Question

A government official is in charge of allocating social programs throughout the city of Vancouver. He will decide where these social outreach programs should be located based on the percentage of residents living below the poverty line in each region of the city. He takes a simple random sample of 126 people living in Gastown and finds that 22 have an annual income that is below the poverty line.

Suppose that the government official wants to re-estimate the population proportion and wishes for his 90% confidence interval to have a margin of error no larger than 0.05. How large a sample should he take to achieve this? Please carry answers to at least six decimal places in intermediate steps, and use at least 3 decimal places in your critical value.

Sample Size =

Answer 1

Solution:

Given:

Previous sample size = n = 126 , x= 22

confidence level = c = 90%

Margin of Error = E = 0.05

We have to find sample size.

Formula:

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

Sample size = n = 156