Question

Use the following information to answer the next two exercises: Five hundred and eleven $\left(511\right)$homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations. One hundred seventy-three $\left(173\right)$of the homes surveyed met the minimum recommendations for earthquake preparedness, and $338$did not.

Find the confidence interval at the $90\%$Confidence Level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness.

a. $(0.2975,0.3796)$

b.$(0.6270,0.6959)$

c. $(0.3041,0.3730)$

d.$(0.6204,0.7025)$

Short answer

The population proportion has a $90\%$confidence interval of$0.627\le p\le 0.696$

Step-by-step solution

Introduction

A confidence interval is a range for assessing an undetermined parameter in likelihood ratio statistics.

From one or both sides, it can be bounded. At a given confidence level, a confidence interval is calculated.

The $95$percent threshold is the most commonly utilized, however other amounts are occasionally used as well.

Explanation

If the proportion of observed in a random sample of size $n$that relate to a class of interest is $\widehat{p}$, an estimate $100(1-\alpha )\mathrm{\%}$confidence interval on the fraction $p$of the population which belongs to this class is

Equation $1$

$\widehat{p}-z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\le p\le \widehat{p}+z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

where $z\frac{\alpha }{2}$denotes the normally distributed distribution's top $\frac{\alpha }{2}$basis point.

Three hundred thirty-eight $(x=338)$of the $511$residences surveyed do not satisfy the minimal disaster preparedness guidelines.

As a result, the population mean of homes that do not meet the minimum earthquake preparedness standards is

$\widehat{p}=\frac{338}{511}=0.661$

For such a two-sided confidence interval of $90\%$

$\begin{array}{c}\frac{\alpha }{2}=\frac{1-0.90}{2}=0.05,\\ \end{array}$

then

$z_{\frac{\alpha }{2}}=1.65.$

As a result of Equations $1$and $2$, the population proportion has a $90\%$two-sided CI.

role="math" localid="1649783275438" $\begin{array}{r}0.661-1.65\sqrt{\frac{0.661(1-0.661)}{511}}\le p\le 0.661+1.65\sqrt{\frac{0.661(1-0.661)}{511},}\\ 0.661-0.0344\le p\le 0.661+0.0344\end{array}$

As a result, $p$has a $90\%$CI.

$0.627\le p\le 0.696$