Expected Counts in Chi-Squared Goodness-of-Fit Tests of Normality

Question

I have a variable with of 200 values that I would like to test for normality using the Chi-square Goodness of Fit test. To do this, I have to calculate, for each value, the expected value in a normal distribution. How would I calculate it?

Answer 1

Shapiro-Wilk test of normality. If you have your choice of goodness-of-fit tests, I think you might get better results with the Shapiro-Wilk test. Here is an example with $n = 200$ observations from $\mathsf{Norm}(\mu=100,\sigma=15).$ This procedure tests whether the data are consistent with some normal distribution.

Using R:

set.seed(2020)           # for reproducibility
x = rnorm(200, 100, 15)  # generate normal data
summary(x);  sd(x)       # data summary
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  54.15   89.03  101.01   99.95  110.91  148.02 
[1] 16.938
shapiro.test(x)          # test for normality

        Shapiro-Wilk normality test

data:  x
W = 0.99546, p-value = 0.8155

Normal probability plot. A normal probability plot (normal quantile-quantile plot) provides an informal way to judge normality of a dataset. The emirical CDF of the sample is transformed so that points for a normal sample should lie approximately in a straight line. Here is an example, using the same data as above.

qqnorm(x);  qqline(x, col="green")

Kolmogorov-Smirnov Test. If you want to know whether data are consistent with the particular normal distribution $\mathsf{Norm}(\mu=100,\sigma=15)$ with the specified population mean and standard deviation, then you might use a Kolmogorov-Smirnov GOF test (as implemented in R) as follows:

ks.test(x, pnorm, 100, 15)  # specified parameters

        One-sample Kolmogorov-Smirnov test

data:  x
D = 0.042461, p-value = 0.8636
alternative hypothesis: two-sided

Results for nonnormal data. We show results of the Shapiro-Wilk test for samples from normal and exponential populations.

set.seed(407)
u = runif(200, 0, 10)
summary(u); sd(u)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
0.02289 2.32162 4.74545 4.94375 7.64946 9.96443 
[1] 3.02981
shapiro.test(u)

        Shapiro-Wilk normality test

data:  u
W = 0.94561, p-value = 7.288e-07

set.seed(408)
v = rexp(200, 1/10)
summary(v); sd(v)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
 0.03374  2.69530  6.46082  9.81964 13.56134 52.08287 
[1] 9.959839
shapiro.test(v)

        Shapiro-Wilk normality test

data:  v
W = 0.81954, p-value = 1.773e-14

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Chi-squared GOF tests. For a chi-squared test on possibly normal data, there are two cases to consider.

(a) Parameters unspecified. Test whether the data belong to some unspecified normal distribution. (This is what the Shapiro-Wilk test does.) See this page for details.

(b) Parameters specified. Test whether the data are specifically $\mathsf{Norm}(100, 15).$ (This is analogous to the Kolmogorov-Smirnov test.)

Then you could sort the data into perhaps $k = 10$ bins (categories) with boundaries at the deciles of $\mathsf{Norm}(100, 15).$ and determine the count $X_i$ in bins $i = 1, 2, \dots, 10.$ Using the deciles to make bins, you would expect $E = 200/10 = 20$ observations in each bin. Then the chi-squared GOF test statistic is $Q = \sum_{i=1}^{10} \frac{(X_i - E)^2}{E},$ which has approximately a chi-squared distribution with degrees of freedom $\mathrm{df} = k - 1.$

Note: In general, you should have $E > 5$ and you would need to choose the number $k$ of bins accordingly. If you choose arbitrary bins (not based on equally spaced percentiles), then you would need to find the probability $p_i$ according to the hypothetical normal distribution in each bin to get expected counts $E_i = np_i.$