Of 1000 randomly selected cases of lung cancer 843 resulted in death within 10 years. Construct a 95\% two-sided confidence interval on the death rate from lung cancer. How large must the sample if we wish to be at least 95\% confident that the error in estimating p is less than 0.03 regardless of the value of p?

UkusakazaL 2021-08-01 Answered
Of 1000 randomly selected cases of lung cancer 843 resulted in death within 10 years. Construct a \(\displaystyle{95}\%\) two-sided confidence interval on the death rate from lung cancer.
a) Construct a \(\displaystyle{95}\%\) two-sides confidence interval on the death rate from lung cancer. Round your answer 3 decimal places.
\(\displaystyle?\leq{p}\leq?\)
b) Using the point estimate of p obtained from the preliminary sample what sample size is needed to be \(\displaystyle{95}\%\) confident that the error in estimatimating the true value of p is less than 0.00?
c) How large must the sample if we wish to be at least \(\displaystyle{95}\%\) confident that the error in estimating p is less than 0.03 regardless of the value of p?

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Expert Answer

facas9
Answered 2021-08-09 Author has 13314 answers

Part (a)
Consider the \(\displaystyle{n}={1000}\) and \(\displaystyle{x}={843}\). Then, the point estimate is,
\(\displaystyle\hat{{{P}}}={1}-{0.843}\)
\(\displaystyle={0.157}\)
Then, at \(\displaystyle{95}\%\) confidence level.
\(\displaystyle\alpha={1}-{0.95}\)
\(\displaystyle\alpha={0.05}\)
\(\displaystyle{\frac{{\alpha}}{{{2}}}}={0.025}\)
\(\displaystyle{Z}_{{{\frac{{\alpha}}{{{2}}}}}}={Z}_{{{0.025}}}\)
\(\displaystyle{Z}_{{{\frac{{\alpha}}{{{2}}}}}}={1.960}\)
Then, the margin of the error is,
\(\displaystyle{E}={Z}_{{{\frac{{\alpha}}{{{2}}}}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)
\(=1.96\sqrt{\frac{0.843(0.157)}{1000}}\)
\(\displaystyle={0.023}\)
Thus, the \(\displaystyle{95}\%\) confidence interval for the population p is,
\(\displaystyle{0.843}-{0.023}\leq{p}\leq{0.843}+{0.023}\)
\(\displaystyle{0.820}\leq{p}\leq{0.866}\)
Part (b)
Consider the margin error is,
\(\displaystyle{E}={0.03}\)
Then, the sample size is,
\(\displaystyle{n}={\left({\frac{{{Z}_{{{\frac{{\alpha}}{{{2}}}}}}}}{{{E}}}}\right)}^{{{2}}}\hat{{{P}}}{\left({1}-\hat{{{P}}}\right)}\)
\(\displaystyle={\left({\frac{{{1.96}}}{{{0.03}}}}\right)}^{{{2}}}{0.843}\times{0.157}\)
\(\displaystyle={564.93}\)

\(\displaystyle={565}\)
Part(c) For p less than 0.03, the sample size is obtained as,
\(\displaystyle{n}={\left({\frac{{{Z}_{{{\frac{{\alpha}}{{{2}}}}}}}}{{{E}}}}\right)}^{{{2}}}\hat{{{P}}}{\left({1}-\hat{{{P}}}\right)}\)
\(\displaystyle={\left({\frac{{{1.96}}}{{{0.03}}}}\right)}^{{{2}}}{0.5}\times{0.5}\)
\(\displaystyle={1067.11}\)
\(\displaystyle={1068}\)

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