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Hypothesis tests for Poisson and geometric models questions
The testing procedure does not change when the distribution does. State hypotheses about the parameter, compute the tail probability of what you saw, and compare it with the significance level.
7 original questions · 25 marks · the hypothesis tests for poisson and geometric models notes · Further Statistics 1
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Write down suitable hypotheses for testing whether a Poisson rate has increased from a known value of 3.
Worked answer
H₀: λ = 3 against H₁: λ > 3, a one-tailed test. B1 B1 for the two hypotheses. The hypotheses name the population parameter, never the observed count.Faults occur at Po(3) per batch. A new batch shows 8 faults. Test at the 5% level whether the rate has increased.
Worked answer
H₀: λ = 3 against H₁: λ > 3, one-tailed at 5%. Under H₀, P(X ≥ 8) = 1 − P(X ≤ 7) = 0.0119. Since 0.0119 < 0.05, reject H₀. There is evidence at the 5% level that the fault rate has risen. M1 for the correct upper tail, A1 for 0.0119, A1 for the comparison and conclusion in context.Accidents at a site follow Po(10) per month. After a safety campaign one month records 4. Test at the 5% level whether the rate has fallen.
Worked answer
H₀: λ = 10, H₁: λ < 10, one-tailed. The observed value is low, so use the lower tail: P(X ≤ 4) = 0.0293. Since 0.0293 < 0.05, reject H₀: there is evidence at the 5% level that the accident rate has fallen since the campaign. B1 for the hypotheses, M1 for using the lower tail, A1 for 0.0293, A1 for the conclusion in context.A process is claimed to succeed with probability 0.4 per attempt. The first success comes on the 8th attempt. Test at the 5% level whether the success probability is lower than claimed.
Worked answer
H₀: p = 0.4 against H₁: p < 0.4. A small p produces a long wait, so the evidence sits in the upper tail of X: P(X ≥ 8) = 0.6⁷ = 0.0280. Since 0.0280 < 0.05, reject H₀: there is evidence that the success probability is below 0.4. B1 for the hypotheses, M1 for the upper tail, A1 for 0.0280, A1 for the conclusion in context.Explain how a two-tailed test at the 5% level differs in execution from a one-tailed test at the same level.
Worked answer
The 5% is split between the two tails, so each carries only 2.5%. The observed tail probability is compared with 0.025 rather than 0.05, which makes rejection harder and can reverse the conclusion of an otherwise identical calculation. B1 for the split into two tails of 2.5% each, B1 for comparing with 0.025 rather than 0.05.A Poisson test has H₀: λ = 4 against H₁: λ > 4 at the 5% level. Given that P(X ≤ 7) = 0.9489 and P(X ≤ 8) = 0.9786, find the critical region and state the actual significance level.
Worked answer
P(X ≥ 8) = 1 − P(X ≤ 7) = 1 − 0.9489 = 0.0511, which exceeds 0.05, so 8 is not extreme enough to reject. P(X ≥ 9) = 1 − P(X ≤ 8) = 1 − 0.9786 = 0.0214, which is below 0.05.
The critical region is therefore X ≥ 9, with actual significance level 0.0214. M1 for forming an upper tail from the tabulated values, A1 for 0.0511 ruling out a count of 8, A1 for the critical region, A1 for the actual significance level. Take the tail one place above the tabulated value: reading P(X ≥ 8) as 1 − P(X ≤ 8) is the standard error and shifts the whole region.Cars pass a checkpoint at a mean rate of 2.5 per minute. To test at the 5% level whether the rate has increased, the number passing in a 4-minute period is recorded. Find the critical region and the actual significance level of the test. In the period observed, 18 cars pass. Carry out the test.
Worked answer
H₀: λ = 2.5 per minute against H₁: λ > 2.5, one-tailed.
Scale the parameter to the interval actually observed. Over 4 minutes the mean is 4 × 2.5 = 10, so under H₀ the count X follows Po(10). Testing an 18 against a mean of 2.5 is the error that wrecks this question.
Work down the upper tail. P(X ≥ 15) = 1 − P(X ≤ 14) = 1 − 0.9165 = 0.0835, which is above 5%, so 15 is not extreme enough. P(X ≥ 16) = 1 − 0.9513 = 0.0487, which is below 5%.
The critical region is X ≥ 16, and the actual significance level is 0.0487, or 4.87%. It undershoots the nominal 5% because X takes only whole values, so the tail cannot be trimmed to an exact 5%.
The observed value 18 lies in the critical region, so reject H₀. There is evidence at the 5% level that the rate at which cars pass the checkpoint has increased. Equivalently P(X ≥ 18) = 0.0143 < 0.05.
B1 for the hypotheses, M1 for scaling the mean to the 4-minute interval, M1 for working down the upper tail, A1 for the critical region, A1 for the actual significance level, A1 for the conclusion in context.
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