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 122 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.

Answer 1

Answer:

The 95% confidence interval for the true proportion of Gastown residents living below the poverty line is (0.133, 0.2766).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence interval $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

He takes a simple random sample of 122 people living in Gastown and finds that 25 have an annual income that is below the poverty line. This means that $n = 122$ and $\pi = \frac{25}{122} = 0.205$

We have $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2}$ = 0.975, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.205 - 1.96\sqrt{\frac{0.205*0.795}{122}} = 0.133$

The upper limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.205 + 1.96\sqrt{\frac{0.205*0.795}{122}} = 0.133 = 0.2766$

The 95% confidence interval for the true proportion of Gastown residents living below the poverty line is (0.133, 0.2766).