Question

Use the following information to answer the next three exercises: According to a Field Poll, $79\%$of California adults (actual results are $400$out of $506$surveyed) feel that “education and our schools” is one of the top issues facing California. We wish to construct a $90\%$confidence interval for the true proportion of California adults who feel that education and the schools is one of the top issues facing California

A $90\%$confidence interval for the population proportion islocalid="1649785149439" $\_\_\_\_\_\_\_.$

a. $(0.761,0.820)$

b. $(0.755,0.826)$

c.$(0.755,0.826)$

d.$(0.130,0.183)$

Short answer

The population proportion for confidence interval of $90\%$is$0.761\le p\le 0.820$

Step-by-step solution

Introduction

A confidence interval is a range for calculating an unknown parameter in bayesian inference statistics.

From one or both sides, it can be bounded. At a given confidence level, a confidence measurement can be done.

The $95$percent threshold is the most commonly utilised, 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 of the population which belongs to this class is

Equation localid="1649844549206" $1$

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

where $z\frac{\alpha }{2}$denotes the standard normal distribution's top $\frac{\alpha }{2}$percentage point.

According to a Field Poll, "education and our schools" is one of the top issues confronting California, with $\widehat{p}=79\mathrm{\%}=0.79$of adults (actual findings are $x=400$out of $n=506$polled) agreeing.

For a two-sided confidence level of $90\%$,

$\frac{\alpha }{2}=\frac{1-0.90}{2}=0.05$

then

Equation $2$

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

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

$\begin{array}{c}0.79-1.65\sqrt{\frac{0.79(1-0.79)}{506}}\le p\le 0.79+1.65\sqrt{\frac{0.79(1-0.79)}{506},}\\ 0.79-0.0298\le p\le 0.79+0.0298\end{array}$

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

$0.761\le p\le 0.820$