Question

You are the program officer for an outreach program for struggling families. Because of very limited funding (welcome to the non profit sector), you are only able to start the program up in zip codes with an average household income of less than $36,500.

Also because of limited funding, you rely on samples obtained by a government agency for each zipcode.

According to a sample of 800, zipcode 00001 has an average household of $39,500 with a standard deviation of $14,000. Using a confidence interval of 95%, is this zipcode immediately disqualified? Why or why not? Demonstrate your work.

Answer 1

Solution:

Given: You are the program officer for an outreach program for struggling families. Because of very limited funding, you are only able to start the program up in zip codes with an average household income of less than $36,500.

Sample size = n = 800

Sample mean =

Standard Deviation = s= 14,000

c = confidence level = 95%

We have to find a 95% confidence interval for an average household income.

Formula:

Since sample size = n = 800 is large, we can use z distribution to find confidence interval even if population standard deviation is unknown.

where

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 we get:

Thus we are 95% confident that an average household income for zipcode 00001 is between : $38,529.85 to $40,470.15

Since this confidence interval is much greater than $36,500,   this zipcode is immediately disqualified.