So we need to build a confidence interval for the percentage of cell phone users who develop cancer of the brain or nervous system. Using P for this value we have that the confidence interval is given by:
[tex]p-Z_{(1-\frac{\alpha}{2})}\cdot\sqrt[]{\frac{p(1-p)}{n}}\leq P\leq p+Z_{(1-\frac{\alpha}{2})}\cdot\sqrt[]{\frac{p(1-p)}{n}}[/tex]
Where P is the percentage we want to estimate, p is the rate of cancer found in previous studies (0.0438%/100 in this case) and n is study sample size (here is 420079). The parameter alpha is given by the following equation:
[tex]\alpha=1-\frac{\text{ confidence wanted}}{100}=1-\frac{90}{100}=0.1[/tex]
Then we have to find the Z value that is giving in tables. Since we have:
[tex]Z_{(1-\frac{\alpha}{2})}=Z_{(1-\frac{0.1}{2})}=Z_{(0.95)}[/tex]
we must look for the value 0.95 in the table:
Which basically means that Z=1.65. Now let's substitute all of the values we get in the equation:
[tex]\begin{gathered} 0.000438-1.65\cdot\sqrt[]{\frac{0.000438(1-0.000438)}{420079}}\leq P\leq0.000438+1.65\cdot\sqrt[]{\frac{0.000438(1-0.000438)}{420079}} \\ 0.000438-5.327\cdot10^{-5}\leq P\leq0.000438+5.327\cdot10^{-5} \\ 0.0003847\leq P\leq0.0004913 \end{gathered}[/tex]
This interval written with percentages is:
[tex]0.03847\leq P\leq0.04913[/tex]
