Question

In a survey of a random sample of 35 households in the Cherry Creek neighborhood of Denver, it was found that 8 households turned out the lights and pretended not to be home on Halloween. Compute a 90% confidence interval for p, the proportion of all households in Cherry Creek that pretend not to be home on Halloween. (Round your answers to four decimal places.)

Answer 1

Answer: (0.1118, 0.3454)

Step-by-step explanation:

Confidence interval for population proportion is given by :-

$\hat{p}\pm z^c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ , where $\hat{p}$= sample proportion , n= sample size, $z^c$ = critical z-value for c confidence level.

Let p = the proportion of all households in Cherry Creek that pretend not to be home on Halloween.

Given : n= 35

$\hat{p}=\dfrac{8}{35}\approx0.2286$

Critical z-value for 90% confidence level =1.645

A 90% confidence interval for p, the proportion of all households in Cherry Creek that pretend not to be home on Halloween:

$0.2286\pm (1.645)\sqrt{\dfrac{0.2286(1-0.2286)}{35}}\\\\=0.2286\pm 0.1168\\\\=(0.2286-0.1168,\ 0.2286+0.1168)\\\\=(0.1118,\ 0.3454)$

∴ Required confidence interval : (0.1118, 0.3454)