Question

Think about a population proportion that you may be interested in and propose a confidence interval problem for this parameter. For example, you may like to estimate the population proportion of adults in the US who own SUVs. The data could be that you researched online by looking at a local dealership to find that 142 of the 432 vehicles sold are SUVs. You want to calculate a 90% (or another level) confidence interval for the population proportion. Assume a random sample.

Answer 1

Sample Proportion = p = Number of adults in the US who own SUVs / Total number of vehicles sold

Total number of vehicles sold = 432

Number of adults in the US who own SUVs = 142

Sample Proportion = p = 142 / 432 = 0.3287

(1-$\alpha$) % confidence interval for the population proportion is given by :

$Upper\ limit = p + Z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}$

$Lower\ limit = p - Z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}$

We have, p = 0.3287

n = 432

$\alpha = 1-0.90 = 0.10$

$Z_{\alpha/2} =Z_{0.10/2}=Z_{0.05} = 1.645$

Putting all the values in the above expression we get :

$Upper\ limit = 0.3287 + 1.645\sqrt{\frac{0.3287(1-0.3287)}{432}} =0.366$

$Lower\ limit = 0.3287 - 1.645\sqrt{\frac{0.3287(1-0.3287)}{432}} =0.292$

So, 90% confidence interval for the population proportion is given by : ( 0.292 , 0.366 )