91影视

In a sample of 420095 Danish cell phone users, 0.0321% of them developed cancer of the brain or nervous system. The rate of cancer was found to be 0.0340% for those not using cell phones.

The sample proportion of cell phone users who developed cancer is computed below:

$$\begin{gathered} \hat{p}=0.0321\% \\ =\frac{0.0321}{100} \\ =0.000321 \end{gathered}$$

The sample proportion of cell phone users who did not develop cancer is computed below:

$$\begin{gathered} \hat{q}=1-\hat{p} \\ =1-0.000321 \\ =0.999679 \end{gathered}$$

a.

The given level of significance is 0.10

Therefore, the value of $z_{\frac{伪}{2}}$from the standard normal table is equal to 1.645.

The margin of error is computed as shown:

$$\begin{gathered} E=z_{\frac{伪}{2}}脳\sqrt{\frac{\hat{p}\hat{q}}{n}} \\ =1.645脳\sqrt{\frac{0.000321脳0.999679}{420095}} \\ =0.0000455 \end{gathered}$$

Therefore, the margin of error is 0.0000455.

a.

The 90% confidence interval is computed as follows:

$$\begin{gathered} \hat{p}-E<p<\hat{p}+E \\ 0.000321-0.0000455<p<0.000321+0.0000455 \\ 0.000276<p<0.000367 \\ 0.0276\%<p<0.0367\% \end{gathered}$$

Thus, the 90% confidence interval is equal to (0.0276%,0.0367%).

b.

Since confidence interval includes the value of 0.034%, there does not appear to be any difference betweenthe rate of cancer of the brain or nervous system that develops in cell phone users as compared to the rate of such cancer among those not using cell phones.