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 25 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.)
95% confidence interval = ( )

Answer 1

The 95% confidence interval would be given by (0.128419;0.267581)

Explanation:

1) Notation and definitions

`X=25` people that have an annual income that is below the poverty line

`n=126` random sample taken

`\hat p=(25)/(126)=0.198` estimated proportion of people that have an annual income that is below the poverty line.

`p` true population proportion of people that have an annual income that is below the poverty line.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

2) Confidence interval

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by
`\alpha=1-0.95=0.05` and
`\alpha/2 =0.025`. And the critical value would be given by:

`t_(1-\alpha/2)=1.96`

The confidence interval for the proportion is given by the following formula:

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

If we replace the values obtained we got:

`0.198 - 1.96\sqrt{(0.198(1-0.198))/(126)}=0.128419`

`0.198 + 1.96\sqrt{(0.198(1-0.198))/(126)}=0.267581`

The 95% confidence interval would be given by (0.128419;0.267581)