Question

(a)Find the margin of error for the estimate of the percentage.
(b) Obtain a sample size that will ensure a margin of error of at most percentage points for a $99\%$ confidence interval without making a guess for the observed value of $\hat{p}$.
(c) Find a confidence interval for $p$if, for a sample of the size determined in part (b),$82.2\%$ of the registered voters sampled favor the creation of standards on CAFO pollution and, in general, view CAFOs unfavorably.
(d) Determine the margin of error for the estimate in part (c) and compare it to the margin of error specified in part (b).
(e) Repeat parts (b)-(d) if you can reasonably presume that the percentage of registered voters sampled who favor the creation of standards on CAFO pollution and, in general, view CAFOs unfavorably will be between $75\%$and$85\%$.
(f) Compare the results you obtained in parts (b)-(d) with those obtained in part (e).

Short answer

(a) Margin error for population proportion is$0.0326$localid="1653304512126" $p=0.0326$

(b) The sample size needed to keep the margin of error to a maximum of localid="1653304548361" $0.015$is localid="1653304561238" $n=7396$.

(c) Confidence level localid="1653304571379">99%is localid="1653304599353">0.815to localid="1653304614861" $0.8335$.

(d) The localid="1653304626887">0.0115margin of error calculated in part (d) is smaller than the localid="1653304636631" $0.015$margin of error stated in part (b).

(e)localid="1653304647213" $n=7396,99\%CI=0.83930$to 0.8607, E=0.107$0.8607,E=0.0107$.

(f) Part (b)(d) results: $n=2401,$localid="1653333404960" $95\%$$CI=0.3408$to $0.3792$, E=0.0192

Part (e) results: n=2401, 95 percent CI=0.3804 to 0.4196, E=0.0196

Step-by-step solution

Part (a) Step 1: Given Information

$1000$registered voters favoured the creation of CAFO pollution standard regulating limitations.

The level of confidence is $99$percent.

Explanation:

According to the data, $80$percent of 1000 registered voters favoured the establishment of CAFO pollution standard regulating limits.

As a result, the number of successes is $x=800$, and the sample population is $n=1000$.

As a result, the sample population proportion is

$\hat{p}=\frac{x}{n}=\frac{800}{1000}=0.8$.

Confidence level is $99\%$

$z_{\frac{\alpha }{2}}=2.575$

The margin of error for the estimation of population percentage, is

$E=z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{\frac{{\displaystyle \hat{p}}(1-{\displaystyle \hat{p}})}{n}}$

$E=2.58\sqrt{\frac{0.8(1-0.8)}{1000}}$

$E=2.580\times 0126$

$E=0.0326$

In percentage,$E=3.2\%$

As a result, the margin of error for the population percentage estimate is

Part (b) Step 1: Given information

Similar to part (a) of this excercise

Explanation

Now,$n=0.25.\sqrt{{\left(\frac{z_{a2}}{E}\right)}^2}$yields $A(1-\alpha )$level confidence interval for a population with a margin of error of at most E, sample size.

$\alpha =0.01$for a confidence level of $99\%$.

So,

$z_{a/2}=z_{0.005}=2.58$.

$E=0.015$.

As a result, the required sample size n is

$$\begin{gathered} n=0.25{\left(\frac{2.58}{0.015}\right)}^2 \\ n=7396 \end{gathered}$$.

As a result, $n=7396$is the minimum sample size required to keep the margin of error to 0.02.

Part (c) Step 1: Given information

Part (a) of this activity is the same.

The sample proportion is $0.822$and the sample size is $7396$.

Explanation

Given,

$n=7396$,$\stackrel{\wedge }{p}=0.822$.

$\alpha =0.01$for a 99 percent confidence level.

So,$z_{a/2}=z_{0.005}=2.58$.

As a result, the $99$percent confidence interval will be written as

$CI=\hat{p}\pm z_{a/2}\sqrt{\frac{{\displaystyle \hat{p}}(1-{\displaystyle \hat{p}})}{n}}$.

$CI=0.822\pm 2.58\sqrt{\frac{0.822(1-0.822)}{7396}}$

$$\begin{gathered} CI=0.822\pm 0.0115 \\ CI=0.8105to0.8335 \end{gathered}$$

As a result, the $95\%$confidence interval is $0.8105$to $0.8335$.

Part (d) Step 1: Given information

Part (a) of this activity is the same.

The sample proportion is $0.822$and the sample size is $739$.

Explanation

The percent confidence interval for portion (c) is to .
Divide the breadth of the confidence interval by two to get the margin of error.
So,
$E=\frac{0.8335-0.8105}{2}$.

This margin of error is , which is lower than themargin of error given in component (b).

Part (e) Step 1: Given information

Similar to part (a) of this excercise

Explanation

For a population with a margin of error of at most $E$, $A(1-\alpha )$level confidence interval is derived by -

$n=0.25\cdot {\left(\frac{z_{a/2}}{E}\right)}^2$

$A(1-\alpha )$for a confidence level of 99 percent.

So,

$z_{a/2}=z_{0.005}=2.58$.

$E=0.015$.

As a result, the required sample size n is$n=0.25{\left(\frac{2.58}{0.015}\right)}^2$

$n=7396$.

As a result, $n=7396$is the minimum sample size required to keep the margin of error to $0.02$.

Step 2: 

Given, $n=7396$,$\hat{p}=0.85$.

$\alpha =0.01$for a $99\%$confidence level.

So, $z_{a/2}=z_{0.005}=2.58$.

As a result, the $99\%$percent confidence interval will be written as ,

$CI=\hat{p}\pm z_{a/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$.

role="math" localid="1653333099352" $CI=0.85\pm 2.58\sqrt{\frac{0.85(1-0.85)}{7396}}$

$CI=0.85\pm 0.0107$

$CI=0.8393to0.8607$

As a result, the $95\%$confidence interval is $0.8393$to $0.8607$.

Divide the breadth of the confidence interval by two to get the margin of error.

So,$E=\frac{0.8607-0.8393}{2}=0.0107$

Part (f) Step 1: Given information

Same as part (a) of this excercise

Explanation

Part (b)(d) results: $n=2401$, $95$percent $CI=0.3408to0.3792$,$E=0.0192$

Part (e) results: $n=2401$, $95$percent $CI=0.3804$to $0.4196$, $E=0.0196$