The previous post reported that a study found a 95% confidence interval for the the area of the Mandelbrot set to be 1.506484 ± 0.000004. What was the sample size that was used to come to that conclusion?

A 95% confidence interval for a proportion is given by

hat{p} pm 1.96 sqrt{frac{hat{p} ,(1 - hat{p})}{n}}

and so if a confidence interval of width w is centered at the proportion estimate p hat, then we can solve

frac{w}{2} = 1.96 sqrt{frac{hat{p} (1 - hat{p})}{n}}

to find

n = 15.37 frac{hat{p}(1 - hat{p})}{w^2}

Now in the example at the top of the post, we’re randomly sampling points from the square [−2, 2] × [−2, 2] and counting how many land inside the Mandelbrot set. The square has area 16, so p hat equals 1.506484 / 16. The width of the confidence interval for the area is 8 × 10−6. This means the width of the confidence interval for the proportion is 5 × 10−7, and so n = 5.244 × 1012.

Doubts regarding Mandelbrot area

The reasoning above is correct for inferring sample size from a confidence interval, but the specific application to the Mandelbrot set is suspect. Did someone really do over a trillion iterations? Apparently not.

As far as I can tell, this was the source of the estimate above. It’s not based on random samples but rather on regular grids of samples, treated as if they were random samples, and then extrapolated to get the estimate above. The result may be correct, but it’s not a simple Monte Carlo estimate.

 

 

The post Inferring sample size from confidence interval first appeared on John D. Cook.

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