# Test fit of discrete data distribution to normal distribution

Here are some grades. (They aren't test scores. They aren't a sample.) I want to test whether a normal distribution derived from the mean and variance of the grades provides a good fit for the grades. What other distribution might fit is, for now, an orthogonal issue. The question is whether the normal distribution provides a valid model.

> grades
 NA 75 70 70 80 75 85 85 75 85 75 80 85 80 75 70 70 85 90 90 NA 85 NA 75 80
 80 85 80 85 85 75 NA 75 70 80 65 80 75 70 80 85 65 85 75 75 85 75 60 85 80
 85 85 70 70 80 75 80 80 NA 80 75 85 80 80 55 75 60 90 80 75 80 70 85 75 80
 75 80 75 75 70 85 85 80 80 75 70 NA 75 70 75 70 65 85 70 70 80 80 85 85 70
 75 70 75 75 65 70 80 75 65 65
> t
55 60 65 70 75 80 85 90
1  2  6 18 27 25 22  3


The grades come from a discrete set of possible values from five to 100 in increments of five. Because of that, a Chi-squared test might be appropriate. First, the normal distribution probabilities that any of the discrete values will occur:

> p <- dnorm(seq(55,90,5), mean(grades, na.rm=TRUE), sd(grades, na.rm=TRUE))
> p
 0.0004433536 0.0031786310 0.0136901824 0.0354207364 0.0550535301
 0.0514034173 0.0288322176 0.0097150119


Why are the values of p so small? Why do they not add-up to one?

> tf <- as.data.frame(t)
> tf
1     55    1
2     60    2
3     65    6
4     70   18
5     75   27
6     80   25
7     85   22
8     90    3
> chisq.test(tf$Freq, p=p, rescale.p=TRUE) Chi-squared test for given probabilities data: tf$Freq
X-squared = 7.0452, df = 7, p-value = 0.4242

Warning message:
In chisq.test(tf$Freq, p = p, rescale.p = TRUE) : Chi-squared approximation may be incorrect  Why does the chisq.test function produce the warning? I need numbers, because I have a lot of these data sets and want fit results for a large sample of them. Visual inspection isn't proof. The closest relevant question I've found is https://stats.stackexchange.com/a/140576/59460 ("Testing Whether a Binomial Distribution Fits Data") • dnorm gives the value of the PDF. The value of the PDF at x is NOT the probability that x will occur. Since the normal distribution is continuous, the probability that any particular value will occur is zero. That’s why the values don’t add up to one. They are not probabilities. The reason they’re so small is that the value of the PDF at$x= \mu$is about$0.4/ \sigma$. – Joe Aug 14 '19 at 1:26 ## 2 Answers Data. You have $$n = 103$$ 'grades' with values $$v = (55. 60, 65, \dots, 90)$$ occurring with respective frequencies $$f = (2, 6, 18, 27, 25, 22, 3).$$ Treating this as a sample, we use $$A =\bar X = \frac 1n \sum_{i=1}^7 f_iv_i,$$ where $$n = \sum_{i=1}^7 f_i = 103,$$ so that the sample mean $$A = \bar X = 77.039$$ estimates the mean $$\mu$$ of a population. v = c(60, 65, 70, 75, 80, 85, 90) f = c( 2, 6, 18, 27, 25, 22, 3) a = sum(v*f)/sum(f); a  77.03883  Similarly, the sample variance is $$S^2 = \frac{1}{n-1}\sum_{i=1}^7 f_i(v_i - a)^2 = 44.822$$ so that the population variance $$\sigma^2$$ is estimated by $$44.822.$$ Also, the population standard deviation is estimated by $$S = \sqrt{44.822} = 6.695.$$ n = sum(f); sum(f*(v-a)^2)/(n-1)  44.82201  Shapiro-Wilk test of normality. The 103 observations are not normally distributed, because they have been rounded to only seven different values. For data not so severely rounded, an excellent test of normality would be the Shapiro-Wilk test. But your 103 observations fail this test with a P-value $$0.00012$$ far below $$0.05.$$ x = rep(v, times=f) shapiro.test(x) Shapiro-Wilk normality test data: x W = 0.9384, p-value = 0.0001217  Chi-squared GOF test. However, one can still ask whether the unrounded values might be consistent with the normal distribution $$\mathsf{Norm}(\mu =77.039, \sigma=6.695).$$ Here is a histogram where the bins are chosen to put the observed values at the bin centers. The fit isn't excellent, but a chi-squared goodness-of-fit (GOF) test may not reject this normal distribution as a possible fit. [See Addendum for R code.] A difficulty with the GOF test is that it requires expected count larger than $$5$$ in each of the intervals, and this normal distribution doesn't assign enough probability to the first and last intervals. Thus we are left with five intervals: $$(-\infty, 67.5],$$ $$(67.5, 72.5],$$ $$(72.5, 77,5],$$ $$(77.5, 82.5],$$ $$(82.5, \infty).$$ We need to know the probability that $$\mathsf{Norm}(77.039, 6.695)$$ assigns to each of the five intervals. That can be found using the normal CDF, called pnorm in R. (This is where you had trouble in your Question.) int.end = c(-Inf, 67.5, 72.5, 77.5, 82.5, Inf) int.prb = diff(pnorm(int.end, 77.04, 6.695)) sum(int.prb)  1  Then, according to the normal model we can find the expected counts $$E_i$$ in each interval by multiplying by $$n = 103.$$ We combine observed counts $$O_i$$ for the first and last intervals. int.exp = n*int.prb int.obs = c(8, 18, 27, 25, 25)  Here is a table of the observed and expected counts, which are used to find the chi-squared GOF statistic. cbind(int.obs, int.exp) int.obs int.exp [1,] 8 7.93993 [2,] 18 17.69146 [3,] 27 28.68967 [4,] 25 27.31845 [5,] 25 21.36049  The chi-squared GOF statistic $$Q = \sum_{i=1}^5 \frac{(O_i - E_i)^2}{E_i} = 0.922$$ has approximately a chi-squared distribution with $$\nu = 5 - 3$$ degrees of freedom. The P-value 0.922 is far above 5% so we do not reject the normal model. Rejection would have required $$Q$$ to exceed the critical value $$c = 5.99.$$ q = sum((int.obs - int.exp)^2/int.exp); q  0.9222296 1 - pchisq(q, 2)  0.6305803 qchisq(.95,2)  5.991465  If you have many of these GOF fit tests to do, you will want to find convenient R code for doing the whole process at once. Some of the code suggested in @Joe's Answer (+1) may be helpful. Just remember that you need to check that expected counts exceed 5, and combine bins of they don't. Approximate methods. Finally, here is an approximate method for using the Shapiro-Wilk test. As we saw at the beginning of this Answer, this test rejected your normal model, likely because the observations are rounded to only a few integers. If you add random noise to these observations, roughly speaking, to 'unround' them, then they pass the Shapiro-Wilk test. In your case random noise from the distribution $$\mathsf{Unif}(-2.5, 2.5)$$ should suffice. Then we have approximate P-value 0.318 and the normal model is not rejected. (You will get a somewhat different answer each time you try this random approximation: subsequent runs gave P-values 0.1856 and 0.4348.) set.seed(1234) shapiro.test(x + runif(n, -2.5, 2.5)) Shapiro-Wilk normality test data: x + runif(n, -2.5, 2.5) W = 0.98538, p-value = 0.318  If you have many of these tests to do, you might try this approximate procedure as a screening device, and then check cases where normality is rejected with the chi-squared GOF test. Looking at histograms and normal probability plots might also give quick view. Here is a normal probability plot of your data.  qqnorm(x); qqline(x, col="red") If the 'clusters' touch the red line (except possibly for the smaller first and last ones), you might take that as some evidence of a fit to normal. Addendum per Comments: Here is R code for the figure above with histogram of data and density curve. The first few lines make sure the data is in memory. v = c(60, 65, 70, 75, 80, 85, 90) f = c( 2, 6, 18, 27, 25, 22, 3) x = rep(v, times=f) cutp = seq(60-2.5, 90+2.5, by = 5) hist(x, prob=T, br=cutp, col="skyblue2", xlim=c(40,110), ylim=c(0,.06)) curve(dnorm(x, 77.03, 6.695), add=T, lwd=2, col="red")  • Perfect. I've reproduced it all, with exception of the "Histogram of x" plot, which beautifully illustrates and I could use it. The suggestion to permute the data with noise, in order to use the continuous tests is a welcome addition. In the Chi-squared result interpretation, how are the degrees of freedom determined for use with the 0.95 certainty to get the rejection threshold, as qchisq(.95,2)? Why was it five minus three? Aug 14 '19 at 16:57 • A chi-sq test for which probabilities in all$k$categories are known has$k-1$degrees of freedom. If you have to use the data to estimate parameters, then each estimation 'costs' 1 df. Here we estimated$\nu$and$\sigma$so that$5 - 1 - 2 = 2$df. Aug 14 '19 at 17:26 • The code for the histogram required some fussing to accommodate both histogram and density curve. I have resurrected it from my R session and posted it as an Addendum. // In previous Comment$\nu$should be$\mu.$Aug 14 '19 at 17:54 I am not experienced enough to comment on the appropriateness of this technique, but here's help with your code: grades <- rep(seq(55,90,5), c(1,2,6,18,27,25,22,3)) t <- table(grades) my.cdf.vals <- c(rep(0,8),1) my.cdf.vals[2:8] <- pnorm(seq(55,85,5)+2.5, mean(grades), sd(grades)) p <- rep(0,8) for (i in 1:8) { p[[i]] <- my.cdf.vals[[i+1]] - my.cdf.vals[[i]] } chisq.test(as.data.frame(t)$Freq, p=p, simulate.p.value =TRUE)


For the PMF, I used the normal distribution probability of a value being below 57.5 (with the mean and sd from the data) as being the probability of a grade equal to 55. That technique might not be preferred if grades can actually be below 55 (such as 50, 45, etc.). Similarly, I used the probability of a grade being above 87.5 from the normal distribution to make the probability of the grade being equal to 90 in the PMF.

The output that I got was:

Chi-squared test for given probabilities
with simulated p-value (based on 2000
replicates)

data:  as.data.frame(t)\$Freq
X-squared = 7.1553, df = NA, p-value = 0.3793

• Some nice code here (+1), but you need to screen for intervals with expected values that are too small. Aug 14 '19 at 5:27
• Thanks, I wasn’t sure if the technique was appropriate. Your answer clearly explains all of the considerations necessary.
– Joe
Aug 14 '19 at 9:37
• I poked-around on CRAN to find a solution for screening the intervals for too small number of expected values. "cluster" almost fit, but not. Ended up writing one. github.com/wbreeze/acd_stats/tree/master/prechi Aug 23 '19 at 21:34