Further Statistics: Hypothesis Testing and Chi-Squared Depth
Hypothesis Testing and Chi-Squared Depth, a Further Statistics topic, combines the chi-squared tests of FP.FS3 with the hypothesis-testing framework of FP.FS4 at greater depth: goodness-of-fit tests where a parameter of the hypothesised distribution (such as a Poisson mean or a geometric p) must first be estimated from the sample data, which costs one extra degree of freedom (v = categories - 1 - number of parameters estimated); and two-tailed hypothesis tests, where the significance level is split between an upper and a lower critical region found separately, with the actual combined significance level equal to the sum of the two actual tail probabilities. Both extensions test whether a student can adapt the standard one-parameter, one-tailed method rather than just quoting it.
Before you start
Make sure you're comfortable with these topics first:
Method
- For a goodness-of-fit test with an unknown parameter, first estimate the parameter from the sample data (e.g. the sample mean as an estimate of a Poisson lambda, or 1/sample mean as an estimate of a geometric p), then use that estimated value to calculate every expected frequency exactly as in a standard test.
- Reduce the degrees of freedom by one for every parameter estimated from the sample: v = (number of categories) - 1 - (number of parameters estimated), pooling any category with expected frequency below 5 as usual before finalising v.
- For a two-tailed hypothesis test (H1: parameter =/= a stated value), split the significance level equally between the two tails (e.g. 2.5% in each tail for an overall 5% test), and find the lower and upper critical regions separately.
- State the actual combined significance level of a two-tailed test as the sum of the actual probability in the lower critical region and the actual probability in the upper critical region.
- To decide the outcome, check whether the observed test statistic lies in EITHER critical region (for a two-tailed test), or compare the calculated X^2 with the critical value at the adjusted degrees of freedom (for a chi-squared test with an estimated parameter).
- Interpret every conclusion in context, and be explicit in the write-up about which extra step (parameter estimation, or splitting into two tails) was applied, since this is exactly the depth a marker is checking for beyond the standard one-parameter, one-tailed method.
Worked example
The number of accidents per week at a factory is recorded over 100 weeks. The observed frequencies are: 0 accidents: 37 weeks, 1 accident: 36 weeks, 2 accidents: 20 weeks, 3 or more accidents: 7 weeks, and the total number of accidents recorded over the 100 weeks was 130. Test, at the 5% significance level, whether a Poisson distribution is a suitable model for the number of accidents per week, using the sample mean as an estimate of lambda. You may use the critical value 5.991 for 2 degrees of freedom at the 5% level.
- Estimate the mean from the data: lambda_hat = (total accidents)/(total weeks) = 130/100 = 1.3.
- Using lambda_hat = 1.3, find P(X=0) = e^-1.3 = 0.2725, P(X=1) = 1.3e^-1.3 = 0.3543, P(X=2) = (1.3^2/2)e^-1.3 = 0.2303, and P(X>=3) = 1 - (0.2725+0.3543+0.2303) = 0.1429.
- Multiply by 100 to find expected frequencies: E = 27.25, 35.43, 23.03, 14.29 (all above 5, so no pooling is needed).
- Calculate (O-E)^2/E for each category: (37-27.25)^2/27.25 = 3.486; (36-35.43)^2/35.43 = 0.009; (20-23.03)^2/23.03 = 0.398; (7-14.29)^2/14.29 = 3.718.
- Sum these contributions: X^2 = 3.486+0.009+0.398+3.718 = 7.61 (3 s.f.). Since one parameter (lambda) was estimated, v = 4 categories - 1 - 1 = 2, matching the given critical value 5.991.
- Final answer: since X^2 = 7.61 is greater than 5.991, reject H0 at the 5% level - there is evidence that a Poisson distribution is not a suitable model for the number of accidents per week.
Practice questions
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Q1Explain why estimating a parameter from the sample data before carrying out a chi-squared goodness-of-fit test reduces the degrees of freedom by an extra 1.Show answer
Answer: Each parameter estimated from the same data used in the test uses up one extra degree of freedom, because the expected frequencies are no longer fully determined by the hypothesised model alone - they also depend on a value calculated from the observed data
Q2A goodness-of-fit test has 6 categories, and the mean of the hypothesised Poisson distribution is estimated from the sample. State the degrees of freedom.Show answer
Answer: v = 6 - 1 - 1 = 4
Q3A two-tailed hypothesis test at the 10% significance level splits the significance level between two tails. State the nominal significance level used in each tail.Show answer
Answer: 5% in each tail (10%/2)
Q4Data from 50 trials gives a sample mean of 2.5 for a variable believed to follow a Poisson distribution. Find the estimated value of lambda used in a goodness-of-fit test.Show answer
Answer: lambda_hat = 2.5 (the sample mean is used directly as the estimate of the Poisson mean)
Q5Explain, in the context of a hypothesis test, why the power of a test generally increases as the true value of the parameter moves further from the value stated in H0.Show answer
Answer: The further the true parameter is from the hypothesised value (in the direction of H1), the more the distribution of the test statistic shifts into the critical region, so a bigger difference is more likely to be detected and correctly lead to rejecting H0
Q6A geometric goodness-of-fit test uses 5 categories, with the parameter p estimated from the sample mean. Find the degrees of freedom.Show answer
Answer: v = 5 - 1 - 1 = 3
Q7A two-tailed test of H0: lambda=20 has actual lower-tail probability 0.018 and actual upper-tail probability 0.021. State the actual combined significance level of the test.Show answer
Answer: 0.018 + 0.021 = 0.039 (3.9%)
Exam-style questions
Written in the style of a A Level Further Maths exam paper, with a full mark scheme.
A call centre receives calls at a mean rate of 10 per hour when operating normally. A manager wants to test, at the 5% significance level (2.5% in each tail), whether the mean rate has changed during a particular hour, using X, the number of calls received in that hour, and testing H0: lambda=10 against H1: lambda=/=10. (a) Find the lower critical region, in the form X<=a. (b) Find the upper critical region, in the form X>=b. (c) State the actual combined significance level of the test. (d) During the hour tested, 4 calls were received. State the conclusion of the test.
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A researcher records the number of attempts needed by 80 people to solve a puzzle for the first time, with results: 1 attempt: 36 people, 2 attempts: 22 people, 3 attempts: 12 people, 4 or more attempts: 10 people, and the total number of attempts summed over all 80 people was 160. Test, at the 5% significance level, whether a geometric distribution is a suitable model, using the sample mean to estimate p. You may use the critical value 5.991 for 2 degrees of freedom (after estimating one parameter) at the 5% level.
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A test of H0: lambda=9 against H1: lambda<9 uses the critical region X<=3, where X ~ Po(lambda) is the number of events observed. (a) Given that the true mean is actually 4, find the power of the test. (b) Given that the true mean is actually 6, find the power of the test. (c) Comment on how the two powers compare, with a reason.
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