Hypothesis testing (Poisson, geometric) Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Hypothesis testing (Poisson, geometric) questions. See exactly how to solve problems on hypothesis-testing, tail-probability, poisson, geometric.

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Further Maths70 questionsStep-by-step solutions

Question 1

2 markseasy

The number of flaws in a randomly chosen

1

m length of copper wire is modelled by a Poisson distribution with mean

\lambda

. A hypothesis test of

H_0

\lambda = 4.5

against

H_1

\lambda > 4.5

is carried out at the

5\%

significance level, using the single observation

X

. This is a one-tailed test and the critical region lies in the upper tail of

X

. The critical region is the largest upper-tail region whose probability does not exceed the significance level: it is

X \ge c

, where

c

is the smallest integer for which

P(X \ge c) \le 0.05

. The observed value is

x = 9

. Calculate the probability, under

H_0

, of obtaining a value at least as extreme as this in the direction of

H_1

, that is

P(X \ge 9)

. Give your answer correct to

4

decimal places.

Worked solution

State the hypotheses, the tail and the distribution under $H_0$
$H_{0}:\lambda=4.5,\ H_{1}:\lambda>4.5\ \text{(upper tail)};\quad X\sim\text{Po}(4.5)$
The alternative hypothesis says that an extreme result means a large value of $X$ , so the upper tail is used.
Write the required probability using the complement
$P(X\ge 9)=1-P(X\le 8)=1-\sum_{k=0}^{8}e^{-4.5}\frac{4.5^{k}}{k!}$
The value $X=8$ is not part of the region $X\ge 9$ , so the cumulative probability stops at $8$ .
Evaluate the probability
$P(X \ge 9)=0.040257$
This is the probability, under $H_0$ , of a result at least as extreme as the one observed.
State the required probability to $4$ decimal places
$P(X \ge 9)=0.0403$
This is the probability asked for, in the tail indicated by $H_1$ .

Answer

0.0403

Question 2

2 markseasy

The number of calls received by a helpline in a randomly chosen

10

-minute interval is modelled by a Poisson distribution with mean

\lambda

. A hypothesis test of

H_0

\lambda = 6

against

H_1

\lambda > 6

is carried out at the

5\%

significance level, using the single observation

X

. This is a one-tailed test and the critical region lies in the upper tail of

X

. The critical region is the largest upper-tail region whose probability does not exceed the significance level: it is

X \ge c

, where

c

is the smallest integer for which

P(X \ge c) \le 0.05

. The observed value is

x = 11

. Calculate the probability, under

H_0

, of obtaining a value at least as extreme as this in the direction of

H_1

, that is

P(X \ge 11)

. Give your answer correct to

4

decimal places.

Worked solution

State the hypotheses, the tail and the distribution under $H_0$
$H_{0}:\lambda=6,\ H_{1}:\lambda>6\ \text{(upper tail)};\quad X\sim\text{Po}(6)$
The alternative hypothesis says that an extreme result means a large value of $X$ , so the upper tail is used.
Write the required probability using the complement
$P(X\ge 11)=1-P(X\le 10)=1-\sum_{k=0}^{10}e^{-6}\frac{6^{k}}{k!}$
The value $X=10$ is not part of the region $X\ge 11$ , so the cumulative probability stops at $10$ .
Evaluate the probability
$P(X \ge 11)=0.042621$
This is the probability, under $H_0$ , of a result at least as extreme as the one observed.
State the required probability to $4$ decimal places
$P(X \ge 11)=0.0426$
This is the probability asked for, in the tail indicated by $H_1$ .

Answer

0.0426

Question 3

2 markseasy

The number of cars passing a remote checkpoint in a randomly chosen minute is modelled by a Poisson distribution with mean

\lambda

. A hypothesis test of

H_0

\lambda = 3.2

against

H_1

\lambda > 3.2

is carried out at the

10\%

significance level, using the single observation

X

. This is a one-tailed test and the critical region lies in the upper tail of

X

. The critical region is the largest upper-tail region whose probability does not exceed the significance level: it is

X \ge c

, where

c

is the smallest integer for which

P(X \ge c) \le 0.1

. The observed value is

x = 6

. Calculate the probability, under

H_0

, of obtaining a value at least as extreme as this in the direction of

H_1

, that is

P(X \ge 6)

. Give your answer correct to

4

decimal places.

Worked solution

State the hypotheses, the tail and the distribution under $H_0$
$H_{0}:\lambda=3.2,\ H_{1}:\lambda>3.2\ \text{(upper tail)};\quad X\sim\text{Po}(3.2)$
The alternative hypothesis says that an extreme result means a large value of $X$ , so the upper tail is used.
Evaluate the probability
$P(X \ge 6)=0.105408$
This is the probability, under $H_0$ , of a result at least as extreme as the one observed.
State the required probability to $4$ decimal places
$P(X \ge 6)=0.1054$
This is the probability asked for, in the tail indicated by $H_1$ .

Answer

0.1054

Question 4

2 markseasy

The number of misprints on a randomly chosen page of a newspaper is modelled by a Poisson distribution with mean

\lambda

. A hypothesis test of

H_0

\lambda = 8

against

H_1

\lambda > 8

is carried out at the

1\%

significance level, using the single observation

X

. This is a one-tailed test and the critical region lies in the upper tail of

X

. The critical region is the largest upper-tail region whose probability does not exceed the significance level: it is

X \ge c

, where

c

is the smallest integer for which

P(X \ge c) \le 0.01

. The observed value is

x = 15

. Calculate the probability, under

H_0

, of obtaining a value at least as extreme as this in the direction of

H_1

, that is

P(X \ge 15)

. Give your answer correct to

4

decimal places.

Worked solution

State the hypotheses, the tail and the distribution under $H_0$
$H_{0}:\lambda=8,\ H_{1}:\lambda>8\ \text{(upper tail)};\quad X\sim\text{Po}(8)$
The alternative hypothesis says that an extreme result means a large value of $X$ , so the upper tail is used.
Write the required probability using the complement
$P(X\ge 15)=1-P(X\le 14)=1-\sum_{k=0}^{14}e^{-8}\frac{8^{k}}{k!}$
The value $X=14$ is not part of the region $X\ge 15$ , so the cumulative probability stops at $14$ .
Evaluate the probability
$P(X \ge 15)=0.017257$
This is the probability, under $H_0$ , of a result at least as extreme as the one observed.
State the required probability to $4$ decimal places
$P(X \ge 15)=0.0173$
This is the probability asked for, in the tail indicated by $H_1$ .

Answer

0.0173

Question 5

2 markseasy

The number of accidents at a road junction in a randomly chosen week is modelled by a Poisson distribution with mean

\lambda

. A hypothesis test of

H_0

\lambda = 5.5

against

H_1

\lambda > 5.5

is carried out at the

5\%

significance level, using the single observation

X

. This is a one-tailed test and the critical region lies in the upper tail of

X

. The critical region is the largest upper-tail region whose probability does not exceed the significance level: it is

X \ge c

, where

c

is the smallest integer for which

P(X \ge c) \le 0.05

. The observed value is

x = 10

. Calculate the probability, under

H_0

, of obtaining a value at least as extreme as this in the direction of

H_1

, that is

P(X \ge 10)

. Give your answer correct to

4

decimal places.

Worked solution

State the hypotheses, the tail and the distribution under $H_0$
$H_{0}:\lambda=5.5,\ H_{1}:\lambda>5.5\ \text{(upper tail)};\quad X\sim\text{Po}(5.5)$
The alternative hypothesis says that an extreme result means a large value of $X$ , so the upper tail is used.
Write the required probability using the complement
$P(X\ge 10)=1-P(X\le 9)=1-\sum_{k=0}^{9}e^{-5.5}\frac{5.5^{k}}{k!}$
The value $X=9$ is not part of the region $X\ge 10$ , so the cumulative probability stops at $9$ .
Evaluate the probability
$P(X \ge 10)=0.053777$
This is the probability, under $H_0$ , of a result at least as extreme as the one observed.
State the required probability to $4$ decimal places
$P(X \ge 10)=0.0538$
This is the probability asked for, in the tail indicated by $H_1$ .

Answer

0.0538

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