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I have the following data:

enter image description here

I would like to use the chi-squared goodness of fit test to test whether it comes from a normal distribution. How shall I go about it?

In particular,

  • I would like to know how to estimate the mean and standard deviation (or variance) of the normal distribution that I use to find expected frequencies for the test,
  • I would also appreciate a reference to a place where I can read more on the justification of the choice of the degree of freedom (that goes beyond the comment "you estimate two parameters, so reduce the degree of freedom by 2")

Ps. this question is from a high school syllabus, I am asking it, because I like to understand (at least partially) what I am teaching.

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    $\begingroup$ Here is one reference itl.nist.gov/div898/handbook/eda/section3/eda35f.htm. $\endgroup$ – user2974951 Sep 12 '19 at 8:08
  • $\begingroup$ Thank you for that link. Even though it mostly discusses what to do and not why it is justified, it led me to the R nortest library. In the documentation I found a comment about why the chi-squared test is not recommended for this type of investigation. This comment references an article by Moore, D.S.(1986) Tests of the chi-squared type. Does someone know an online copy of this article? $\endgroup$ – Ferenc Beleznay Sep 12 '19 at 8:53
  • $\begingroup$ The piece by Moore is a book chapter (in the 1986 book by D'Agostino and Stephens, Goodness of Fit Techniques). I don't know of a place you can legally access it, sorry, but one obvious reason the test is not recommended is its terrible power against most alternatives people tend to be interested in. $\endgroup$ – Glen_b Sep 12 '19 at 12:28
  • $\begingroup$ I did not mention in my previous comment, but apparently the Moore article also discusses, that the reduction of the degree of freedom by 2 (which, as I understand is routinely used as a rule of thumb) is not even correct. $\endgroup$ – Ferenc Beleznay Sep 12 '19 at 14:28
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    $\begingroup$ @whuber Thanks, it's a fair question. When the raw data are unavailable (which is rarer these days but it does sometimes happen), and only binned data is to hand, then typically interest is in more-or-less 'smooth' alternatives to the normal, in which case one might consider tests of the Neyman-Barton type. One may use orthogonal polynomials (for the normal, this would indicate Hermite polynomials) to partition a chi-squared statistic into lower order and higher terms, and use lower order terms. It's also possible to split off the lowest two (mean-variance) out of test statistic. ...ctd $\endgroup$ – Glen_b Sep 14 '19 at 3:28
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The best estimates of normal mean $\mu$ and variance $\sigma^2$ are the sample mean $\bar X$ and variance $S^2.$ In elementary courses, a common way to estimate these parameters from grouped data is to assume all observations in an interval are located at the center.

Thus, if frequencies are $f = (45, 55, 38, 27, 25, 10)$ and corresponding midpoints are $m = (5, 12.5, 17.5, 22.5, 27.5, 35),$ then the sample mean is $$A = \bar X = \frac{\sum_{j=1}^6 f_jm_j}{\sum_{j=1}^6 f_j} = \frac{\sum_{j=1}^6 f_jm_j}{n} = 16.1125$$

and the sample standard deviation is $$S = \sqrt{\frac{\sum_{j=1}^6 f_i(m_i - \bar X)^2}{n-1}} = 8.4647.$$

Computations from R statistical software are:

m = c(5,12.5,17.5, 22.5,27.5, 35)
f = c(45,55,38,27,25,10)
a = sum(f*v)/sum(f); a             # average
[1] 16.1125
sum(f)
[1] 200
s = sqrt(sum(f*(v - a)^2)/199); s  # standard deviation
[1] 8.464742

The $n = 200$ 'observations' (approximated as midpoints) can be be expressed in R as shown below and then the sample mean and standard deviation can be computed directly, with the same numerical results as above.

x = rep(m, f)
a = mean(x);  s = sd(x)
a; s
[1] 16.1125
[1] 8.464742

summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   5.00   12.50   15.00   16.11   22.50   35.00 

Using the interval boundaries, we can make a frequency histogram of the 200 observations. We also show the density function (black curve) of the 'best fitting' normal distribution with $\mu = \bar X, \sigma = S.$

As you can see, the fit is not very good. The data are noticeably skewed to the right. Specifically, the approximate mode is 12.5, smaller than the median 15, which is in turn smaller than the mean 16.11. Also, the median is nearer to the first quartile than to the third quartile.

cutp = c(0,10,15,20,25,30,40)
hist(x, br=cutp, col="skyblue2")
  curve(dnorm(x, a, s), add=T, lwd=2)
  curve(dgamma(x, 3.62, 0.2249), add=T, col="red")

enter image description here

For what it's worth, the density curve (red) of the skewed distribution $\mathsf{Gamma}(\alpha=3.62, \lambda = 0.2249)$ happens to fit the histogram somewhat better.

In order to do a chi-squared goodness-of-fit test of the data to the best fitting normal distribution, we need to find probabilities corresponding to the histogram bars, and from them expected counts for the various intervals. Below, I have used the convention of letting the lowest and highest intervals extend to $-\infty$ and $\infty,$ respectively, in order to capture the full probability $1$ under the normal curve.

The chi-squared statistic is $Q = 8.713,$ which exceeds the critical value $c = 7.9147$ for a test at the 5% level of significance. The P-value of the test is 0.333 < 0.05. So the null hypothesis that the data are from a normal distribution is rejected.

cutp1 = c(-Inf,10,15,20,25,30,Inf)
diff(pnorm(cutp1, a,  s))
[1] 0.23511251 0.21260606 0.22925697 0.17615241 0.09643483 0.05043724
exp = diff(pnorm(cutp1,16.1124,8.4547))*200;  exp
[1] 46.97052 42.56178 45.90282 35.24831 19.26978 10.04679 # Expected counts
q = sum((f-exp)^2/exp)
[1] 8.712543         # Chi-squared statistic
1 - pchisq(q, 3)
[1] 0.0333673        # P-value < 0.05; Reject null hypothesis (normality)
qchisq(.95, 3)
[1] 7.814728         # Critical value of test at 5% level

Notes: (1) You asked about the computation of the degrees of freedom for the chi-squared distribution. It is $6 - 1 - 2 = 3.$ As you say a degree of freedom is subtracted for each parameter estimated. @Glen_b has discussed this briefly, and @whuber has provided a relevant link.

(2) Also, @whuber has mentioned the possible inflation in the estimated variance that results from using binned, instead of original, data. However, for the particular problem at hand, it seems to me that rejecting $H_0$ is correct, especially on account of the skewness of your data, mentioned earlier.

Here is a brief simulation illustrating that rounding data to the nearest 10 can increase standard deviation:

set.seed(913)
m = 10^5; n = 200;  s = s.r = numeric(m)
for(i in 1:m) {
 x = rnorm(n, 50, 10)
 s[i] = sd(x);  s.r[i] = sd(round(x/10)*10) }
mean(s); mean(s.r)
[1] 9.989076   # mean of actual SDs
[1] 10.39652   # mean of SDs for rounded data

(3) Here are histograms of nine samples of size $n=200$ from $\mathsf{Norm}(16,7)$ along with the normal density curve. Generally speaking, such normal samples seem to fit the normal density better than your sample fits the normal density in your example above. Also, all of them have negative observations.

enter image description here

set.seed(123)
par(mfrow=c(3,3))
for(i in 1:9){
  x = rnorm(200, 16, 7)
  hist(x, prob=T, col="skyblue2", ylim=c(0,.06))
  curve(dnorm(x, 16, 8), add=T) }
par(mfrow=c(1,1))
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    $\begingroup$ The use of the chi-squared distribution in this application is theoretically justified when you use maximum likelihood estimation of the parameters, rather than using the unadjusted center values. (By "unadjusted" I mean you haven't applied Sheppard's corrections.) The degrees of freedom may be (quite) wrong when you don't follow this procedure carefully. See stats.stackexchange.com/a/17148/919 for details. (In this case it doesn't make a material difference: I obtain a chi-squared statistic of 8.71.) $\endgroup$ – whuber Sep 13 '19 at 18:09
  • $\begingroup$ @whuber. Thanks for link and mention of Sheppard's correction. Because bin widths vary in this example, it is difficult to know how to implement the correction. $\endgroup$ – BruceET Sep 13 '19 at 18:47
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    $\begingroup$ Good point. We might take that as a little bit more motivation to carry out the MLE. It's not hard, BTW. Here's R code to estimate $\mu$ and $\log\sigma:$ x <- c(0,10,15,20,25,30,40); n <- c(45,55,38,27,25,10); f <- function(theta) {-sum(log(diff(pnorm(x, theta[1], exp(theta[2])))) * n)}; fit <- nlm(f, c(20, log(10))) $\endgroup$ – whuber Sep 13 '19 at 19:09

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