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 123 people living in Gastown and finds that 25 have an annual income that is below the poverty line.
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.)

95% confidence interval = ( )

Answer 1

Answer: (0.132132, 0.274368)

Step-by-step explanation:

Given : A simple random sample of 123 people living in Gastown and finds that 25 have an annual income that is below the poverty line.

i.e. n= 123

$\hat{p}=\dfrac{25}{123}\approx0.203252$

Critical value for 95% confidence interval : $z_{\alpha/2}=1.96$

Confidence interval for population :

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

i.e. $0.203252\pm (1.96)\sqrt{\dfrac{0.203252(1-0.203252)}{123}}$

$0.203252\pm (1.96)\sqrt{\dfrac{0.203252(1-0.203252)}{123}}\\\\\0.203252\pm0.071118\\\\=(0.20325-0.071118,\ 0.203252+0.071118)\\\\=(0.132132,\ 0.274368)$

Hence, the 95% confidence interval for the true proportion of Gastown residents living below the poverty line : (0.132132, 0.274368)