1.3 Estimation of \( \pi \) From the probability \( p \) of free beer in \( \S 1.2 \), we can also estimate \( \pi \), because the relation \( p=\pi / 4 \) implies \( \pi=4 \times p \). (a) For sample size (or replications) \( n=100 \), find \( 95 \% \) confidence interval and the mean for pi estimation. (b) Repeat (a) for sample size \( n=500,1000,5000 \); (c) Plot mean, and confidence interval against n . Is the width of the interval increasing or decreasing as n increases? Hints For sample size \( n=100 \), we first generate the sample/simulation as in the dart game \( \S 1.2 \). Then we add another column "pi" for \( 4 * \) p. From the column 'pi', we can compute the statistics: mean, and \( 95 \% \) confidence interval (LB, UB). Repeat the above procedure for sample size \( n=500,1000,5000 \), and find the associated statistics (mean, LB, UB). Now we can put these data into a table, with four columns \( n \), mean, LB, UB. Use the first-column \( n \) as \( x \)-axis, and other columns as \( y \)-axis, we can plot the desired results.
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We need to estimate \(\pi\) using the relation \(p = \pi / 4\), which implies \(\pi = 4 \times p\). We will perform simulations to estimate \(p\) and then calculate \(\pi\) for different sample sizes \(n = 100, 500, 1000, 5000\). Show more…
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