Factory Farming Funk. The U.S. Environmental Protection Agency reported that concentrated animal feeding operations (CAFOs) dump $2$trillion pounds of waste into the environment annually, contaminating the groundwater in 17 states and polluting more than $35,000$miles of our nation's rivers. In a survey of $1000$registered voters by Snell, Perry, and Associates, $80\%$favored the creation of standards to limit such pollution and, in general, viewed CAFOs unfavorably. Find and interpret a $99\%$confidence interval for the percentage of all registered voters who favor the creation of standards on CAFO pollution and, in general, view CAFOs unfavorably.

(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 $99\%$localid="1653240664989" $99\%$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), 828*localid="1653240627989" $82.2\%$localid="1653240689333" $82.2\%$ 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).

To find the margin error for the estimate of $p$.

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 $\stackrel{\wedge }{p}=\frac{x}{n}\frac{800}{1000}=0.8$.$p=xn=8001000=0.8$.

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

localid="1653272065930" $za/2=z0.005=2.58$.

localid="1653272127341" $E=za/2p\left(1p\right)nE=2.580.8(10.8)1000E=2.580.0126E=0.0326$is the margin of error for the estimation of population percentagelocalid="1653272151952" $p$.

As a result, the margin of error for the population percentage p estimate is localid="1653272095119" $0.0326$.

The sample size needed to keep the margin of error to a maximum of $0.015$.

$n=0.25\left(za2E\right)2$yields A(1-a) level confidence interval for a population with a margin of error of at most$E$, sample size.

$=0.01$for a confidence level of $99$percent.

So, $za/2=z0.005=2.58$. $E=0.015$.

As a result, the required sample size n is $n=0.25(2.580.015)2n=7396$.

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

Part an of this activity is the same.

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

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

$=0.01$at a $95\%$confidence level. So, $z\alpha /2=z0.005=2.58$.

As a result, the $99$percent confidence interval will be $CI=pz/2p\left(1p\right)$.

localid="1653225180846">nCI=0.8222.580.822(10.822)7396CI=0.8220.0115(0.8105to0.8335)

As a result, the $99$percent confidence interval ranges from $0.8105$to $0.8335$.

Part an of this activity is the same.

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

The $99$percent confidence interval for portion (c) is $0.8105$to $0.8335$.

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

So,

localid="1653237066493">E=0.8335−0.81052=0.0115.

This margin of error is $0.0115$, which is lower than the $0.015$margin of error given in component (b).

Similar to part (a) of this excersice

$n=0.25\left(za2E\right)2$yields a (1) level confidence interval for a population with a margin of error of at most E, sample size.

$=0.01$for a confidence level of $99$percent.

So,

localid="1653236901458">za/2=z0.005=2.58.

localid="1653237032520">E=0.015.

Given, n=7396,p^=0.85.

=0.01 with a 95% degree of confidence.

So,

$n=7396,p=0.85$$z\alpha /2=z0.005=2.58$.

$CI=pz/2p\left(1p\right)$will be the $99$percent confidence interval.

localid="1653236845785">n⇒CI=0.85±2.580.85(1−0.85)

$7396\Rightarrow CI=0.85\pm 0.0107\Rightarrow CI=0.8393$to $0.8607$

As a result, the confidence interval for the $99$th percentile is $0.8393$to 0.8607.

Divide the width of the confidence interval by $2$to calculate the margin of error.

So, $E=0.8607-0.83932=0.0107$.

Similar to part (a) of this excersice

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 $0.4196,E=0.0196$