91Ó°ÊÓ

When dealing with proportions in statistics, a sample proportion gives us an idea of what fraction of the sample has a particular characteristic. In this exercise, the characteristic in question is developing cancer of the brain or nervous system among Danish cell phone users. The sample proportion, denoted as \( \hat{p} \), is calculated by dividing the number of users who developed cancer by the total number of users surveyed. For instance, the sample proportion of 0.0321% is first converted into a decimal, resulting in \( \hat{p} = 0.000321 \). This proportion tells us that out of all the surveyed cell phone users, approximately 0.0321% have developed this type of cancer.

The standard error (SE) is a measure of the variability or spread of a sampling distribution. It helps us understand how much we expect our sample proportion to fluctuate around the true population proportion. The formula for standard error when estimating a proportion is given by:
\[ SE = \sqrt{ \frac{ \hat{p}(1 - \hat{p}) }{n} } \]
In our problem, \( \hat{p} = 0.000321 \) and the sample size \( n \) is 420,095. Plugging these values in, we get:
\[ SE = \sqrt{ \frac{0.000321(1 - 0.000321)}{420,095} } \approx 0.0000277 \]
This small SE indicates that there is very little variability in our sampling distribution, as we'd expect with such a large sample size. It provides the basis for calculating our confidence interval.

A z-score is a measure that describes the position of a sample proportion in terms of standard deviations away from the mean of the population. For constructing confidence intervals, we use the z-score to determine how many standard errors we need to move away from the sample proportion to capture the desired level of confidence. For a 90% confidence interval, the z-score is typically about 1.645.
The margin of error (ME) is calculated as:
\[ ME = z \cdot SE \]
Substituting our specific values, we have:
\[ ME = 1.645 \cdot 0.0000277 \approx 0.0000456 \]
This margin of error is then used to construct the confidence interval for the population proportion. The confidence interval is found by:
\[ CI = \hat{p} \pm ME \]
Substituting \( \hat{p} \) and ME, we find:
\[ CI = 0.000321 \pm 0.0000456 \]
Thus, our 90% confidence interval is approximately \[ 0.0002754 \leq p \leq 0.0003666 \], confirming that the true proportion of cell phone users who develop this type of cancer is likely within this range.