z critical value for c = 90% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.90) / 2 = 1.90 / 2 = 0.9500

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_c = 1.645

Thus we get:

Thus a 90% confidence interval is:

or

Thus a 90​% confidence interval for  the proportion of all teenagers who want more family discussions about school is :   

or