# 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?

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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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,
$$\displaystyle{n}={\left({\frac{{{Z}_{{{\frac{{\alpha}}{{{2}}}}}}}}{{{E}}}}\right)}^{{{2}}}\hat{{{P}}}{\left({1}-\hat{{{P}}}\right)}$$
$$\displaystyle={\left({\frac{{{1.96}}}{{{0.03}}}}\right)}^{{{2}}}{0.5}\times{0.5}$$
$$\displaystyle={1067.11}$$
$$\displaystyle={1068}$$