Question

1. in a poll taken in march of 2007, gallup asked 1006 national adults whether they were baseball fans. 36% said they were. a year previously 37% of a smaller size sample had reported being baseball fans.

a) find the margin of error for the 2007 poll if we want 90% confidence in our estimate of the percent of national adults who are baseball fans.
b) explain what the margin of error means.
c) if we wanted to be 99% confident, would the margin of error be larger or smaller?
d) find the margin of error for 99% confidence level.

Explanation:

Step1: Recall the formula for margin of error

The formula for margin of error \(E = z\sqrt{\frac{p(1 - p)}{n}}\) where \(p\) is the sample proportion, \(n\) is the sample size and \(z\) is the z - score corresponding to the confidence level.
For a \(90\%\) confidence level, the z - score \(z_{0.90}\) (using standard normal distribution tables or a calculator) is \(z = 1.645\). The sample proportion \(p=0.36\) and \(n = 1006\).

Step2: Calculate the margin of error for \(90\%\) confidence level

Substitute the values into the formula:

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Step3: Explain the meaning of margin of error

The margin of error means that we are \(90\%\) confident that the true proportion of national adults who are baseball fans is within \(0.0249\) (or \(2.49\%\)) of the sample proportion of \(0.36\) (or \(36\%\)).

Step4: Analyze the effect of confidence level on margin of error

The formula for margin of error \(E = z\sqrt{\frac{p(1 - p)}{n}}\). As the confidence level increases, the z - score increases. For a \(99\%\) confidence level, the z - score \(z_{0.99}=2.576\) (which is larger than \(z_{0.90} = 1.645\)). Since \(E\) is directly proportional to \(z\) (when \(p\) and \(n\) are constant), the margin of error will be larger for a \(99\%\) confidence level.

Step5: Calculate the margin of error for \(99\%\) confidence level

Using \(z = 2.576\), \(p = 0.36\) and \(n=1006\)

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Answer:

a) The margin of error for the \(90\%\) confidence level is approximately \(0.0249\) (or \(2.49\%\)).
b) We are \(90\%\) confident that the true proportion of national adults who are baseball fans is within \(2.49\%\) of the sample proportion of \(36\%\).
c) The margin of error would be larger for a \(99\%\) confidence level.
d) The margin of error for the \(99\%\) confidence level is approximately \(0.0389\) (or \(3.89\%\)).