Goodness-of-fit tests

The chi-squared statistic

Multiply each class's model probability by the total to get its expected frequency, then accumulate:

Pearson Edexcel 9FM0 formulae booklet: included, under Non-parametric tests, for goodness-of-fit and contingency tables alike, with percentage points of χ² tabulated separately. Squaring makes over- and under-shoots count alike. Dividing by E makes a gap of 3 in a class expecting 5 count for far more than the same gap in a class expecting 500. A large value means the model fits badly. The test never proves anything either way, since it only compares the statistic with a critical value and reports evidence.

Degrees of freedom and pooling

Start with the number of classes. Subtract 1, because the frequencies must total the sample size. Then subtract 1 more for every parameter estimated from the data. Fitting a Poisson model with λ taken from the sample mean costs a degree of freedom. Fitting one with λ specified in advance does not. A binomial with both n and p given costs nothing extra; a binomial with p estimated costs one.

Small expected frequencies weaken the chi-squared approximation, so small classes are pooled with their neighbours before the statistic is computed. Check the expected-frequency condition your specification states; Pearson Edexcel 9FM0 uses a minimum expected frequency of 5. Combine classes only where the combined class remains meaningful; otherwise the approximation may be unsuitable and a different test may be needed. Pooling is a separate step from estimating parameters: the class count used for the degrees of freedom is the pooled one, and each estimated parameter then removes one further degree of freedom.

A Poisson fit from start to finish

A complete fit runs in four steps: estimate λ from the data, build the expected frequencies, pool the tail, then test. Work down the table in one pass and keep the expected values to two decimal places until the final statistic.

Read the last row of the observed table before totalling anything. A table that stops at a stated value, four flaws here, records exactly how many events there were, so the sample mean is available and the fitted λ is that mean. A table whose last row reads '4 or more' does not. Counting each of those observations as a bare 4 gives a lower bound for the total and a lower bound for λ, never the mean, and every expected frequency built on it inherits that bias. In that case the total or λ itself must be supplied. The model's final class is 'x or more' either way, because probability leaks past the largest value seen and the expected frequencies must total n.

Describe the drawing

Five pairs of bars compare the observed flaw counts with the expected frequencies from a fitted Poisson of mean one point two, and the pairs are almost the same height everywhere. The last class expects only three point four, below five, so a coral bracket pools it with its neighbour into a class expecting twelve point nought five. Each surviving cell's squared gap over its expectation is printed beneath it, and the four contributions add to the tiny statistic nought point nought nine seven.

When the model is continuous

The fits above are all to discrete models, where a class is a single value and its probability is read straight off the distribution. A continuous model attaches no probability to a value, so a class has to be an interval and its probability is the area under the density across it, which is F(b) − F(a) for a class running from a to b. Multiply that by the total for the expected frequency, and the rest of the test proceeds unchanged: pool anything under 5, count the classes that survive, subtract for estimated parameters.

Two points of care. The outer classes are opened out to run to −∞ and to +∞, or the probabilities total less than 1 and the expected frequencies total less than n. And the boundaries used are the true ones: masses recorded to the nearest gram in a class labelled 45 to 49 occupy 44.5 to 49.5, and the model probability is taken between those. A normal model whose mean and standard deviation are both estimated from the grouped data costs two degrees of freedom, one for each.

Opening the two outer classes to −∞ and +∞ is what makes the five class probabilities total 1. Each observed bar is close to the expected frequency under the fitted curve, which is why the statistic is small.

Describe the drawing

A normal curve of mean fifty and standard deviation five drawn over the 200 masses, with the three middle classes outlined as histogram bars of fifty-two, seventy-eight and forty-six. The area under the curve is divided at forty-two point five, forty-seven point five, fifty-two point five and fifty-seven point five, and the two outer areas are left open, running off the ends of the axis so that the five probabilities total one. Multiplying them by two hundred gives the expected frequencies printed beneath: thirteen point three six, forty-eight point three five, seventy-six point five nine, forty-eight point three five and thirteen point three six, which total two hundred. The bars and the areas are close everywhere, and the test statistic of one point two nine falls far short of nine point four eight eight.