Question

A random sample of 500 people is taken. 260 of them rent (as oppose to own) their residence. Compute a 95% confidence interval for the true proportion

Answer 1

Answer: A 95% confidence interval for the true proportion (p) will be (0.4762136, 0.5637864).

Step-by-step explanation:

Let p = True proportion of people rent their residence.

Given: n= 500

People rent their residence = 260

$p=\dfrac{260}{500}=0.52$

z-value for 95% confidence level = 1.96

A 95% confidence interval for the true proportion will be :

$p\pm z^*\sqrt{\dfrac{p(1-p)}{n}}$

$0.52\pm (1.96)\sqrt{\dfrac{0.52(1-0.52)}{500}}\\\\=0.52\pm(1.96\sqrt{\dfrac{0.2496}{500}})\\\\=0.52\pm 1.96\sqrt{0.0004992}\\\\=0.52\pm (1.96)(0.02234)\\\\=0.52\pm0.0437864\\\\=(0.52-0.0437864,\ 0.52+0.0437864)\\\\=(0.4762136,\ 0.5637864)$

A 95% confidence interval for the true proportion (p) will be (0.4762136, 0.5637864).