Worksheet 17: The Normal Approximation#
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For the distribution shown on the slides, a. What is the confidence level for the interval \([\hat{\mu} - 1, \hat{\mu} + 1]\)? That is, what is the probability that \(|\hat{\mu} - \mu| \le 1\)?
b. What is the size \(a\) of the confidence interval with confidence \(0.8\)? That is, how large do we need to take \(a\) so that \(\Pr[|\hat{\mu}-\mu| \le a] \ge 0.8\)?
The 68-95-99 rule: So as long as \(n\) is large enough: \(\Pr[|\hat{\mu_n}-\mu|\le C \frac{\sigma}{\sqrt{n}}] \ge \begin{cases}.68 & C = 1\\ .95 & C = 2\\ .997 & C = 3\end{cases}\)
In the poll scenario, how large should I take \(n\) to be 95% sure that my estimate is within \(\frac{1}{10}\) of the truth?
I was only able to poll \(n = 49\) people. How confident am I that I am within \(\frac{1}{10}\) of the truth?
The 68-95-99 rule: So as long as \(n\) is large enough: \(\Pr[|\hat{\mu_n}-\mu|\le C \frac{\sigma}{\sqrt{n}}] \ge \begin{cases}.68 & C = 1\\ .95 & C = 2\\ .997 & C = 3\end{cases}\)
I only polled \(n = 49\) people. What is the error \(a\) for my \(95\%\) confidence interval?
In the microplastics scenario, suppose I know \(\sigma_x = 10\).
a. How large should I take \(n\) to be 99% sure that my sample mean is within \(\frac{1}{10}\) of the true microplastics concentration?
b. Suppose I only took \(n = 25\) samples. What is the size \(a\) of the error for my \(68\%\) confidence interval?