# Twifficiency Scores

**John Myles White » Statistics**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

Neil Kodner wrote a great post this morning about yesterday’s Twifficiency scores outbreak. He grabbed all the auto-tweeted scores he could find and plotted their distribution. I was struck by the asymmetry of the resulting distribution, which you can see below:

Thankfully, Neil handed me the raw data for his plot, so I was able to run a K-S test for normality, which rejected normality pretty easily, though I’m coming up with a tie that I’m surprised by:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | scores <- read.csv('twifficiencyscores.txt', header = FALSE) scores <- scores[,1] m <- mean(scores) s <- sd(scores) ks.test(scores, 'pnorm', m, s) # # One-sample Kolmogorov-Smirnov test # #data: scores #D = 0.0616, p-value < 2.2e-16 #alternative hypothesis: two-sided # #Warning message: #In ks.test(scores, "pnorm", m, s) : # cannot compute correct p-values with ties |

I suppose that I’m a bit worried that the p-value is simply a reflection of sample size here, since there are 7089 measurements. Would it be more compelling to bootstrap the D score from the K-S test on samples of 500 scores at a time to confirm that the non-normality is present even in small groups of scores?

Assuming that the data really has a skewed distribution, does anyone understand the scoring system well enough to say what produces the asymmetry?

**leave a comment**for the author, please follow the link and comment on their blog:

**John Myles White » Statistics**.

R-bloggers.com offers

**daily e-mail updates**about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.