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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, 256said that they were sitisfied and the remaining 769said they were not. Construct and interpret a90%confidence interval for the proportion of adults who are satisfied with how things are going. Follow the four-step process.

Short Answer

Expert verified

90%confident that the true proportion of adults who are satisfied with how things are going is between0.2276and0.2720or between22.76%and27.20%

Step by step solution

01

Given Information

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 proportionpat the 99%confidence level.

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

02

Explanation

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

Large counts condition: Satisfied, because np^=Number of successes =25610and n(1-p^)=Number of failures=76910

We note that all three conditions are satisfied.

localid="1650246693041" p^=xn=256256+769=25610250.2498

For confidence level 1-α=90%=0.90,

determine zα/2=z0.05using table A (look up 0.05in the table, the z-score is then the found z-score with the opposite sign:

localid="1650246840901" zα/2=z0.05=1.645

The margin of error is then:

localid="1650246864036" E=zα/2·p^(1-p^)n=1.645·0.2498(1-0.2498)10250.0222

The confidence interval is then:


localid="1650246674872" 0.2276=0.2498-0.0222=p^-E<p<p^+E=0.2498+0.0222=0.2720

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