220k views
0 votes
Can you please help me solve this question thank you

Can you please help me solve this question thank you-example-1

1 Answer

4 votes

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:


p-Z_{(1-(\alpha)/(2))}\cdot\sqrt[]{(p(1-p))/(n)}\leq P\leq p+Z_{(1-(\alpha)/(2))}\cdot\sqrt[]{(p(1-p))/(n)}

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:


\alpha=1-\frac{\text{ confidence wanted}}{100}=1-(90)/(100)=0.1

Then we have to find the Z value that is giving in tables. Since we have:


Z_{(1-(\alpha)/(2))}=Z_{(1-(0.1)/(2))}=Z_((0.95))

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:


\begin{gathered} 0.000438-1.65\cdot\sqrt[]{(0.000438(1-0.000438))/(420079)}\leq P\leq0.000438+1.65\cdot\sqrt[]{(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}

This interval written with percentages is:


0.03847\leq P\leq0.04913

Can you please help me solve this question thank you-example-1
User Neekoy
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories