Risk Modelling and Survival Analysis · Graduation and graduation tests

Graduation Tests: Goodness of Fit for Graduated Mortality Rates

Q: What is the difference between the cumulative deviations test and the chi-squared test?

The chi-squared test sums squared standardised deviations, so it detects deviations that are large overall. The cumulative deviations test sums raw deviations, so positive and negative cancel. It therefore detects consistent bias in one direction, which the chi-squared test can miss.

Q: What does the grouping of signs test detect?

It detects clumping, where positive and negative deviations come in long runs instead of mixing randomly. You count the positive groups and compare with the expected number given n1 positives and n2 negatives. Too few groups suggests clumping.

Q: How many degrees of freedom does the chi-squared test for graduation use?

Use the number of age groups minus the number of parameters fitted to the data. If the rates come from a standard table and no parameters were fitted, use the number of groups. Always state this in your answer.

Q: Is the signs test one-sided or two-sided?

It is two-sided. Bias can make too many or too few deviations positive. Under H0 the number of positives is Binomial(m, ½), and you reject if it is too far from the middle in either direction.

Q: If a graduation passes all tests, is it a good graduation?

It means there is no evidence of poor fit, bias or clumping from these tests. Smoothness must be tested separately, for example with the serial correlation and smoothness tests. You also need to judge whether the graduation is suitable for its purpose.

Updated 11 October 2026 · Fact-checked

Graduation tests check whether graduated mortality rates fit the crude rates well enough. Compute standardised deviations zx = (actual deaths − expected deaths) ÷ √(expected deaths). Then apply the chi-squared test for overall fit, the signs and cumulative deviations tests for bias, and the grouping of signs test for clumping.

Understand Graduation Tests: Goodness of Fit

A graduated mortality table smooths crude rates. It should do two things at once: stay close to the data and stay smooth. Graduation tests of goodness of fit deal with the first aim. Smoothness is tested separately.

Every test starts from the standardised deviation at each age x. You compare actual deaths θx with expected deaths Ex × μ̂x (or Ex × q̂x), where Ex is the central exposed to risk. Dividing by the square root of the expected deaths puts every age on the same scale. If the graduation is right, each zx is approximately N(0,1), and the zx are approximately independent.

Different tests look for different faults. The chi-squared test looks for overall poor fit, where deviations are too large in total. The individual standardised deviations test looks for single ages with a large deviation. The signs test and cumulative deviations test look for bias, where the graduation is mostly above or below the data. The grouping of signs test looks for clumping, where deviations of one sign come in long runs.

No single test finds every fault. A graduation can pass the chi-squared test and still be biased or clumped. That is why you apply several tests and state what each one can and cannot detect.

All these tests assume the zx are independent N(0,1) under the null hypothesis that the graduated rates are the true rates. Since the graduation uses the data, you reduce the degrees of freedom for fitted parameters. The tests also cannot say whether the graduation is smooth.

Key rules to remember

Standardised deviation: zx = (θx − Ex·μ̂x) ÷ √(Ex·μ̂x); θx is actual deaths, Ex·μ̂x is expected deaths. Under the null hypothesis zx is approximately N(0,1).
Chi-squared statistic: χ² = Σ zx²; Compare with χ² on m − (number of fitted parameters) degrees of freedom, where m is the number of age groups. Reject if the statistic exceeds the upper critical value. It is a one-sided test.
Individual standardised deviations test: Expect about 5% of |zx| > 1.96 and about 68% within ±1; Check the count outside ±1.96 and the shape of the distribution. Look for large single deviations, and for deviations that are too large or too small overall.
Signs test: P = number of positive zx, out of m. Under H0, P ~ Binomial(m, ½); Two-sided test for bias. Zero deviations are normally excluded from m. For large m, use the normal approximation.
Cumulative deviations test: Σ(θx − Ex·μ̂x) ÷ √(Σ Ex·μ̂x) ~ N(0,1) approximately; Two-sided test for overall bias. Cancellation of positive and negative deviations is the point, so it detects consistent bias that the chi-squared test can miss.
Grouping of signs test: Given n1 positive and n2 negative deviations, let G = number of positive runs. Use the null distribution of G given n1 and n2; Detects clumping. Too few groups means long runs of one sign. It is a one-sided test using tables, or a normal approximation for large n1 and n2.

How to solve Graduation Tests: Goodness of Fit questions

Use this order for any question that asks you to test a graduation. It keeps the working tidy and earns method marks.

1Write the null hypothesis: the graduated rates are the true underlying rates, so deaths are as expected and zx ~ N(0,1) independently.
2Compute expected deaths Ex·μ̂x for each age or group, then zx = (actual − expected) ÷ √expected.
3Apply the chi-squared test: sum zx², state the degrees of freedom after allowing for fitted parameters, and compare with the critical value.
4Apply the individual standardised deviations test: count how many |zx| exceed 1.96 and check the spread, comparing with the expected 5%.
5Apply the signs test and the cumulative deviations test for bias: count positives, and compute the cumulative statistic.
6Apply the grouping of signs test: count positive and negative signs and the number of positive runs, then compare with the critical value.
7Conclude for each test in words, and state which fault each test targets. Suggest an improvement if the graduation fails, such as changing the formula or the number of parameters.

Quickest way: Table-first approach for tests

When to use it: Use this when you are given a table of ages with actual and expected deaths and have limited time.

Add columns for zx and zx² straight away, and a sign column.
Total the zx² column and compare with the χ² critical value immediately.
Count positives for the signs test, using the count of nonzero deviations as m.
Total actual and expected deaths for the cumulative test without extra columns.
Write the sign string and count positive runs by eye.
Write one sentence per test: reject or do not reject, and what it means.

Common mistakes in Graduation Tests: Goodness of Fit

Using the wrong degrees of freedom for the chi-squared test
Students use the number of age groups and forget that graduation used the data to fit parameters.
Fix: Use m minus the number of parameters fitted. If rates are taken from a standard table with no fitting, use m.
Treating the chi-squared test as proof the graduation is good
A pass feels conclusive.
Fix: State that it only checks overall size of deviations. It can miss bias, clumping and a few large outliers.
Using a one-sided signs test
Students copy the chi-squared set-up.
Fix: The signs test is two-sided because bias can be either way. Reject if P is too small or too large.
Confusing the cumulative deviations test with the chi-squared test
Both use sums over ages.
Fix: The chi-squared test squares deviations, so no cancellation. The cumulative test sums raw deviations so positive and negative cancel, which shows bias.
Counting runs of both signs in the grouping of signs test
The word 'groups' is read loosely.
Fix: Count only positive runs, as the test statistic is G, the number of positive groups. State n1 and n2 before using the tables.
Forgetting to say what a failed test means
Students stop at reject or do not reject.
Fix: Add one line: overall poor fit, bias, an outlier at a particular age, or clumping, and say what you would change.

Worked examples

Example 1

A graduation covers 5 age groups. Actual deaths are 12, 18, 25, 30, 20 and expected deaths under the graduated rates are 10, 20, 24, 36, 16 respectively. No parameters were fitted. Carry out the chi-squared test at 5% significance. The upper 5% point of χ² with 5 degrees of freedom is 11.07.

Show the solution

H0: graduated rates are the true rates, so zx ~ N(0,1).
z1 = (12 − 10) ÷ √10 = 2 ÷ 3.1623 = 0.632, so z1² = 0.4.
z2 = (18 − 20) ÷ √20 = −2 ÷ 4.4721 = −0.447, so z2² = 0.2.
z3 = (25 − 24) ÷ √24 = 1 ÷ 4.899 = 0.204, so z3² = 1 ÷ 24 = 0.0417.
z4 = (30 − 36) ÷ √36 = −1, so z4² = 1.
z5 = (20 − 16) ÷ √16 = 1, so z5² = 1.
Sum: 0.4 + 0.2 + 0.0417 + 1 + 1 = 2.6417.
Degrees of freedom = 5. Critical value 11.07. 2.6417 < 11.07, so do not reject H0.

Answer: χ² = 2.64 on 5 degrees of freedom, below 11.07. There is no evidence of poor overall fit. This does not rule out bias or clumping, which other tests must check.

Example 2

For the graduation above, apply the signs test and the cumulative deviations test, both at 5% significance. For the signs test, the number of positives P ~ Binomial(5, ½) under H0, and P(P = 0 or P = 5) = 0.0625. Use the normal 5% two-sided critical value 1.96.

Show the solution

Signs of zx: +, −, +, −, +. So P = 3 positives out of 5.
Under H0, P ~ Binomial(5, ½). The two-sided tail: P(P ≥ 3) = (10 + 5 + 1) ÷ 32 = 0.5, so the two-sided p-value is well above 5% (even the most extreme outcomes, 0 or 5 positives, have total probability 0.0625, which is above 0.05). Do not reject H0.
Cumulative test: total actual = 12 + 18 + 25 + 30 + 20 = 105.
Total expected = 10 + 20 + 24 + 36 + 16 = 106.
Statistic = (105 − 106) ÷ √106 = −1 ÷ 10.296 = −0.097.
|−0.097| < 1.96, so do not reject H0.

Answer: Signs test: 3 positives out of 5, no evidence of bias. Cumulative deviations test: statistic −0.097, no evidence of bias. With only five groups, both tests have little power, so the conclusion is weak.

Exam tips

Write the null hypothesis and the distribution of zx once at the top. Every test then follows from it and you avoid repeating yourself.
For each test, name the fault it detects: overall fit, outliers, bias or clumping. Examiners often give marks for this alone.
Always state the degrees of freedom and the critical value you use, even if the numbers are given.
Where a question asks you to compare tests, say that the chi-squared test squares deviations and cannot see bias, while the cumulative deviations test lets signs cancel.
In computer-based papers, put zx in a column, then use built-in functions for sums and counts. Show the formula and the result in your answer.

Practice questions from Graduation and graduation tests

Graduation Tests: Goodness of Fit: frequently asked questions

What is the difference between the cumulative deviations test and the chi-squared test?

The chi-squared test sums squared standardised deviations, so it detects deviations that are large overall. The cumulative deviations test sums raw deviations, so positive and negative cancel. It therefore detects consistent bias in one direction, which the chi-squared test can miss.

What does the grouping of signs test detect?

It detects clumping, where positive and negative deviations come in long runs instead of mixing randomly. You count the positive groups and compare with the expected number given n1 positives and n2 negatives. Too few groups suggests clumping.

How many degrees of freedom does the chi-squared test for graduation use?

Use the number of age groups minus the number of parameters fitted to the data. If the rates come from a standard table and no parameters were fitted, use the number of groups. Always state this in your answer.

Is the signs test one-sided or two-sided?

It is two-sided. Bias can make too many or too few deviations positive. Under H0 the number of positives is Binomial(m, ½), and you reject if it is too far from the middle in either direction.

If a graduation passes all tests, is it a good graduation?

It means there is no evidence of poor fit, bias or clumping from these tests. Smoothness must be tested separately, for example with the serial correlation and smoothness tests. You also need to judge whether the graduation is suitable for its purpose.