Question

In: Economics

Since1935,the Gallup organization–best known for the GallupPoll–has been a leader in the measurement and analysis of...

Since1935,the Gallup organization–best known for the GallupPoll–has been a leader in the measurement and analysis of people’s attitudes, opinions, and behavior. In a September 2003 survey, 1,200 registered voters were interviewed and asked: “California is facing a record 38-billion deficit. Do you think California should or should not raise sales taxes to reduce the state budget deficit?” Among those polled, 498 indicated they were in favor of raising sales taxes. *show work*

A) Compute a 99% confidence interval for the estimate of the proportion of voters who were in favor of raising sales taxes. Indicate both low-end and high-end estimates.

B)A smaller sample of 700 voters identified themselves as independent. Among the independent voters, 364 said they would support higher sales taxes. The survey cited a margin of sampling error of no more than 5 percentage points. Is this claim correct? Use the same confidence level as in part a.

Solutions

Expert Solution

a)

Sample size, n=1200

Observed value of p=fraction of respondents in favor of raising sales tax=498/1200=0.415

Observed value of q=fraction of respondents not in favor of raising sales tax=1-0.415=0.585

Now we calculate the standard error of the proportion,

A 99% confidence interval includes 49.5% of the area on the either side of mean in sampling distribution. If we look at the normal distribution are under the curve tables for 0.4950 area, we get

z=2.58

Lower end estimate =

Upper end estimate=

Confidence Interval

We can say that with 99% confidence level that proportion of population which is in favor raising the sales tax lies between 0.3783 and 0.4517.

b)

Sample size, n=700

Observed value of p=fraction of respondents in favor of raising sales tax=364/700=0.52

Observed value of q=fraction of respondents not in favor of raising sales tax=1-0.52=0.48

Now we calculate the standard error of the proportion,

A 99% confidence interval includes 49.5% of the area on the either side of mean in sampling distribution. If we look at the normal distribution are under the curve tables for 0.4950 area, we get

z=2.58

Margin of sampling error=

Margin of sampling error is less than the cited margin of up to 5%,

So, claim is incorrect.


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