Questions: A study of 420,095 cell phone users found that 0.0319% of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0332% for those not using cell phones. Complete parts (a) and (b). a. Use the sample data to construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system. % < p < % (Do not round until the final answer. Then round to three decimal places as needed.)

Solution Steps

We have a sample of \( n = 420095 \) cell phone users, and the proportion of users who developed cancer of the brain or nervous system is given as \( \hat{p} = 0.0319\% = 0.000319 \).

We are tasked with constructing a \( 90\% \) confidence interval for the proportion of cell phone users who develop cancer. The corresponding significance level is \( \alpha = 1 - 0.90 = 0.10 \).

Using the formula for the confidence interval for a single population proportion:

\[ \hat{p} \pm z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

where \( z \) is the z-score corresponding to the \( 90\% \) confidence level, which is approximately \( 1.645 \).

Substituting the values:

\[ \hat{p} \pm 1.645 \sqrt{\frac{0.000319(1 - 0.000319)}{420095}} \]

Calculating the margin of error:

\[ \sqrt{\frac{0.000319(1 - 0.000319)}{420095}} \approx \sqrt{\frac{0.000319 \cdot 0.999681}{420095}} \approx \sqrt{\frac{0.0003187}{420095}} \approx \sqrt{7.577 \times 10^{-9}} \approx 0.0000870 \]

Thus, the margin of error is:

\[ 1.645 \cdot 0.0000870 \approx 0.000143 \]

Now, we can calculate the confidence interval:

\[ (0.000319 - 0.000143, 0.000319 + 0.000143) = (0.000176, 0.000462) \]

Converting the confidence interval back to percentage form:

\[ (0.000176 \times 100\%, 0.000462 \times 100\%) = (0.0176\%, 0.0462\%) \]

Final Answer