Edexcel S2 2021 June — Question 1 14 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2021
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeMultiple binomial probability calculations
DifficultyStandard +0.3 This is a standard S2 binomial hypothesis testing question with routine calculations: basic binomial probabilities, independence application, and normal approximation to binomial. All steps follow textbook procedures with no novel insight required, making it slightly easier than average for A-level.
Spec5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean

  1. Spany sells seeds and claims that \(5 \%\) of its pansy seeds do not germinate. A packet of pansy seeds contains 20 seeds. Each seed germinates independently of the other seeds.
    1. Find the probability that in a packet of Spany's pansy seeds
      1. more than 2 but fewer than 5 seeds do not germinate,
      2. more than 18 seeds germinate.
    Jem buys 5 packets of Spany's pansy seeds.
  2. Calculate the probability that all of these packets contain more than 18 seeds that germinate. Jem believes that Spany's claim is incorrect. She believes that the percentage of pansy seeds that do not germinate is greater than 5\%
  3. Write down the hypotheses for a suitable test to examine Jem's belief. Jem planted all of the 100 seeds she bought from Spany and found that 8 did not germinate.
  4. Using a suitable approximation, carry out the test using a \(5 \%\) level of significance.

Question 1:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(X \sim B(20, 0.05)\) or \(Y \sim B(20, 0.95)\)B1 Writing or using \(B(20,0.05)\). Allow \(Y \sim B(20, 0.95)\) if \(Y\) is clearly defined. Implied by 1 correct probability.
Part (a)(i)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(P(X \leqslant 4) - P(X \leqslant 2)\) or \(\binom{20}{3}0.05^3 \times 0.95^{17} + \binom{20}{4}0.05^4 \times 0.95^{16}\)M1 For \(P(X \leqslant 4) - P(X \leqslant 2)\) and one correct prob, or \(P(X=3) + P(X=4)\) and 1 correct prob
\(= 0.072909...\) awrt \(\mathbf{0.0729}\)A1
Part (a)(ii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(P(X \leqslant 1)\) or \(P(Y \geqslant 19) = 20 \times (0.95)^{19}(0.05) + (0.95)^{20}\)M1 For \(P(X \leqslant 1)\) or \([20]\times(0.95)^{19}(0.05)+(0.95)^{20}\) – condone missing 20
\(= 0.7358\) awrt \(\mathbf{0.736}\)A1 (5 marks total for (i) and (ii))
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Let \(W\) = no. of packets where \(Y > 18\); \(P(W=5) = (0.7358...)^5\)M1 For \((\text{their}(a)(ii))^5\)
\(= 0.21573...\) awrt \(\mathbf{0.216}\)A1 (2 marks)
Part (c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(H_0: p = 0.05 \quad H_1: p > 0.05\)B1 Both hypotheses correct with \(p\) or \(\pi\) (1 mark)
Part (d)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(V \sim B(100, 0.05)\) approximated by \(V \sim \text{Po}(5)\)M1A1 1st M1: realising Poisson approximation appropriate. NB Po(95) is M0A0. 1st A1: correct mean for Poisson
\(P(V \geqslant 8) = 1 - P(V \leqslant 7) = 1 - 0.8666 = 0.1334\)M1 2nd M1: for \(1 - P(V \leqslant 7)\) or \(P(V \leqslant 7) = 0.8666\); or writing \(P(V \geqslant 10) = 0.0318\) or \(P(V \geqslant 9) = 0.0681\) or \(P(V \geqslant 11) = 0.0137\) leading to a CR
CR for 1-tail: \(V \geqslant 10\) oe; CR for 2-tail: \(V \geqslant 11\) oeA1 2nd A1: awrt 0.133 or \(V \geqslant 10\) oe (e.g. \(V > 9\)) or \(V \geqslant 11\) oe; allow any letter but CR must match part (c)
Accept \(H_0\) or not significant or 8 does not lie in the critical regiondM1 3rd dM1: dep on 2nd M1; ft their CR or probability. Correct statement comparing 8 with their CR or their prob with 0.05 or 0.025
Data consistent with Spany's claim or Insufficient evidence for Jem's belief or insufficient evidence that percentage of seeds not germinating is more than 5%A1cso All previous marks must be awarded. Correct statement in context. Need Bold words. NB award M1A1 for correct contextual statement on its own. (6 marks)
Total: 14 marks
Special Cases:
AnswerMarks
SCGuidance
SC1Sight of \(N(5\) or \(95,\ \sqrt{4.75}^2)\) M1A1; probability awrt 0.125/6 M1A1; Correct contextual conclusion dM1A1
SC2No approximation: Use of \(B(100, 0.05)\) M0A0; probability awrt 0.128 or \(CR \geqslant 10\) M1A1; then M0A0
# Question 1:

## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim B(20, 0.05)$ or $Y \sim B(20, 0.95)$ | B1 | Writing or using $B(20,0.05)$. Allow $Y \sim B(20, 0.95)$ if $Y$ is clearly defined. Implied by 1 correct probability. |

## Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X \leqslant 4) - P(X \leqslant 2)$ or $\binom{20}{3}0.05^3 \times 0.95^{17} + \binom{20}{4}0.05^4 \times 0.95^{16}$ | M1 | For $P(X \leqslant 4) - P(X \leqslant 2)$ and one correct prob, or $P(X=3) + P(X=4)$ and 1 correct prob |
| $= 0.072909...$ awrt $\mathbf{0.0729}$ | A1 | |

## Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X \leqslant 1)$ or $P(Y \geqslant 19) = 20 \times (0.95)^{19}(0.05) + (0.95)^{20}$ | M1 | For $P(X \leqslant 1)$ or $[20]\times(0.95)^{19}(0.05)+(0.95)^{20}$ – condone missing 20 |
| $= 0.7358$ awrt $\mathbf{0.736}$ | A1 | (5 marks total for (i) and (ii)) |

## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $W$ = no. of packets where $Y > 18$; $P(W=5) = (0.7358...)^5$ | M1 | For $(\text{their}(a)(ii))^5$ |
| $= 0.21573...$ awrt $\mathbf{0.216}$ | A1 | (2 marks) |

## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.05 \quad H_1: p > 0.05$ | B1 | Both hypotheses correct with $p$ or $\pi$ (1 mark) |

## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V \sim B(100, 0.05)$ approximated by $V \sim \text{Po}(5)$ | M1A1 | 1st M1: realising Poisson approximation appropriate. NB Po(95) is M0A0. 1st A1: correct mean for Poisson |
| $P(V \geqslant 8) = 1 - P(V \leqslant 7) = 1 - 0.8666 = 0.1334$ | M1 | 2nd M1: for $1 - P(V \leqslant 7)$ or $P(V \leqslant 7) = 0.8666$; or writing $P(V \geqslant 10) = 0.0318$ or $P(V \geqslant 9) = 0.0681$ or $P(V \geqslant 11) = 0.0137$ leading to a CR |
| CR for 1-tail: $V \geqslant 10$ oe; CR for 2-tail: $V \geqslant 11$ oe | A1 | 2nd A1: awrt 0.133 or $V \geqslant 10$ oe (e.g. $V > 9$) or $V \geqslant 11$ oe; allow any letter but CR must match part (c) |
| Accept $H_0$ or not significant or 8 does not lie in the critical region | dM1 | 3rd dM1: dep on 2nd M1; ft their CR or probability. Correct statement comparing 8 with their CR or their prob with 0.05 or 0.025 |
| Data consistent with **Spany's claim** or Insufficient evidence for **Jem's belief** or insufficient evidence that **percentage of seeds not germinating** is more than **5%** | A1cso | All previous marks must be awarded. **Correct** statement in context. Need **Bold words**. NB award M1A1 for correct contextual statement on its own. (6 marks) |

**Total: 14 marks**

### Special Cases:
| SC | Guidance |
|---|---|
| SC1 | Sight of $N(5$ or $95,\ \sqrt{4.75}^2)$ M1A1; probability awrt 0.125/6 M1A1; Correct contextual conclusion dM1A1 |
| SC2 | No approximation: Use of $B(100, 0.05)$ M0A0; probability awrt 0.128 or $CR \geqslant 10$ M1A1; then M0A0 |
\begin{enumerate}
  \item Spany sells seeds and claims that $5 \%$ of its pansy seeds do not germinate. A packet of pansy seeds contains 20 seeds. Each seed germinates independently of the other seeds.\\
(a) Find the probability that in a packet of Spany's pansy seeds\\
(i) more than 2 but fewer than 5 seeds do not germinate,\\
(ii) more than 18 seeds germinate.
\end{enumerate}

Jem buys 5 packets of Spany's pansy seeds.\\
(b) Calculate the probability that all of these packets contain more than 18 seeds that germinate.

Jem believes that Spany's claim is incorrect. She believes that the percentage of pansy seeds that do not germinate is greater than 5\%\\
(c) Write down the hypotheses for a suitable test to examine Jem's belief.

Jem planted all of the 100 seeds she bought from Spany and found that 8 did not germinate.\\
(d) Using a suitable approximation, carry out the test using a $5 \%$ level of significance.

\hfill \mbox{\textit{Edexcel S2 2021 Q1 [14]}}