Hard Further Maths Hypothesis testing (Poisson, geometric) Questions

Challenging, exam-style Further Maths Hypothesis testing (Poisson, geometric) questions with worked solutions. Stretch yourself on the hardest hypothesis-testing, actual-significance-level, one-tailed, poisson problems.

hypothesis-testingactual-significance-levelone-tailedpoissongeometriccritical-region
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
A drug trial recruits volunteers one at a time and each volunteer reports a side effect with probability pp, independently of the others. Let XX be the number of volunteers recruited up to and including the first one reporting a side effect, so that P(X=x)=(1−p)x−1pP(X=x)=(1-p)^{x-1}p for x=1, 2, 3, …x=1,\ 2,\ 3,\ \ldots A hypothesis test of H0H_0: p=0.05p = 0.05 against H1H_1: p>0.05p > 0.05 is carried out at the 10%10\% significance level, using the single observation XX. This is a one-tailed test and the critical region lies in the lower tail of XX. The critical region is the largest lower-tail region whose probability does not exceed the significance level: it is X≤cX \le c, where cc is the largest integer for which P(X≤c)≤0.1P(X \le c) \le 0.1. The actual significance level of the test is the probability, calculated under H0H_0, that XX lies in the critical region. Find the actual significance level of this test, giving your answer as a probability correct to 44 decimal places.
Show worked solution

Worked solution

  1. State the hypotheses and the tail of the test

    H0:p=0.05,H1:p>0.05(one-tailed, lower tail)H_{0}:p=0.05,\quad H_{1}:p>0.05\quad \text{(one-tailed, lower tail)}

    The alternative hypothesis fixes which tail (or tails) of XX forms the critical region.

  2. State the distribution of the test statistic under H0H_0

    X∼Geo(0.05)X\sim\text{Geo}(0.05)

    All probabilities below are calculated assuming H0H_0 is true.

  3. State the critical-region rule for the lower tail

    c=max⁡{k:P(X≤k)≤0.1}c=\max\{k:P(X\le k)\le 0.1\}

    The critical region is the largest lower-tail region whose probability does not exceed 0.10.1.

  4. Recall the geometric probability function

    P(X=x)=(1−p)x−1p,x=1, 2, 3, …P(X=x)=\left(1-p\right)^{x-1}p,\quad x=1,\ 2,\ 3,\ \ldots

    The first x−1x-1 trials fail and the xxth succeeds.

  5. Recall the geometric upper-tail probability

    P(X≥c)=(1−p)c−1P(X\ge c)=\left(1-p\right)^{c-1}

    The event X≥cX\ge c says the first c−1c-1 trials all fail.

  6. Recall the geometric cumulative probability

    P(X≤c)=1−(1−p)cP(X\le c)=1-\left(1-p\right)^{c}

    The complement of X≤cX\le c is that the first cc trials all fail.

  7. Distinguish the strict and non-strict inequalities

    P(X>c)=P(X≥c+1)≠P(X≥c)P(X>c)=P(X\ge c+1)\neq P(X\ge c)

    Confusing these two shifts the critical value by one, which is the commonest error in this topic.

  8. Test k=1k=1 in the lower tail

    P(X≤1)=0.050000≤0.1P(X\le 1)=0.050000\le 0.1

    The condition holds, so X≤1X\le 1 is a permissible region.

  9. Test k=2k=2 in the lower tail

    P(X≤2)=0.097500≤0.1P(X\le 2)=0.097500\le 0.1

    The condition holds, so X≤2X\le 2 is a permissible region.

  10. Test k=3k=3 in the lower tail

    P(X≤3)=0.142625>0.1P(X\le 3)=0.142625> 0.1

    The region X≤3X\le 3 has probability greater than 0.10.1, so it is too large; the previous value of kk is the critical value.

  11. State the critical value

    c=2c=2

    This is the integer picked out by the stated convention.

  12. State the critical region

    X≤2X \le 2

    The test rejects H0H_0 exactly when XX lies in this region.

  13. Write the actual significance level as a probability

    P(critical region)=P(X≤2)P(\text{critical region})=P(X \le 2)

    The actual significance level is the probability of the critical region under H0H_0, not the nominal level.

  14. Evaluate that probability

    P(X≤2)=0.097500P(X \le 2)=0.097500

    This is computed from the distribution of XX under H0H_0.

  15. Compare the actual level with the nominal level

    0.097500<0.10.097500 < 0.1

    Because XX is discrete the actual level is strictly less than the nominal level of 0.10.1.

  16. State the actual significance level to 44 decimal places

    P(critical region)=0.0975P(\text{critical region})=0.0975

    This is the true probability of a Type I error for this test.

Answer
0.09750.0975
Question 2
9 markschallenging
Batteries are taken from a large bin one at a time and each battery is flat with probability pp, independently of the others. Let XX be the number of batteries taken up to and including the first flat one, so that P(X=x)=(1−p)x−1pP(X=x)=(1-p)^{x-1}p for x=1, 2, 3, …x=1,\ 2,\ 3,\ \ldots A hypothesis test of H0H_0: p=0.08p = 0.08 against H1H_1: p<0.08p < 0.08 is carried out at the 5%5\% significance level, using the single observation XX. This is a one-tailed test and the critical region lies in the upper tail of XX. The critical region is the largest upper-tail region whose probability does not exceed the significance level: it is X≥cX \ge c, where cc is the smallest integer for which P(X≥c)≤0.05P(X \ge c) \le 0.05. The actual significance level of the test is the probability, calculated under H0H_0, that XX lies in the critical region. Find the actual significance level of this test, giving your answer as a probability correct to 44 decimal places.
Show worked solution

Worked solution

  1. State the hypotheses and the tail of the test

    H0:p=0.08,H1:p<0.08(one-tailed, upper tail)H_{0}:p=0.08,\quad H_{1}:p<0.08\quad \text{(one-tailed, upper tail)}

    The alternative hypothesis fixes which tail (or tails) of XX forms the critical region.

  2. State the distribution of the test statistic under H0H_0

    X∼Geo(0.08)X\sim\text{Geo}(0.08)

    All probabilities below are calculated assuming H0H_0 is true.

  3. State the critical-region rule for the upper tail

    c=min⁡{k:P(X≥k)≤0.05}c=\min\{k:P(X\ge k)\le 0.05\}

    The critical region is the largest upper-tail region whose probability does not exceed 0.050.05.

  4. Recall the geometric probability function

    P(X=x)=(1−p)x−1p,x=1, 2, 3, …P(X=x)=\left(1-p\right)^{x-1}p,\quad x=1,\ 2,\ 3,\ \ldots

    The first x−1x-1 trials fail and the xxth succeeds.

  5. Test k=33k=33 in the upper tail

    P(X≥33)=0.069376>0.05P(X\ge 33)=0.069376> 0.05

    The region X≥33X\ge 33 has probability greater than 0.050.05, so it is too large to be the critical region.

  6. Test k=34k=34 in the upper tail

    P(X≥34)=0.063826>0.05P(X\ge 34)=0.063826> 0.05

    The region X≥34X\ge 34 has probability greater than 0.050.05, so it is too large to be the critical region.

  7. Test k=35k=35 in the upper tail

    P(X≥35)=0.058720>0.05P(X\ge 35)=0.058720> 0.05

    The region X≥35X\ge 35 has probability greater than 0.050.05, so it is too large to be the critical region.

  8. Test k=36k=36 in the upper tail

    P(X≥36)=0.054022>0.05P(X\ge 36)=0.054022> 0.05

    The region X≥36X\ge 36 has probability greater than 0.050.05, so it is too large to be the critical region.

  9. Test k=37k=37 in the upper tail

    P(X≥37)=0.049701≤0.05P(X\ge 37)=0.049701\le 0.05

    The condition holds, and no smaller kk works, so this is the critical value.

  10. State the critical value

    c=37c=37

    This is the integer picked out by the stated convention.

  11. State the critical region

    X≥37X \ge 37

    The test rejects H0H_0 exactly when XX lies in this region.

  12. Write the actual significance level as a probability

    P(critical region)=P(X≥37)P(\text{critical region})=P(X \ge 37)

    The actual significance level is the probability of the critical region under H0H_0, not the nominal level.

  13. Evaluate that probability

    P(X≥37)=0.049701P(X \ge 37)=0.049701

    This is computed from the distribution of XX under H0H_0.

  14. Compare the actual level with the nominal level

    0.049701<0.050.049701 < 0.05

    Because XX is discrete the actual level is strictly less than the nominal level of 0.050.05.

  15. State the actual significance level to 44 decimal places

    P(critical region)=0.0497P(\text{critical region})=0.0497

    This is the true probability of a Type I error for this test.

Answer
0.04970.0497
Question 3
9 markschallenging
The number of breakdowns of a machine in a randomly chosen month is modelled by a Poisson distribution with mean λ\lambda. A hypothesis test of H0H_0: λ=6.5\lambda = 6.5 against H1H_1: λ<6.5\lambda < 6.5 is carried out at the 5%5\% significance level, using the single observation XX. This is a one-tailed test and the critical region lies in the lower tail of XX. The critical region is the largest lower-tail region whose probability does not exceed the significance level: it is X≤cX \le c, where cc is the largest integer for which P(X≤c)≤0.05P(X \le c) \le 0.05. The actual significance level of the test is the probability, calculated under H0H_0, that XX lies in the critical region. Find the actual significance level of this test, giving your answer as a probability correct to 44 decimal places.
Show worked solution

Worked solution

  1. State the hypotheses and the tail of the test

    H0:λ=6.5,H1:λ<6.5(one-tailed, lower tail)H_{0}:\lambda=6.5,\quad H_{1}:\lambda<6.5\quad \text{(one-tailed, lower tail)}

    The alternative hypothesis fixes which tail (or tails) of XX forms the critical region.

  2. State the distribution of the test statistic under H0H_0

    X∼Po(6.5)X\sim\text{Po}(6.5)

    All probabilities below are calculated assuming H0H_0 is true.

  3. State the critical-region rule for the lower tail

    c=max⁡{k:P(X≤k)≤0.05}c=\max\{k:P(X\le k)\le 0.05\}

    The critical region is the largest lower-tail region whose probability does not exceed 0.050.05.

  4. Recall the Poisson probability function

    P(X=x)=e−λλxx!,x=0, 1, 2, …P(X=x)=e^{-\lambda}\frac{\lambda^{x}}{x!},\quad x=0,\ 1,\ 2,\ \ldots

    Every probability in this test is obtained from this formula.

  5. Note how a cumulative probability is built up

    P(X≤k)=∑x=0ke−λλxx!P(X\le k)=\sum_{x=0}^{k}e^{-\lambda}\frac{\lambda^{x}}{x!}

    The lower tail is a finite sum of Poisson probabilities.

  6. Relate the upper tail to a cumulative probability

    P(X≥c)=1−P(X≤c−1)P(X\ge c)=1-P(X\le c-1)

    Note the c−1c-1: the value X=c−1X=c-1 is not part of the region X≥cX\ge c.

  7. Test k=0k=0 in the lower tail

    P(X≤0)=0.001503≤0.05P(X\le 0)=0.001503\le 0.05

    The condition holds, so X≤0X\le 0 is a permissible region.

  8. Test k=1k=1 in the lower tail

    P(X≤1)=0.011276≤0.05P(X\le 1)=0.011276\le 0.05

    The condition holds, so X≤1X\le 1 is a permissible region.

  9. Test k=2k=2 in the lower tail

    P(X≤2)=0.043036≤0.05P(X\le 2)=0.043036\le 0.05

    The condition holds, so X≤2X\le 2 is a permissible region.

  10. Test k=3k=3 in the lower tail

    P(X≤3)=0.111850>0.05P(X\le 3)=0.111850> 0.05

    The region X≤3X\le 3 has probability greater than 0.050.05, so it is too large; the previous value of kk is the critical value.

  11. State the critical value

    c=2c=2

    This is the integer picked out by the stated convention.

  12. State the critical region

    X≤2X \le 2

    The test rejects H0H_0 exactly when XX lies in this region.

  13. Write the actual significance level as a probability

    P(critical region)=P(X≤2)P(\text{critical region})=P(X \le 2)

    The actual significance level is the probability of the critical region under H0H_0, not the nominal level.

  14. Evaluate that probability

    P(X≤2)=0.043036P(X \le 2)=0.043036

    This is computed from the distribution of XX under H0H_0.

  15. Compare the actual level with the nominal level

    0.043036<0.050.043036 < 0.05

    Because XX is discrete the actual level is strictly less than the nominal level of 0.050.05.

  16. State the actual significance level to 44 decimal places

    P(critical region)=0.0430P(\text{critical region})=0.0430

    This is the true probability of a Type I error for this test.

Answer
0.04300.0430
Question 4
9 markschallenging
The number of bacteria colonies on a randomly chosen agar plate is modelled by a Poisson distribution with mean λ\lambda. A hypothesis test of H0H_0: λ=13\lambda = 13 against H1H_1: λ>13\lambda > 13 is carried out at the 5%5\% significance level, using the single observation XX. This is a one-tailed test and the critical region lies in the upper tail of XX. The critical region is the largest upper-tail region whose probability does not exceed the significance level: it is X≥cX \ge c, where cc is the smallest integer for which P(X≥c)≤0.05P(X \ge c) \le 0.05. The actual significance level of the test is the probability, calculated under H0H_0, that XX lies in the critical region. Find the actual significance level of this test, giving your answer as a probability correct to 44 decimal places.
Show worked solution

Worked solution

  1. State the hypotheses and the tail of the test

    H0:λ=13,H1:λ>13(one-tailed, upper tail)H_{0}:\lambda=13,\quad H_{1}:\lambda>13\quad \text{(one-tailed, upper tail)}

    The alternative hypothesis fixes which tail (or tails) of XX forms the critical region.

  2. State the distribution of the test statistic under H0H_0

    X∼Po(13)X\sim\text{Po}(13)

    All probabilities below are calculated assuming H0H_0 is true.

  3. State the critical-region rule for the upper tail

    c=min⁡{k:P(X≥k)≤0.05}c=\min\{k:P(X\ge k)\le 0.05\}

    The critical region is the largest upper-tail region whose probability does not exceed 0.050.05.

  4. Recall the Poisson probability function

    P(X=x)=e−λλxx!,x=0, 1, 2, …P(X=x)=e^{-\lambda}\frac{\lambda^{x}}{x!},\quad x=0,\ 1,\ 2,\ \ldots

    Every probability in this test is obtained from this formula.

  5. Test k=16k=16 in the upper tail

    P(X≥16)=0.236393>0.05P(X\ge 16)=0.236393> 0.05

    The region X≥16X\ge 16 has probability greater than 0.050.05, so it is too large to be the critical region.

  6. Test k=17k=17 in the upper tail

    P(X≥17)=0.164507>0.05P(X\ge 17)=0.164507> 0.05

    The region X≥17X\ge 17 has probability greater than 0.050.05, so it is too large to be the critical region.

  7. Test k=18k=18 in the upper tail

    P(X≥18)=0.109535>0.05P(X\ge 18)=0.109535> 0.05

    The region X≥18X\ge 18 has probability greater than 0.050.05, so it is too large to be the critical region.

  8. Test k=19k=19 in the upper tail

    P(X≥19)=0.069833>0.05P(X\ge 19)=0.069833> 0.05

    The region X≥19X\ge 19 has probability greater than 0.050.05, so it is too large to be the critical region.

  9. Test k=20k=20 in the upper tail

    P(X≥20)=0.042669≤0.05P(X\ge 20)=0.042669\le 0.05

    The condition holds, and no smaller kk works, so this is the critical value.

  10. State the critical value

    c=20c=20

    This is the integer picked out by the stated convention.

  11. State the critical region

    X≥20X \ge 20

    The test rejects H0H_0 exactly when XX lies in this region.

  12. Write the actual significance level as a probability

    P(critical region)=P(X≥20)P(\text{critical region})=P(X \ge 20)

    The actual significance level is the probability of the critical region under H0H_0, not the nominal level.

  13. Evaluate that probability

    P(X≥20)=0.042669P(X \ge 20)=0.042669

    This is computed from the distribution of XX under H0H_0.

  14. Compare the actual level with the nominal level

    0.042669<0.050.042669 < 0.05

    Because XX is discrete the actual level is strictly less than the nominal level of 0.050.05.

  15. State the actual significance level to 44 decimal places

    P(critical region)=0.0427P(\text{critical region})=0.0427

    This is the true probability of a Type I error for this test.

Answer
0.04270.0427
Question 5
9 markschallenging
Emails are opened one at a time and each email contains an attachment with probability pp, independently of the others. Let XX be the number of emails opened up to and including the first one containing an attachment, so that P(X=x)=(1−p)x−1pP(X=x)=(1-p)^{x-1}p for x=1, 2, 3, …x=1,\ 2,\ 3,\ \ldots A hypothesis test of H0H_0: p=0.01p = 0.01 against H1H_1: p≠0.01p \neq 0.01 is carried out at the 10%10\% significance level, using the single observation XX. This is a two-tailed test, so a probability of at most 0.050.05 (that is 5%5\%) is used in each tail. The critical region is the largest region of each tail whose probability does not exceed 0.050.05: it is X≤c1X \le c_1 or X≥c2X \ge c_2, where c1c_1 is the largest integer with P(X≤c1)≤0.05P(X \le c_1) \le 0.05 and c2c_2 is the smallest integer with P(X≥c2)≤0.05P(X \ge c_2) \le 0.05. State the critical region for this test.
Show worked solution

Worked solution

  1. State the hypotheses and the tail of the test

    H0:p=0.01,H1:p≠0.01(two-tailed)H_{0}:p=0.01,\quad H_{1}:p\neq0.01\quad \text{(two-tailed)}

    The alternative hypothesis fixes which tail (or tails) of XX forms the critical region.

  2. State the distribution of the test statistic under H0H_0

    X∼Geo(0.01)X\sim\text{Geo}(0.01)

    All probabilities below are calculated assuming H0H_0 is true.

  3. Split the significance level between the two tails

    0.12=0.05 in each tail\frac{0.1}{2}=0.05\text{ in each tail}

    For a two-tailed test each tail is allowed a probability of at most 0.050.05.

  4. Recall the geometric probability function

    P(X=x)=(1−p)x−1p,x=1, 2, 3, …P(X=x)=\left(1-p\right)^{x-1}p,\quad x=1,\ 2,\ 3,\ \ldots

    The first x−1x-1 trials fail and the xxth succeeds.

  5. State the critical-region rule for the lower tail

    c1=max⁡{k:P(X≤k)≤0.05}c_{1}=\max\{k:P(X\le k)\le 0.05\}

    The lower part of the critical region is the largest lower-tail region with probability at most 0.050.05.

  6. Test k=4k=4 in the lower tail

    P(X≤4)=0.039404≤0.05P(X\le 4)=0.039404\le 0.05

    The condition holds, so X≤4X\le 4 is a permissible region.

  7. Test k=5k=5 in the lower tail

    P(X≤5)=0.049010≤0.05P(X\le 5)=0.049010\le 0.05

    The condition holds, so X≤5X\le 5 is a permissible region.

  8. Test k=6k=6 in the lower tail

    P(X≤6)=0.058520>0.05P(X\le 6)=0.058520> 0.05

    The region X≤6X\le 6 has probability greater than 0.050.05, so it is too large; the previous value of kk is the critical value.

  9. State the lower critical value

    c1=5c_{1}=5

    This is the largest integer whose lower tail has probability at most 0.050.05.

  10. State the critical-region rule for the upper tail

    c2=min⁡{k:P(X≥k)≤0.05}c_{2}=\min\{k:P(X\ge k)\le 0.05\}

    The upper part of the critical region is the largest upper-tail region with probability at most 0.050.05.

  11. Test k=298k=298 in the upper tail

    P(X≥298)=0.050542>0.05P(X\ge 298)=0.050542> 0.05

    The region X≥298X\ge 298 has probability greater than 0.050.05, so it is too large to be the critical region.

  12. Test k=299k=299 in the upper tail

    P(X≥299)=0.050037>0.05P(X\ge 299)=0.050037> 0.05

    The region X≥299X\ge 299 has probability greater than 0.050.05, so it is too large to be the critical region.

  13. Test k=300k=300 in the upper tail

    P(X≥300)=0.049536≤0.05P(X\ge 300)=0.049536\le 0.05

    The condition holds, and no smaller kk works, so this is the critical value.

  14. State the upper critical value

    c2=300c_{2}=300

    This is the smallest integer whose upper tail has probability at most 0.050.05.

  15. State the critical region

    X≤5 or X≥300X \le 5 \text{ or } X \ge 300

    The test rejects H0H_0 exactly when XX lies in this region.

Answer
X≤5 or X≥300X \le 5 \text{ or } X \ge 300

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