2K10 begins In January $2010$, a Gallup Poll asked a random sample of adults, "In general, are you satisfied or dissatisfied with the way things are going in the United States at this time?" In all, $256$said that they were sitisfied and the remaining $769$said they were not. Construct and interpret a$90\%$confidence interval for the proportion of adults who are satisfied with how things are going. Follow the four-step process.

x= Number of successes $=256$

n=Sample size $=256+769=1025$

c=Confidence interval $=90\%=0.90$

State We are interested in estimating the population proportion$p$at the $99\%$confidence level.

$p=$The proportion of adults who are satisfied with how things are going.

For a population proportion p, we intend to calculate a one-sample z-interval.

Because the adults come from a random sample, the random requirement is satisfied.

Because the sample of $1025$adults represents less than $10\%$of the adult population, the $10\%$condition is satisfied.

Large counts condition: Satisfied, because $n\hat{p}=$Number of successes $=256\ge 10$and $n(1-\hat{p})=$Number of failures$=769\ge 10$

We note that all three conditions are satisfied.

localid="1650246693041" $$\begin{gathered} \hat{p}=\frac{x}{n}=\frac{256}{256+769} \\ =\frac{256}{1025} \\ \approx 0.2498 \end{gathered}$$

For confidence level $1-\alpha =90\%=0.90$,

determine $z_{\alpha /2}=z_{0.05}$using table A (look up $0.05$in the table, the z-score is then the found z-score with the opposite sign:

localid="1650246840901" $$\begin{gathered} z_{\alpha /2}=z_{0.05} \\ =1.645 \end{gathered}$$

The margin of error is then:

localid="1650246864036" $$\begin{gathered} E=z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ =1.645·\sqrt{\frac{0.2498(1-0.2498)}{1025}} \\ \approx 0.0222 \end{gathered}$$

The confidence interval is then:

localid="1650246674872" $$\begin{gathered} 0.2276=0.2498-0.0222 \\ =\hat{p}-E<p<\hat{p}+E \\ =0.2498+0.0222 \\ =0.2720 \end{gathered}$$