A surveyor is measuring the height of a cliff known to be about 1000 feet. He assumes his instrument is properly calibrated and that his measurement errors are independent, with mean \(\mu = 0\) and variance \(\sigma^2 = 10\). He plans to take \(n\) measurements and form the average. Estimate, using (a) Chebyshev’s inequality and (b) the normal approximation, how large \(n\) should be if he wants to be 95 percent sure that his average falls within 1 foot of the true value. Now estimate, using (a) and (b), what value should \(\sigma^2\) have if he wants to make only 10 measurements with the same confidence?
Chebyeshev’s inequality which will be used in the answers is: \[ \begin{align} P\left(\left|\frac{S_n}{n} - \mu\right|\geq \epsilon\right)&\leq\frac{\sigma^2}{n\epsilon^2}\\ \end{align} \]
The normal approximation to the mean which will be used in the answers is: \[ \begin{align} P\left(\left|\frac{S_n}{n} - \mu\right| < \epsilon\right) = 2\Phi\left(\frac{\sqrt{n}\epsilon}{\sigma}\right) \end{align} \]
\[ \begin{align} P\left(\left|\frac{S_n}{n}\right|\geq 1\right)&\leq\frac{10}{n}\\ 0.05 &\leq \frac{10}{n}\\ n &\leq \frac{10}{0.05}\\ n &= \mathbf{200} \end{align} \]
\[ \begin{align} P\left(\left|\frac{S_n}{n} - \mu\right| < \epsilon\right) &= 2\Phi\left(\frac{\sqrt{n}\epsilon}{\sigma}\right)\\ 0.05 &= 2\Phi\left(\sqrt{\frac{n}{10}}\right)\\ 0.025 &= \Phi\left(\frac{\sqrt{n}}{\sqrt{10}}\right)\\ \Phi^{-1}\left(0.025\right) = 1.96 &= \frac{\sqrt{n}}{\sqrt{10}}\\ n &= 10\times 1.96^2 = 38.416 \Rightarrow\mathbf{39} \end{align} \]
\[ \begin{align} P\left(\left|\frac{S_n}{n}\right|\geq 1\right)&\leq\frac{\sigma^2}{10}\\ 0.05 &\leq \frac{\sigma^2}{10}\\ \sigma^2 &= \mathbf{0.5} \end{align} \]
\[ \begin{align} P\left(\left|\frac{S_n}{n} - \mu\right| < \epsilon\right) &= 2\Phi\left(\frac{\sqrt{n}\epsilon}{\sigma}\right)\\ 0.05 &= 2\Phi\left(\sqrt{\frac{10}{\sigma}}\right)\\ 0.025 &= \Phi\left(\sqrt{\frac{10}{\sigma}}\right)\\ 1.96 &= \frac{\sqrt{10}}{\sqrt{\sigma}}\\ \sqrt{\sigma} &= \frac{\sqrt{10}}{1.96}\\ \sigma^2 &= \mathbf{2.6} \end{align} \]