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 129 people living in Gastown and finds that 24 have an annual income that is below the poverty line.

Part ii) Use the sample data to compute a 95% confidence interval for the true proportion of Gastown residents living below the poverty line. (Please carry answers to at least six decimal places in intermediate steps. Give your final answer to the nearest four decimal places).

Answer 1

Solution :

Given that,

n = 129

x = 24

Point estimate = sample proportion = = x / n = 24/129=0.186

1 -   = 1- 0.186 =0.814

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2   * ((( * (1 - )) / n)

= 1.96 (((0.186*0.814) / 129)

E = 0.0671

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.186- 0.0671< p < 0.186+0.0671

0.1189< p < 0.2531

The 95% confidence interval for the population proportion p is : 0.1189,0.2531