Binomial and Poisson tests: mixed practice Problem set B Worked solutions

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

1. (a) A call centre receives an average of 8.5 calls per hour. The manager believes that it has become quieter since a new website opened.

   • The mean is the one for a single hour: λ = 8.5.
   • A reduction gives a lower tail.
   • H0: λ = 8.5, H1: λ < 8.5, where λ is the mean number of calls per hour.

2. (b) A junction has had an average of 1.3 accidents per week. Residents think that there are more since the speed limit was raised.

   • The mean is the one for a single week: λ = 1.3.
   • An increase gives an upper tail.
   • H0: λ = 1.3, H1: λ > 1.3, where λ is the mean number of accidents per week.

A web server logs an average of 0.45 errors per minute. After an update the server is watched for 15 minutes and logs 11 errors.

1. (a) State the distribution of the number of errors in 15 minutes, if the mean has not changed.

   • The mean for 15 minutes is 0.45 × 15 = 6.75.
   • X ~ Po(6.75)

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

   • H0: λ = 0.45, H1: λ > 0.45, where λ is the mean number of errors per minute.
   • From the calculator: Poisson cumulative probability with mean 6.75 gives P(X ≤ 10) = 0.91827
   • p-value = P(X ≥ 11) = 1 − 0.91827 = 0.0817
   • 0.0817 > 0.05, so do not reject H0. There is insufficient evidence at the 5% level that the mean number of errors per minute has increased.

A mountain pass has an average of 41 days per year with snow on the road. Last year there were 29. Is there evidence, at the 5% significance level, that the mean number of snowy days per year has decreased?

• H0: λ = 41, H1: λ < 41, where λ is the mean number of snowy days per year.
• Under H0, X ~ Po(41)
• From the calculator: Poisson cumulative probability with mean 41 gives P(X ≤ 29) = 0.0312
• 0.0312 < 0.05, so reject H0. There is sufficient evidence at the 5% level that the mean number of snowy days per year has decreased.

A weather station records an average of 0.9 thunderstorms per week in summer. A meteorologist will look at the next 20 weeks and test, at the 5% significance level, whether the mean number of thunderstorms per week has increased.

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

   • H0: λ = 0.9, H1: λ > 0.9, where λ is the mean number of thunderstorms per week.
   • Under H0 the count in 20 weeks is X ~ Po(0.9 × 20) = Po(18).
   • From the calculator: P(X ≥ 25) = 0.0683 > 0.05 and P(X ≥ 26) = 0.0446 ≤ 0.05
   • Critical region: X ≥ 26

2. (b) In the 20 weeks there are 25 thunderstorms. State the conclusion of the test.

   • 25 < 26, so it is not in the critical region.
   • Do not reject H0: there is insufficient evidence that the mean number of thunderstorms per week has increased.

A shop says that 40% of its customers pay by card. The owner thinks that the true percentage is lower. He asks a random sample of 60 customers and tests his belief at the 5% significance level. What is the greatest percentage of the sample who could pay by card for the shop's statement to be rejected?

• H0: p = 0.4, H1: p < 0.4. Under H0, X ~ B(60, 0.4).
• From the calculator: P(X ≤ 17) = 0.0413 ≤ 0.05 and P(X ≤ 18) = 0.0719 > 0.05
• Critical region: X ≤ 17
• 17 ÷ 60 × 100 = 28.3%

The number of faults found in a batch of circuit boards per day is modelled by a Poisson distribution with mean 2.1. After a change of supplier, the manager wants to test, at the 5% significance level, whether the mean number of faults per day has decreased. The test uses the count on a single day for parts (a) and (b), and the count over 5 days for part (c).

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

   • H0: λ = 2.1, H1: λ < 2.1, where λ is the mean number of faults per day.
   • X ~ Po(2.1). The most extreme value is X = 0.
   • From the calculator: P(X ≤ 0) = 0.122 > 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 12.2%.
   • The smallest whole number is 13%.

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

   • Over 5 days, X ~ Po(2.1 × 5) = Po(10.5)
   • From the calculator: P(X ≤ 4) = 0.0211 ≤ 0.05 and P(X ≤ 5) = 0.0504 > 0.05
   • Critical region: X ≤ 4

A gym receives an average of 2.4 complaints per week. After new equipment is installed, it receives 10 complaints in 8 weeks. The manager says that the mean number of complaints per week has decreased.

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

   • H0: λ = 2.4, H1: λ < 2.4, where λ is the mean number of complaints per week.

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

   • Over 8 weeks, X ~ Po(2.4 × 8) = Po(19.2)
   • From the calculator: Poisson cumulative probability with mean 19.2 gives P(X ≤ 10) = 0.0165
   • 0.0165 < 0.05, so reject H0. There is sufficient evidence at the 5% level that the mean number of complaints per week has decreased.

A 1-in-30-years drought is taken to mean that the probability of a drought in any year is 0.035. In the last 12 years there have been 2 droughts. A farmer claims that droughts have become more frequent.

1. (a) Use the binomial model with p = 0.035 to test the claim at the 5% significance level.

   • H0: p = 0.035, H1: p > 0.035. X ~ B(12, 0.035)
   • From the calculator: P(X ≤ 1) = 0.93594
   • p-value = P(X ≥ 2) = 0.0641
   • 0.0641 > 0.05, so do not reject H0. There is insufficient 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 12 years: 0.035 × 12 = 0.42, so X ~ Po(0.42)
   • p-value = P(X ≥ 2) = 1 − 0.93301 = 0.0670
   • 0.0670 > 0.05, so do not reject H0. There is insufficient 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 0.0641 and 0.0670.
   • A level between them rejects H0 under one model only: any level between 6.41% and 6.70%.

A bakery says that 35% of its customers buy a loaf. The owner thinks that the true percentage is lower and tests this at the 5% significance level, using a random sample of customers.

1. (a) Find the greatest percentage of a sample of 50 that gives sufficient evidence to reject the claim.

   • X ~ B(50, 0.35). From the calculator: P(X ≤ 11) = 0.0342 ≤ 0.05 and P(X ≤ 12) = 0.0661 > 0.05
   • 11 ÷ 50 × 100 = 22.0%

2. (b) Find the greatest percentage of a sample of 200 that gives sufficient evidence to reject the claim.

   • X ~ B(200, 0.35). From the calculator: P(X ≤ 58) = 0.0426 ≤ 0.05 and P(X ≤ 59) = 0.0584 > 0.05
   • 58 ÷ 200 × 100 = 29.0%

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 35% is enough to be unlikely under H0.

The number of typing errors per page in a report is modelled by a Poisson distribution with mean 0.45 per page. After a training course, an editor will count the errors over n pages and test, at the 5% significance level, whether the mean number per page 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(0.45n).
• e^(−0.45n) ≤ 0.05
• From the calculator: for n = 6, P(X ≤ 0) = 0.0672 > 0.05; for n = 7, P(X ≤ 0) = 0.0429 ≤ 0.05
• The smallest n is 7.