Hypothesis Testing

Have a go first. Then open the working one line at a time.

1. Scale the rate

   4.5 per month × 3 months = 13.5 per 3 months

   The test counts events in a fixed period. Scale λ to the length of the period you observed.

2. Hypotheses

   Let X be the number of false alarms in 3 months. H0: λ = 13.5, H1: λ < 13.5, where λ is the mean number of false alarms per 3 months.

   The manager thinks the rate has decreased: lower tail. Under H0, X ~ Po(13.5).

3. p-value

   P(X ≤ 7) = 0.0415

   From the calculator, Poisson cumulative probability with mean 13.5 and upper limit 7.

4. Compare

   0.0415 < 0.05, so reject H0.

   Compare p with 0.05.

5. In context

   There is sufficient evidence, at the 5% significance level, that the mean rate of false alarms has decreased.

   Say what it means for the building.

Have a go first. Then open the working one line at a time.

1. Hypotheses

   Let X be the number of customers in one hour. H0: λ = 6, H1: λ > 6, where λ is the mean number of customers per hour.

   The period observed is one hour, the same as the rate, so no scaling is needed. Under H0, X ~ Po(6).

2. p-value

   P(X ≥ 10) = 1 − P(X ≤ 9) = 0.0839

   From the calculator, Poisson cumulative probability with mean 6 and upper limit 9. Take it from 1.

3. Compare

   0.0839 > 0.05, so do not reject H0.

   10 customers is above the average of 6, but not unusual enough.

4. In context

   There is insufficient evidence, at the 5% significance level, that the mean rate of customers has increased.

   Insufficient evidence does not show the rate is unchanged.

Have a go first. Then open the working one line at a time.

1. Hypotheses

   Let X be the number of rescues in one week. H0: λ = 9.4, H1: λ < 9.4

   Under H0, X ~ Po(9.4). The critical region is in the lower tail.

2. Cumulative table

   From the calculator, a table of P(X ≤ k): P(X ≤ 3) = 0.0160, P(X ≤ 4) = 0.0429, P(X ≤ 5) = 0.0935

   Find where the cumulative probability crosses 0.05.

3. Test adjacent values

   P(X ≤ 5) = 0.0935 > 0.05; P(X ≤ 4) = 0.0429 < 0.05

   The largest c with P(X ≤ c) at most 0.05.

4. Critical region

   Critical region: X ≤ 4

   4 or fewer rescues gives sufficient evidence that the rate has decreased.

Have a go first. Then open the working one line at a time.

1. Hypotheses

   Let X be the number of power cuts in one month. H0: λ = 3.6, H1: λ > 3.6

   Under H0, X ~ Po(3.6). The critical region is in the upper tail.

2. Upper tail

   P(X ≥ c) = 1 − P(X ≤ c − 1). P(X ≤ 8) = 0.988, P(X ≤ 9) = 0.996

   From the calculator, cumulative table. Look for where 1 − P(X ≤ c − 1) drops below 0.01.

3. Test adjacent values

   P(X ≥ 9) = 0.0117 > 0.01; P(X ≥ 10) = 0.00402 < 0.01

   The smallest c with P(X ≥ c) at most 0.01.

4. Critical region

   Critical region: X ≥ 10

   10 or more power cuts gives sufficient evidence that the rate has increased.

Have a go first. Then open the working one line at a time.

1. Hypotheses

   Let X be the number of faulty bulbs out of 60. H0: p = 0.15, H1: p > 0.15

   Under H0, X ~ B(60, 0.15). The mean is 9. A proportion uses the binomial, an average rate uses the Poisson.

2. Table

   From the calculator: P(X ≤ 13) = 0.942, P(X ≤ 14) = 0.971

   Cumulative binomial probabilities near the 5% mark.

3. Test adjacent values

   P(X ≥ 14) = 0.0578 > 0.05; P(X ≥ 15) = 0.0291 < 0.05

   The smallest c with P(X ≥ c) at most 0.05.

4. Critical region

   Critical region: X ≥ 15

   The method is the same as for the Poisson distribution: only the distribution changes.