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AI HL · 13.4

Hypothesis Testing

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13.4.18

Binomial and Poisson tests: mixed practice

13.4.18

Binomial and Poisson tests: mixed practice

Target: I can write H0 and H1 for a Poisson mean, defining the parameter. (+5 more)

By the end of this section

Key points

Poisson tests

  • Write the hypotheses about the mean for the unit in the question: H0: λ = 4.2 per month. Define λ in words.
  • Scale the mean to the period of the data: 4.2 per month over 6 months is Po(25.2). The scaling is for finding probabilities. The hypotheses stay about one month.
  • Upper tail: p-value = P(X ≥ k) = 1 − P(X ≤ k − 1). Lower tail: p-value = P(X ≤ k). Compare with the significance level.
  • Critical region: the largest set of values at one end with probability at most the significance level.

When there is no critical region

  • A critical region needs P(most extreme value) ≤ level. For a lower tail test that is P(X ≤ 0) ≤ level.
  • A small mean over a short period can make P(X = 0) larger than the level. Then H0 can never be rejected.
  • To get a critical region, raise the level or collect data over a longer period, which raises the mean.

Binomial tests and rare events

  • A binomial critical region can be turned into a percentage of the sample: X ≤ c in a sample of n is c ÷ n × 100 per cent.
  • A larger sample gives a percentage closer to the value in H0. A smaller departure is then enough to reject H0.
  • A rare event with small p can be modelled by B(n, p) or by Po(np). The p-values are close but not equal. They can give different conclusions at a level between them.

On the calculator

Poisson cumulative probability

  • Choose the Poisson cumulative distribution. Enter the mean for the whole period and the value.
  • The calculator gives P(X ≤ k). For an upper tail use 1 − P(X ≤ k − 1).

Screenshot and video to come

Binomial cumulative probability

  • Choose the binomial cumulative distribution. Enter n, p and the value.
  • Try values near the expected count to find where the probability crosses the level.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
One observation

A call centre receives on average 23 complaints per week. After a change of supplier, it receives 31 complaints in one week. Test, at the 5% significance level, whether the mean number of complaints per week has increased.

Working
  1. Hypotheses

    H0: λ = 23, H1: λ > 23, where λ is the mean number of complaints per week.

    The question asks about an increase, so H1 is λ > 23.

  2. Distribution

    Under H0, X ~ Po(23)

    One week is the period of the mean, so there is no scaling.

  3. p-value

    From the calculator: Poisson cumulative probability with mean 23 gives P(X ≤ 30) = 0.93598; p-value = P(X ≥ 31) = 1 − 0.93598 = 0.0640

    For an upper tail, subtract the cumulative probability up to one below the observed value.

  4. Conclusion

    0.0640 > 0.05, so do not reject H0. There is insufficient evidence at the 5% level that the mean number of complaints per week has increased.

    State it in context.

Video walkthrough coming
Scaling the mean

A pottery produces on average 0.6 flawed mugs per hour. After a new kiln is fitted, 3 flawed mugs are found in a 12-hour shift. Test, at the 5% significance level, whether the mean number of flawed mugs per hour has decreased.

Working
  1. Hypotheses

    H0: λ = 0.6, H1: λ < 0.6, where λ is the mean number of flawed mugs per hour.

    A decrease: H1 is λ < 0.6. The hypotheses use the rate for one hour.

  2. Scale the mean

    Over 12 hours, X ~ Po(0.6 × 12) = Po(7.2)

    The observed count is for 12 hours, so scale the mean to 12 hours.

  3. p-value

    From the calculator: Poisson cumulative probability with mean 7.2 gives P(X ≤ 3) = 0.0719; p-value = P(X ≤ 3) = 0.0719

    Lower tail: the cumulative probability up to the observed value.

  4. Conclusion

    0.0719 > 0.05, so do not reject H0. There is insufficient evidence at the 5% level that the mean number of flawed mugs per hour has decreased.

    The scaling is only for finding the probability. The conclusion is about the rate per hour.

Video walkthrough coming
A critical region

A seaside town has had an average of 4.2 complaints about gulls per month. The council will look at the next 6 months and test, at the 5% significance level, whether the mean number of complaints per month has increased. Find the critical region for the number of complaints in the 6 months.

Working
  1. Hypotheses

    H0: λ = 4.2, H1: λ > 4.2, where λ is the mean number of complaints per month.

    An increase: upper tail.

  2. Scale the mean

    Over 6 months, X ~ Po(4.2 × 6) = Po(25.2)

    Under H0 the count in 6 months has this mean.

  3. Search

    From the calculator: P(X ≥ 34) = 0.0543 > 0.05, P(X ≥ 35) = 0.0371 ≤ 0.05

    Use the cumulative probability on values near the mean. Take the smallest value whose upper tail is at most 5%.

  4. Critical region

    X ≥ 35

    35 or more complaints in 6 months is sufficient evidence of an increase.

Video walkthrough coming
No critical region

The number of faults per day in a machine is modelled by a Poisson distribution with mean 1.9. An engineer tests, using the number of faults on one day, whether the mean has decreased. Show that there is no critical region at the 5% level, and find the smallest whole-number percentage level at which a critical region exists.

Working
  1. Hypotheses

    H0: λ = 1.9, H1: λ < 1.9, where λ is the mean number of faults per day.

    A decrease: lower tail. The smallest possible critical region is X = 0.

  2. Test X = 0

    From the calculator: P(X ≤ 0) = 0.150 > 0.05

    Even the most extreme value is more likely than 5%, so no value can be in the critical region.

  3. So

    At the 5% level there is no critical region: H0 can never be rejected.

    One day of data is too little to show a decrease.

  4. Smallest level

    Need 15.0% ≤ level, so the level is 15%

    The level must be at least 15.0%, and the next whole number is 15.

Video walkthrough coming
A critical region as a percentage

A council states that 30% of residents use the bus. A campaigner thinks the true percentage is lower and takes a random sample of 50 residents. She tests this at the 5% significance level. What is the greatest percentage of the sample that could use the bus for the council's statement to be rejected? Then repeat with a sample of 150.

Working
  1. Hypotheses

    H0: p = 0.3, H1: p < 0.3

    p is the proportion of residents who use the bus. A lower proportion: lower tail.

  2. Sample of 50

    X ~ B(50, 0.3). From the calculator: P(X ≤ 9) = 0.0402 ≤ 0.05 and P(X ≤ 10) = 0.0789 > 0.05

    The critical region is X ≤ 9.

  3. As a percentage

    9 ÷ 50 × 100 = 18.0%

    Reject H0 if at most this percentage of the sample use the bus.

  4. Sample of 150

    X ~ B(150, 0.3): X ≤ 35, which is 35 ÷ 150 × 100 = 23.3%

    With a bigger sample, the sample proportion can be closer to 30% and still be significant.

Video walkthrough coming
Binomial or Poisson

A 1-in-40-years storm is taken to mean that the probability of such a storm in any year is 0.025. In the last 10 years there have been 2 such storms. Test at the 5% significance level whether the storms are now more frequent. Use the binomial model B(10, 0.025) and then the Poisson model with the mean for 10 years, and compare the two p-values.

Working
  1. Binomial

    H0: p = 0.025, H1: p > 0.025. X ~ B(10, 0.025): P(X ≥ 2) = 1 − P(X ≤ 1) = 0.0246

    From the calculator: binomial cumulative probability.

  2. Poisson

    Rate per year 0.025, so for 10 years X ~ Po(0.25): P(X ≥ 2) = 1 − P(X ≤ 1) = 0.0265

    The Poisson mean is n × p.

  3. Compare

    0.0246 and 0.0265 are close, and both are below 0.05

    For a rare event the two models agree closely. Both reject H0.

  4. Conclusion

    There is sufficient evidence at the 5% level that these storms are more frequent than 1 in 40 years.

    Either model leads to the same conclusion here.

Video walkthrough coming