Question

A random sample found that thirty-eight percent of 50 Americans were satisfied with the gun control laws in 2017. Compute a 90% confidence interval for the true proportion of Americans who were satisifed with the gun control laws in 2017. Fill in the blanks appropriately. A 90% for the true proportion of Americans who were satisfied with the gun control laws in 2017 is ( , ) (Keep 3 decimal places)

Answer 1

Solution:

Given: A random sample found that thirty-eight percent of 50 Americans were satisfied with the gun control laws in 2017.

Thus  

n = sample size = 50

Sample Proportion =

We have compute a 90% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017.

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

Thus

Thus a 90% for the true proportion of Americans who were satisfied with the gun control laws in 2017 is: