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An article in the Chicago Tribune (August 29,1999 ) reported that in a poll of residents of the Chicago suburbs, \(43 \%\) felt that their financial situation had improved during the past year. The following statement is from the article: "The findings of this Tribune poll are based on interviews with 930 randomly selected suburban residents. The sample included suburban Cook County plus DuPage, Kane, Lake, McHenry, and Will Counties. In a sample of this size, one can say with \(95 \%\) certainty that results will differ by no more than 3 percent from results obtained if all residents had been included in the poll." Comment on this statement. Give a statistical argument to justify the claim that the estimate of \(43 \%\) is within \(3 \%\) of the true proportion of residents who feel that their financial situation has improved.

Short Answer

Expert verified
The statistical argument is that the margin of error in the given poll, based on a 95% confidence level and a sample size of 930, indeed ensures that the estimate of 43% is within 3% of the true proportion. However, interpreting this, it signifies that if the same polling process was repeated many times, about 95% of the times the result would lie within the specified 3% margin of the true population proportion.

Step by step solution

01

Define the parameters

The sample proportion (p̂) is given as 0.43, and the sample size (n) is 930.
02

Compute the Standard Error

The standard error (SE) can be calculated using the formula SE = sqrt((p̂*(1-p̂))/n), which measures the variability or dispersion of the sample proportion from the population proportion.
03

Calculate the Confidence Interval

A confidence interval can be found by the formula: p̂ ± Z*(SE), where Z is the z-score corresponding to the desired level of confidence. In this case, for a 95% confidence interval, the Z value is approximately 1.96.
04

Verify the margin of error

The margin of error is the amount by which the estimate from the sample can really differ from the population. It is a multiple of the standard error and in this case needs to be within 0.03 (or 3%)

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