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 125 people living in Gastown and finds that 25 have an annual income that is below the poverty line.
Part i) The proportion of the 125 people who are living below the poverty line, 25/125, is which of the following:______.
A. variable of interest.
B. parameter.
C. statistic.
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 three decimal places).
95% confidence interval = ( _ , ___ )

Part I: The proportion of the 125 people who are living below the poverty line, 25/125, is which of the following: statistic, as it is a measure taken from the sample.

Part II:

We have to calculate a 95% confidence interval for the proportion.

The sample proportion is p=0.2.

[tex]p=X/n=25/125=0.2[/tex]

The standard error of the proportion is:

[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.2*0.8}{125}}\\\\\\ \sigma_p=\sqrt{0.00128}=0.035777[/tex]

The critical z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

[tex]MOE=z\cdot \sigma_p=1.96 \cdot 0.035777=0.070122[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL=p-z \cdot \sisgma_p = 0.2-0.070122=0.129878\\\\UL=p+z \cdot \sisgma_p = 0.2+0.070122=0.270122[/tex]

The 95% confidence interval for the population proportion is (0.130, 0.270).