Further Maths Hypothesis testing (Poisson, geometric) Practice Questions

Free Further Maths Hypothesis testing (Poisson, geometric) practice questions with full step-by-step worked solutions. Covers hypothesis-testing, tail-probability, poisson, geometric. Practise exam-style problems and check your method.

hypothesis-testingtail-probabilitypoissongeometriccritical-valuecritical-region
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The number of flaws in a randomly chosen 11 m length of copper wire is modelled by a Poisson distribution with mean λ\lambda. A hypothesis test of H0H_0: λ=4.5\lambda = 4.5 against H1H_1: λ>4.5\lambda > 4.5 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 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 observed value is x=9x = 9. Calculate the probability, under H0H_0, of obtaining a value at least as extreme as this in the direction of H1H_1, that is P(X≥9)P(X \ge 9). Give your answer correct to 44 decimal places.
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Worked solution

  1. State the hypotheses, the tail and the distribution under H0H_0

    H0:λ=4.5, H1:λ>4.5 (upper tail);X∼Po(4.5)H_{0}:\lambda=4.5,\ H_{1}:\lambda>4.5\ \text{(upper tail)};\quad X\sim\text{Po}(4.5)

    The alternative hypothesis says that an extreme result means a large value of XX, so the upper tail is used.

  2. Write the required probability using the complement

    P(X≥9)=1−P(X≤8)=1−∑k=08e−4.54.5kk!P(X\ge 9)=1-P(X\le 8)=1-\sum_{k=0}^{8}e^{-4.5}\frac{4.5^{k}}{k!}

    The value X=8X=8 is not part of the region X≥9X\ge 9, so the cumulative probability stops at 88.

  3. Evaluate the probability

    P(X≥9)=0.040257P(X \ge 9)=0.040257

    This is the probability, under H0H_0, of a result at least as extreme as the one observed.

  4. State the required probability to 44 decimal places

    P(X≥9)=0.0403P(X \ge 9)=0.0403

    This is the probability asked for, in the tail indicated by H1H_1.

Answer
0.04030.0403
Question 2
2 markseasy
Eggs are examined one at a time and each egg is cracked with probability pp, independently of the others. Let XX be the number of eggs examined up to and including the first cracked 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=0.05p = 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 observed value is x=2x = 2. Calculate the probability, under H0H_0, of obtaining a value at least as extreme as this in the direction of H1H_1, that is P(X≤2)P(X \le 2). Give your answer correct to 44 decimal places.
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Worked solution

  1. State the hypotheses, the tail and the distribution under H0H_0

    H0:p=0.05, H1:p>0.05 (lower tail);X∼Geo(0.05)H_{0}:p=0.05,\ H_{1}:p>0.05\ \text{(lower tail)};\quad X\sim\text{Geo}(0.05)

    The alternative hypothesis says that an extreme result means a small value of XX, so the lower tail is used.

  2. Evaluate the probability

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

    This is the probability, under H0H_0, of a result at least as extreme as the one observed.

  3. State the required probability to 44 decimal places

    P(X≤2)=0.0975P(X \le 2)=0.0975

    This is the probability asked for, in the tail indicated by H1H_1.

Answer
0.09750.0975
Question 3
4 marksintermediate
Rock samples are drilled one at a time and each sample contains a trace of gold with probability pp, independently of the others. Let XX be the number of samples drilled up to and including the first one containing a trace of gold, 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=0.04p = 0.04 against H1H_1: p>0.04p > 0.04 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 observed value is x=1x = 1. Which of the following is the correct conclusion of this test?
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Worked solution

  1. State the hypotheses and the tail of the test

    H0:p=0.04,H1:p>0.04(one-tailed, lower tail)H_{0}:p=0.04,\quad H_{1}:p>0.04\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(0.04)X\sim\text{Geo}(0.04)

    All probabilities below are calculated assuming H0H_0 is true.

  3. Write down the probability of a result at least as extreme as the observation

    P(X≤1)P(X \le 1)

    This is calculated under H0H_0 in the tail indicated by H1H_1.

  4. Evaluate that probability

    P(X≤1)=0.040000P(X \le 1)=0.040000

    The value is obtained from the distribution of XX under H0H_0.

  5. Compare the probability with the significance level for that tail

    0.0400≤0.050.0400 \le 0.05

    The observation therefore lies in the critical region.

  6. State the decision

    Reject H0\text{Reject }H_{0}

    The decision follows directly from the comparison above.

  7. Select the conclusion stated in context

    Reject H0: sufficient evidence at the 5% level that the probability that a sample contains a trace of gold has increased\text{Reject }H_{0}\text{: sufficient evidence at the 5\% level that the probability that a sample contains a trace of gold has increased}

    The conclusion must be expressed in the context of the question.

Answer
Reject H0: sufficient evidence at the 5% level that the probability that a sample contains a trace of gold has increased\text{Reject }H_{0}\text{: sufficient evidence at the 5\% level that the probability that a sample contains a trace of gold has increased}
Question 4
6 markshard
Applicants are interviewed one at a time and each applicant is offered a post with probability pp, independently of the others. Let XX be the number of applicants interviewed up to and including the first one offered a post, 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=0.03p = 0.03 against H1H_1: p≠0.03p \neq 0.03 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. The observed value is x=110x = 110. The observed value is judged in the tail in which it lies, and its probability in that tail is compared with 0.050.05. Which of the following is the correct conclusion of this test?
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Worked solution

  1. State the hypotheses and the tail of the test

    H0:p=0.03,H1:p≠0.03(two-tailed)H_{0}:p=0.03,\quad H_{1}:p\neq0.03\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(0.03)X\sim\text{Geo}(0.03)

    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. Decide which tail the observed value lies in

    x=110,E(X)=33.3333⇒upper tailx=110,\quad \text{E}(X)=33.3333\Rightarrow \text{upper tail}

    The observation is compared with the tail on the side of the mean on which it falls.

  5. 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.

  6. 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.

  7. Write down the probability of a result at least as extreme as the observation

    P(X≥110)P(X \ge 110)

    This is calculated under H0H_0 in the tail indicated by H1H_1.

  8. Evaluate that probability

    P(X≥110)=0.036151P(X \ge 110)=0.036151

    The value is obtained from the distribution of XX under H0H_0.

  9. Compare the probability with the significance level for that tail

    0.0362≤0.050.0362 \le 0.05

    The observation therefore lies in the critical region.

  10. Confirm the decision using the critical region

    critical region: X≤1 or X≥100;x=110 ∈CR\text{critical region: }X \le 1 \text{ or } X \ge 100;\quad x=110\ \in \text{CR}

    Membership of the critical region gives the same decision as the tail probability, as it must.

  11. State the decision

    Reject H0\text{Reject }H_{0}

    The decision follows directly from the comparison above.

  12. Select the conclusion stated in context

    Reject H0: sufficient evidence at the 10% level that the probability that an applicant is offered a post has changed\text{Reject }H_{0}\text{: sufficient evidence at the 10\% level that the probability that an applicant is offered a post has changed}

    The conclusion must be expressed in the context of the question.

Answer
Reject H0: sufficient evidence at the 10% level that the probability that an applicant is offered a post has changed\text{Reject }H_{0}\text{: sufficient evidence at the 10\% level that the probability that an applicant is offered a post has changed}
Question 5
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.
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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

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