Question

A study of 420,026 cell phone users found that 136 of them developed cancer of the brain or nervous system. Prior to their study of cell phone use, the rate of such cancer was found to be 0.0251 % 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. (Round to three decimal places as needed). b. Do cell phone users appear to have a rate of cancer of the brain or nervous system that is different from the rate of such cancer among those not using cell phones? Why or why not?

Answer 1

Answer:

a) The 90% confidence interval is:

$0.0003237 \leq \pi \leq 0.0003239$

b) Yes. Because the proportion of cancer rate for the cell phone users population is bigger than 0.00324%, which is over the rate of non-cell phone users (0.0251%).

Step-by-step explanation:

a) We will construct the 90% confidence interval based on the information given by the sample taken in this study.

The sample proportion is:

$p=\frac{X}{n}=\frac{136}{420,026} =0.000324$

The standard deviation is estimated as:

$\sigma=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.000324*0.999676}{420,026}}=\sqrt{7.7\cdot 10^{-10}} = 0.000028$

As the sample size is big enough, we use the z-score. For a 90% CI, the value of z is z=1.645.

The margin of error of the CI is:

$E=z\sigma/\sqrt{n}=1.645* 0.000028 /\sqrt{420,026}\\\\E= 0.000046 /648= 0.00000007$

The 90% CI is:

$p-z\sigma/\sqrt{n}\leq \pi\leq p+z\sigma/\sqrt{n}\\\\0.000324- 0.00000007 \leq\pi\leq 0.000324+ 0.00000007\\\\ 0.0003237 \leq \pi \leq 0.0003239$