Edexcel FS1 2020 June — Question 1 13 marks

Exam BoardEdexcel
ModuleFS1 (Further Statistics 1)
Year2020
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of a Poisson distribution
TypeOne-tailed test (increase or decrease)
DifficultyStandard +0.8 This is a Further Maths Statistics question requiring multiple Poisson hypothesis test techniques including Type II error calculation, which goes beyond standard A-level. While the individual calculations are methodical, the multi-part structure, parameter scaling across different time periods, and Type II error probability (requiring understanding of both null and alternative distributions) make this moderately challenging for Further Maths students.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.05c Hypothesis test: normal distribution for population mean

  1. The number of customers entering Jeff's supermarket each morning follows a Poisson distribution.
Past information shows that customers enter at an average rate of 2 every 5 minutes.
Using this information,
    1. find the probability that exactly 26 customers enter Jeff's supermarket during a randomly selected 1-hour period one morning,
    2. find the probability that at least 21 customers enter Jeff's supermarket during a randomly selected 1-hour period one morning. A rival supermarket is opened nearby. Following its opening, the number of customers entering Jeff's supermarket over a randomly selected 40-minute period is found to be 10
  1. Test, at the 5\% significance level, whether or not there is evidence of a decrease in the rate of customers entering Jeff's supermarket. State your hypotheses clearly. A further randomly selected 20 -minute period is observed and the hypothesis test is repeated. Given that the true rate of customers entering Jeff's supermarket is now 1 every 5 minutes,
  2. calculate the probability of a Type II error.

Question 1:
Part (a)(i)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(X \sim Po(24)\)B1 For realising the distribution is \(Po(24)\) (May be seen or implied in part (ii))
\(P(X = 26) = 0.071912\ldots\) awrt \(\mathbf{0.0719}\)B1 awrt 0.0719
Part (a)(ii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(P(X \geq 21) = 1 - P(X \leq 20)\ [= 1 - 0.24263\ldots]\)M1 Writing or using \(1 - P(X \leq 20)\)
\(= 0.75736\ldots\) awrt \(\mathbf{0.757}\)A1 awrt 0.757
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(H_0: \lambda = 2\ [\mu = 16]\), \(H_1: \lambda < 2\ [\mu < 16]\)B1 Both hypotheses correct (must use \(\mu\) or \(\lambda\))
\(P(Y \leq 10 \mid Y \sim Po(16)) = 0.077396\ldots\) awrt \(\mathbf{0.0774}\)B1 awrt 0.0774; allow awrt 0.08 from correct probability statement; allow CR: \(X \leq 9\)
Not significant / Do not reject \(H_0\) / 10 is not in the CRM1 Correct non-contextual conclusion (may be implied by correct contextual conclusion). Allow f.t. comparison of 'their \(p\)' with 0.05. Ignore contradictory contextual comments for this mark
There is not sufficient evidence to suggest a decrease/change in the rate of customers entering Jeff's supermarketA1 Fully correct solution drawing correct inference in context with all previous marks in (b) scored
Part (c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Use of \(Po(8)\) to attempt critical regionM1 \([P(Y \leq 3) = 0.0423\ldots\ P(Y \leq 4) = 0.0996\ldots]\)
Critical region is \(Y \leq 3\); \(H_0\) is not rejected when \(Y \geq 4\)A1 Finding critical region for the test \(Y \leq 3\) which must come from \(Po(8)\)
True distribution is \(W \sim Po(4)\)B1 Identifying the need to use \(Po(4)\) as the true distribution. Allow \(Po(4)\) seen or used for this mark
\(P(W \geq 4 \mid W \sim Po(4)) = 1 - P(W \leq 3)\ [= 1 - 0.43347\ldots]\)M1 Writing or using \(P(W \geq \text{`}4\text{'})\) or \(1 - P(W \leq \text{`}3\text{'})\) from \(Po(4)\). Allow f.t. on their identified CR but must be using \(Po(4)\)
\(= 0.56652\ldots\) awrt \(\mathbf{0.567}\)A1 awrt 0.567
## Question 1:

### Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim Po(24)$ | B1 | For realising the distribution is $Po(24)$ (May be seen or implied in part (ii)) |
| $P(X = 26) = 0.071912\ldots$ awrt $\mathbf{0.0719}$ | B1 | awrt 0.0719 |

### Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X \geq 21) = 1 - P(X \leq 20)\ [= 1 - 0.24263\ldots]$ | M1 | Writing or using $1 - P(X \leq 20)$ |
| $= 0.75736\ldots$ awrt $\mathbf{0.757}$ | A1 | awrt 0.757 |

### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \lambda = 2\ [\mu = 16]$, $H_1: \lambda < 2\ [\mu < 16]$ | B1 | Both hypotheses correct (must use $\mu$ or $\lambda$) |
| $P(Y \leq 10 \mid Y \sim Po(16)) = 0.077396\ldots$ awrt $\mathbf{0.0774}$ | B1 | awrt 0.0774; allow awrt 0.08 from correct probability statement; allow CR: $X \leq 9$ |
| Not significant / Do not reject $H_0$ / 10 is not in the CR | M1 | Correct non-contextual conclusion (may be implied by correct contextual conclusion). Allow f.t. comparison of 'their $p$' with 0.05. Ignore contradictory contextual comments for this mark |
| There is not sufficient evidence to suggest a decrease/change in the rate of customers entering Jeff's supermarket | A1 | Fully correct solution drawing correct inference in context **with all previous marks in (b) scored** |

### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $Po(8)$ to attempt critical region | M1 | $[P(Y \leq 3) = 0.0423\ldots\ P(Y \leq 4) = 0.0996\ldots]$ |
| Critical region is $Y \leq 3$; $H_0$ is not rejected when $Y \geq 4$ | A1 | Finding critical region for the test $Y \leq 3$ which must come from $Po(8)$ |
| True distribution is $W \sim Po(4)$ | B1 | Identifying the need to use $Po(4)$ as the true distribution. Allow $Po(4)$ seen or used for this mark |
| $P(W \geq 4 \mid W \sim Po(4)) = 1 - P(W \leq 3)\ [= 1 - 0.43347\ldots]$ | M1 | Writing or using $P(W \geq \text{`}4\text{'})$ or $1 - P(W \leq \text{`}3\text{'})$ from $Po(4)$. Allow f.t. on their **identified** CR but must be using $Po(4)$ |
| $= 0.56652\ldots$ awrt $\mathbf{0.567}$ | A1 | awrt 0.567 |

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\begin{enumerate}
  \item The number of customers entering Jeff's supermarket each morning follows a Poisson distribution.
\end{enumerate}

Past information shows that customers enter at an average rate of 2 every 5 minutes.\\
Using this information,\\
(a) (i) find the probability that exactly 26 customers enter Jeff's supermarket during a randomly selected 1-hour period one morning,\\
(ii) find the probability that at least 21 customers enter Jeff's supermarket during a randomly selected 1-hour period one morning.

A rival supermarket is opened nearby. Following its opening, the number of customers entering Jeff's supermarket over a randomly selected 40-minute period is found to be 10\\
(b) Test, at the 5\% significance level, whether or not there is evidence of a decrease in the rate of customers entering Jeff's supermarket. State your hypotheses clearly.

A further randomly selected 20 -minute period is observed and the hypothesis test is repeated. Given that the true rate of customers entering Jeff's supermarket is now 1 every 5 minutes,\\
(c) calculate the probability of a Type II error.

\hfill \mbox{\textit{Edexcel FS1 2020 Q1 [13]}}