Questions on hypothesis tests for Poisson and geometric models

Question 1

Write down suitable hypotheses for testing whether a Poisson rate has increased from a known value of 3.

Accepted 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.

Question 2

Faults occur at Po(3) per batch. A new batch shows 8 faults. Test at the 5% level whether the rate has increased.

Accepted 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.

Question 3

A manufacturer claims that each item it tests passes with probability 0.25, and an inspector suspects the probability is lower. Write down suitable hypotheses for a test based on the number of the item on which the first pass occurs.

Accepted Answer

H₀: p = 0.25 against H₁: p < 0.25, a one-tailed test. B1 B1 for the two hypotheses. The parameter is the probability of a pass on a single item, so the hypotheses are written in terms of p, not in terms of the trial number that is observed.

Question 4

A test of H₀: λ = 6 against H₁: λ < 6 or λ > 6 is carried out at the 5% level. The observed count is 1, and P(X ≤ 1) = 0.0174 when λ = 6. State the value with which 0.0174 should be compared, and give the conclusion.

Accepted Answer

The test is two-tailed, so the lower tail carries only half the 5%: compare with 0.025. Since 0.0174 < 0.025, reject H₀: there is evidence at the 5% level that the rate is not 6. B1 for 0.025, B1 for the conclusion. Comparing with 0.05 would give the same verdict here, but it would be the verdict of a different test.

Question 5

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.

Accepted 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.

Question 6

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.

Accepted 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.

Question 7

Explain how a two-tailed test at the 5% level differs in execution from a one-tailed test at the same level.

Accepted 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.

Question 8

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.

Accepted 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.

Question 9

Each item tested passes independently with probability p, and X is the number of the item on which the first pass occurs. To test H₀: p = 0.25 against H₁: p < 0.25 at the 5% level, find the critical region in terms of X, and state the actual significance level of the test.

Accepted Answer

A small p makes the first pass arrive late, so the critical region is an upper tail X ≥ n. Under H₀, P(X ≥ n) = 0.75ⁿ⁻¹, and this must be below 0.05. Taking logarithms, n − 1 > log 0.05/log 0.75 = 10.41, so n − 1 = 11 and the critical region is X ≥ 12. Check: 0.75¹⁰ = 0.0563, which is too big, and 0.75¹¹ = 0.0422, which is not. The actual significance level is 0.0422. M1 for P(X ≥ n) = 0.75ⁿ⁻¹, M1 for solving 0.75ⁿ⁻¹ < 0.05 by logarithms or by trial, A1 for X ≥ 12, A1 for 0.0422. The power is n − 1, not n: X ≥ 12 means the first eleven items all failed.

Question 10

Calls to a helpline arrive at a mean rate of 6 per hour. In one particular hour 12 calls arrive. Test at the 5% level whether the mean rate for that hour differs from 6, given that P(X ≤ 11) = 0.9799 when X ~ Po(6).

Accepted Answer

H₀: λ = 6 against H₁: λ < 6 or λ > 6, a two-tailed test with 0.025 in each tail. The observed value 12 is high, so use the upper tail: P(X ≥ 12) = 1 − P(X ≤ 11) = 1 − 0.9799 = 0.0201. Since 0.0201 < 0.025, reject H₀: there is evidence at the 5% level that the mean rate in that hour differs from 6. B1 for the two-tailed hypotheses, M1 for 1 − P(X ≤ 11), A1 for 0.0201, A1 for the comparison with 0.025 and the conclusion in context. The question says 'differs', so the alternative is two-sided even though only the upper tail is needed for the arithmetic.

Question 11

A lottery scratchcard is claimed to win with probability 0.1. A player suspects the probability is higher, obtains cards one at a time, and wins on the very first card. Test the claim at the 5% level, and explain what this shows about a test based on a single first success.

Accepted Answer

H₀: p = 0.1 against H₁: p > 0.1. A larger p makes the first win arrive early, so the evidence sits in the lower tail: P(X ≤ 1) = P(X = 1) = 0.1. Since 0.1 > 0.05, do not reject H₀: there is insufficient evidence at the 5% level that the probability exceeds 0.1. Winning on the first card is the most extreme result possible in this direction, and even that has probability 0.1 under H₀, so no outcome could reject the claim at the 5% level. B1 for the hypotheses, M1 for P(X ≤ 1) = 0.1 as the lower tail, A1 for the conclusion with the explanation that the critical region is empty.

Question 12

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.

Accepted 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).
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.

Question 13

Customers enter a shop at a mean rate of 1.5 per ten minutes. The manager wants to test, at the 5% level, whether the rate has changed, using the number of customers entering in a 40-minute period. Find the critical region and the actual significance level of the test. In the period observed, 13 customers enter; carry out the test. For X ~ Po(6): P(X ≤ 1) = 0.0174, P(X ≤ 2) = 0.0620, P(X ≤ 10) = 0.9574, P(X ≤ 11) = 0.9799, P(X ≤ 12) = 0.9912.

Accepted Answer

H₀: λ = 1.5 per ten minutes against H₁: λ < 1.5 or λ > 1.5, two-tailed with 0.025 in each tail.
Over 40 minutes the mean is 4 × 1.5 = 6, so under H₀ the count X follows Po(6).
Lower tail: P(X ≤ 1) = 0.0174 < 0.025 but P(X ≤ 2) = 0.0620 > 0.025, so the lower part of the region is X ≤ 1. Upper tail: P(X ≥ 12) = 1 − 0.9799 = 0.0201 < 0.025 but P(X ≥ 11) = 1 − 0.9574 = 0.0426 > 0.025, so the upper part is X ≥ 12. The critical region is X ≤ 1 or X ≥ 12, with actual significance level 0.0174 + 0.0201 = 0.0375.
The observed 13 lies in the region, so reject H₀: there is evidence at the 5% level that the rate has changed. Equivalently P(X ≥ 13) = 1 − 0.9912 = 0.0088 < 0.025. B1 for the two-tailed hypotheses, M1 for scaling to Po(6), M1 for testing each tail against 0.025, A1 for the critical region, A1 for 0.0375, A1 for the conclusion in context. Each tail is trimmed separately to 2.5%, so the two actual tails are unequal and their sum, not either one, is the level of the test.

Question 14

A stall sells an average of 2 items per day, the daily counts being independent Poisson variables. After a change of layout the stall sells 17 items in a five-day week. Test at the 5% level whether the mean daily rate has increased. The manager instead proposes to test each day separately at the 5% level and to declare an increase if any one day gives a significant result. Find the probability that this procedure declares an increase in a week when the rate has not changed, and explain what is wrong with it. For X ~ Po(10), P(X ≤ 16) = 0.9730; for X ~ Po(2), P(X ≤ 4) = 0.9473 and P(X ≤ 5) = 0.9834.

Accepted Answer

H₀: λ = 2 per day against H₁: λ > 2, one-tailed.
Over five days the mean is 5 × 2 = 10, so the weekly count follows Po(10) under H₀. P(X ≥ 17) = 1 − 0.9730 = 0.0270 < 0.05, so reject H₀: there is evidence at the 5% level that the daily rate has increased.
For a single day under H₀ the count is Po(2). P(X ≥ 5) = 1 − 0.9473 = 0.0527 > 0.05 and P(X ≥ 6) = 1 − 0.9834 = 0.0166 < 0.05, so the daily critical region is X ≥ 6 with size 0.0166.
The five days are independent, so the probability that at least one of them falls in its region when the rate is still 2 is 1 − (1 − 0.0166)⁵ = 1 − 0.9834⁵ = 0.080. The manager's procedure rejects a true H₀ about 8% of the time, not 5%: giving the data five chances to look unusual raises the chance of a false alarm, so the procedure is not a 5% test. Pooling the week into one count keeps the level at 5% and uses all the data at once. B1 for the hypotheses, M1 for Po(10), A1 for 0.0270 and the conclusion in context, B1 for the daily region X ≥ 6 with size 0.0166, M1 for 1 − 0.9834⁵, A1 for 0.080 with the explanation.

Question 15

A supplier claims that 2% of its components are defective, independently of one another. An inspector tests components one at a time and finds that the first defective one is the 150th tested. Explain why a geometric model, rather than a binomial one, describes what the inspector recorded. Test at the 5% level whether the proportion defective is lower than the supplier claims, and comment on how secure the conclusion is.

Accepted Answer

The inspector did not fix the number of components in advance and count defectives; the number tested was decided by when the first defective appeared. The trial number of a first success is what the geometric distribution counts, with p the probability that a single component is defective.
H₀: p = 0.02 against H₁: p < 0.02. A smaller p makes the first defective arrive later, so the evidence sits in the upper tail: P(X ≥ 150) = 0.98¹⁴⁹ = 0.0493, the probability that the first 149 components are all sound. Since 0.0493 < 0.05, reject H₀: there is evidence at the 5% level that the proportion defective is below 2%.
The tail probability is only just below 0.05: had the first defective been the 149th component, 0.98¹⁴⁸ = 0.0503 would not have rejected, and at the 1% level 0.0493 does not reject either. The conclusion rests on a single wait and would be reversed by one component, so it is weak evidence and a second run of testing would be worth having. B1 for the geometric model with the reason, B1 for the hypotheses, M1 for the upper tail 0.98¹⁴⁹, A1 for 0.0493, A1 for the conclusion in context, B1 for the comment on the margin.