Binomial and Poisson tests: mixed practice Problem set A One question at a time

Write down H0 and H1 for each test. Define the parameter you use.

1. (a) A hospital ward has an average of 3.2 admissions per night. The manager thinks that a new rota has reduced this.

   • The mean is the one for a single night: λ = 3.2.
   • A reduction gives a lower tail.
   • H0: λ = 3.2, H1: λ < 3.2, where λ is the mean number of admissions per night.

2. (b) A beekeeper counts an average of 7.4 wasps per hive per day. She suspects that the number has increased.

   • The mean is the one for a single day: λ = 7.4.
   • An increase gives an upper tail.
   • H0: λ = 7.4, H1: λ > 7.4, where λ is the mean number of wasps per hive per day.

A weaver finds an average of 0.35 flaws per metre of cloth. After a change of loom she checks 20 metres and finds 12 flaws.

1. (a) State the distribution of the number of flaws in 20 metres, if the mean has not changed.

   • The mean for 20 metres is 0.35 × 20 = 7.
   • X ~ Po(7)

2. (b) Test, at the 5% significance level, whether the mean number of flaws per metre has increased.

   • H0: λ = 0.35, H1: λ > 0.35, where λ is the mean number of flaws per metre.
   • From the calculator: Poisson cumulative probability with mean 7 gives P(X ≤ 11) = 0.94665
   • p-value = P(X ≥ 12) = 1 − 0.94665 = 0.0533
   • 0.0533 > 0.05, so do not reject H0. There is insufficient evidence at the 5% level that the mean number of flaws per metre has increased.

A stretch of coast has an average of 24 foggy mornings per year. This year there were 33. Is there evidence, at the 5% significance level, that the mean number of foggy mornings per year has increased?

• H0: λ = 24, H1: λ > 24, where λ is the mean number of foggy mornings per year.
• Under H0, X ~ Po(24)
• From the calculator: Poisson cumulative probability with mean 24 gives P(X ≤ 32) = 0.95330
• p-value = P(X ≥ 33) = 1 − 0.95330 = 0.0467
• 0.0467 < 0.05, so reject H0. There is sufficient evidence at the 5% level that the mean number of foggy mornings per year has increased.

A wildlife camera records an average of 1.7 foxes per night in a wood. A ranger will watch for the next 14 nights and test, at the 5% significance level, whether the mean number of foxes per night has increased.

1. (a) Find the critical region for the test.

   • H0: λ = 1.7, H1: λ > 1.7, where λ is the mean number of foxes per night.
   • Under H0 the count in 14 nights is X ~ Po(1.7 × 14) = Po(23.8).
   • From the calculator: P(X ≥ 32) = 0.0623 > 0.05 and P(X ≥ 33) = 0.0426 ≤ 0.05
   • Critical region: X ≥ 33

2. (b) In the 14 nights there are 34 foxes. State the conclusion of the test.

   • 34 ≥ 33, so it is in the critical region.
   • Reject H0: there is sufficient evidence that the mean number of foxes per night has increased.

A council says that 25% of residents cycle to work. A campaigner thinks that the true percentage is lower. She asks a random sample of 80 residents and tests her belief at the 5% significance level. What is the greatest percentage of the sample who could cycle to work for the council's statement to be rejected?

• H0: p = 0.25, H1: p < 0.25. Under H0, X ~ B(80, 0.25).
• From the calculator: P(X ≤ 13) = 0.0421 ≤ 0.05 and P(X ≤ 14) = 0.0740 > 0.05
• Critical region: X ≤ 13
• 13 ÷ 80 × 100 = 16.3%

The number of breakdowns of a lift per day is modelled by a Poisson distribution with mean 1.5. After servicing, the engineer wants to test, at the 5% significance level, whether the mean number of breakdowns per day has decreased. The test uses the count on a single day for parts (a) and (b), and the count over 6 days for part (c).

1. (a) Show that there is no critical region.

   • H0: λ = 1.5, H1: λ < 1.5, where λ is the mean number of breakdowns per day.
   • X ~ Po(1.5). The most extreme value is X = 0.
   • From the calculator: P(X ≤ 0) = 0.223 > 0.05
   • No value can be in the critical region at 5%, so H0 can never be rejected.

2. (b) Find the smallest whole-number percentage level at which the test has a critical region.

   • The level must be at least 22.3%.
   • The smallest whole number is 23%.

3. (c) Find the critical region at the 5% level for the count over 6 days.

   • Over 6 days, X ~ Po(1.5 × 6) = Po(9)
   • From the calculator: P(X ≤ 3) = 0.0212 ≤ 0.05 and P(X ≤ 4) = 0.0550 > 0.05
   • Critical region: X ≤ 3

A printer jams on average 0.8 times per day. After maintenance it jams 3 times in 14 days. The technician says that the mean number of jams per day has decreased.

1. (a) Write down suitable hypotheses, defining the parameter.

   • H0: λ = 0.8, H1: λ < 0.8, where λ is the mean number of jams per day.

2. (b) Test the claim at the 5% significance level.

   • Over 14 days, X ~ Po(0.8 × 14) = Po(11.2)
   • From the calculator: Poisson cumulative probability with mean 11.2 gives P(X ≤ 3) = 0.00423
   • 0.00423 < 0.05, so reject H0. There is sufficient evidence at the 5% level that the mean number of jams per day has decreased.

A 1-in-60-years heatwave is taken to mean that the probability of such a heatwave in any year is 0.0167. In the last 15 years there have been 2 such heatwaves. A climate analyst claims that these heatwaves have become more frequent.

1. (a) Use the binomial model with p = 0.0167 to test the claim at the 5% significance level.

   • H0: p = 0.0167, H1: p > 0.0167. X ~ B(15, 0.0167)
   • From the calculator: P(X ≤ 1) = 0.97465
   • p-value = P(X ≥ 2) = 0.0253
   • 0.0253 < 0.05, so reject H0. There is sufficient evidence at the 5% level that the events have become more frequent.

2. (b) Use a Poisson model with the mean for the whole period to test the claim at the 5% significance level.

   • Mean for 15 years: 0.0167 × 15 = 0.2505, so X ~ Po(0.2505)
   • p-value = P(X ≥ 2) = 1 − 0.97340 = 0.0266
   • 0.0266 < 0.05, so reject H0. There is sufficient evidence at the 5% level that the events have become more frequent.

3. (c) Find the range of significance levels for which the two models give different conclusions.

   • The p-values are 0.0253 and 0.0266.
   • A level between them rejects H0 under one model only: any level between 2.53% and 2.66%.

A school says that 20% of its students walk to school. A governor thinks that the true percentage is lower and tests this at the 5% significance level, using a random sample of students.

1. (a) Find the greatest percentage of a sample of 40 that gives sufficient evidence to reject the claim.

   • X ~ B(40, 0.2). From the calculator: P(X ≤ 3) = 0.0285 ≤ 0.05 and P(X ≤ 4) = 0.0759 > 0.05
   • 3 ÷ 40 × 100 = 7.50%

2. (b) Find the greatest percentage of a sample of 160 that gives sufficient evidence to reject the claim.

   • X ~ B(160, 0.2). From the calculator: P(X ≤ 23) = 0.0425 ≤ 0.05 and P(X ≤ 24) = 0.0655 > 0.05
   • 23 ÷ 160 × 100 = 14.4%

3. (c) Explain why the two percentages are different.

   • The sample proportion varies less from sample to sample when the sample is larger.
   • So a smaller departure from 20% is enough to be unlikely under H0.

The number of cracks per day found in a bridge's concrete is modelled by a Poisson distribution with mean 0.8 per day. After repairs, an inspector will count the cracks over n days and test, at the 5% significance level, whether the mean number per day has decreased. Find the smallest value of n for which the test has a critical region.

• A critical region exists if P(X ≤ 0) ≤ 0.05, where X ~ Po(0.8n).
• e^(−0.8n) ≤ 0.05
• From the calculator: for n = 3, P(X ≤ 0) = 0.0907 > 0.05; for n = 4, P(X ≤ 0) = 0.0408 ≤ 0.05
• The smallest n is 4.