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137 lines
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5.1 KiB
Markdown
137 lines
No EOL
5.1 KiB
Markdown
---
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id: 2095
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title: Uniformity of Math.random()
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date: 2017-03-07T21:50:52+00:00
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author: Brandon Rozek
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aliases:
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- /2017/03/uniformity-math-random/
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permalink: /2017/03/uniformity-math-random/
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medium_post:
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mf2_syndicate-to:
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- 'a:1:{i:0;s:4:"none";}'
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mf2_cite:
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- 'a:4:{s:9:"published";s:25:"0000-01-01T00:00:00+00:00";s:7:"updated";s:25:"0000-01-01T00:00:00+00:00";s:8:"category";a:1:{i:0;s:0:"";}s:6:"author";a:0:{}}'
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format: aside
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tags: ["Statistics"]
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---
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There are many cases where websites use random number generators to influence some sort of page behavior. One test to ensure the quality of a random number generator is to see if after many cases, the numbers produced follow a uniform distribution.
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<!--more-->
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Today, I will compare Internet Explorer 11, Chrome, and Firefox on a Windows 7 machine and report my results.
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## Hypothesis
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H0: The random numbers outputted follow the uniform distribution
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HA: The random numbers outputted do not follow the uniform distribution
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## Gathering Data
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I wrote a small [website](http://share.zeropointshift.com/files/2017/03/random.html) and obtained my data by getting the CSV outputted when I use IE11, Firefox, and Chrome.
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The website works by producing a random number using <code class='language-javascript'>Math.random()</code> between 1 and 1000 inclusive and calls the function 1,000,000 times. Storing it’s results in a file
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This website produces a file with all the numbers separated by a comma. We want these commas to be replaced by newlines. To do so, we can run a simple command in the terminal
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```bash
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grep -oE '[0-9]+' Random.csv > Random_corrected.csv
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```
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Do this with all three files and make sure to keep track of which is which.
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Here are a copy of my files for [Firefox](https://brandonrozek.com/wp-content/uploads/2017/03/Firefox_corrected.csv), [Chrome](https://brandonrozek.com/wp-content/uploads/2017/03/Chrome_corrected-1.csv), and [IE11](https://brandonrozek.com/wp-content/uploads/2017/03/IE11_corrected.csv)
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## Check Conditions
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Since we’re interested in if the random values occur uniformly, we need to perform a Chi-Square test for Goodness of Fit. With every test comes some assumptions
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<u>Counted Data Condition:</u> The data can be converted from quantatative to count data.
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<u>Independence Assumption:</u> One random value does not affect another.
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<u>Expected Cell Frequency Condition:</u> The expected counts are going to be 10000
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Since all of the conditions are met, we can use the Chi-square test of Goodness of Fit
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## Descriptive Statistics
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For the rest of the article, we will use R for analysis. Looking at the histograms for the three browsers below. The random numbers all appear to occur uniformly
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```R
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rm(list=ls())
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chrome = read.csv("~/Chrome_corrected.csv", header = F)
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firefox = read.csv("~/Firefox_corrected.csv", header = F)
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ie11 = read.csv("~/IE11_corrected.csv", header = F)
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```
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```R
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hist(ie11$V1, main = "Distribution of Random Values for IE11", xlab = "Random Value")
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```
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```R
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hist(firefox$V1, main = "Distribution of Random Values for Firefox", xlab = "Random Value")
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```
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```R
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hist(chrome$V1, main = "Distribution of Random Values for Chrome", xlab = "Random Value")
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```
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## Chi-Square Test
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Before we run our test, we need to convert the quantatative data to count data by using the plyr package
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```R
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#Transform to count data
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library(plyr)
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chrome_count = count(chrome)
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firefox_count = count(firefox)
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ie11_count = count(ie11)
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```
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Run the tests
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```R
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# Chi-Square Test for Goodness-of-Fit
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chrome_test = chisq.test(chrome_count$freq)
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firefox_test = chisq.test(firefox_count$freq)
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ie11_test = chisq.test(ie11_count$freq)
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# Test results
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chrome_test
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```
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As you can see in the test results below, we fail to reject the null hypothesis at a 5% significance level because all of the p-values are above 0.05.
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##
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## Chi-squared test for given probabilities
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##
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## data: chrome_count$freq
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## X-squared = 101.67, df = 99, p-value = 0.4069
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`firefox_test`
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##
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## Chi-squared test for given probabilities
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##
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## data: firefox_count$freq
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## X-squared = 105.15, df = 99, p-value = 0.3172
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`ie11_test`
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##
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## Chi-squared test for given probabilities
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##
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## data: ie11_count$freq
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## X-squared = 78.285, df = 99, p-value = 0.9384
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## Conclusion
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At a 5% significance level, we fail to obtain enough evidence to suggest that the distribution of random number is not uniform. This is a good thing since it shows us that our random number generators give all numbers an equal chance of being represented. We can use `Math.random()` with ease of mind. |