Edexcel S2 2001 June — Question 3 7 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2001
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeTwo-tailed test critical region
DifficultyModerate -0.3 This is a straightforward one-tailed binomial hypothesis test with clear setup (n=20, x=2, p=1/4, 10% significance). It requires standard procedure: stating H₀ and H₁, calculating P(X≤2) under H₀, and comparing to 0.05 (one-tail). The only mild challenge is recognizing this is one-tailed (testing for decrease only, since x=2 is below expected 5), but the calculation itself is routine S2 material requiring no novel insight.
Spec2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail

3. In a sack containing a large number of beads \(\frac { 1 } { 4 }\) are coloured gold and the remainder are of different colours. A group of children use some of the beads in a craft lesson and do not replace them. Afterwards the teacher wishes to know whether or not the proportion of gold beads left in the sack has changed. He selects a random sample of 20 beads and finds that 2 of them are coloured gold. Stating your hypotheses clearly test, at the \(10 \%\) level of significance, whether or not there is evidence that the proportion of gold beads has changed.

Question 3:
Hypotheses:
AnswerMarks
\(H_0: p = \frac{1}{4}\), \(H_1: p \neq \frac{1}{4}\)B1; B1
\(X\) = no. of gold beads in sample of 20. Under \(H_0\), \(X \sim B(20, \frac{1}{4})\)
Critical Region:
\(P(X \leq 1) = 0.0243\), \(P(X \leq 8) = 0.9591\)
AnswerMarks
C.R.: \(X \leq 1\) or \(X \geq 9\)M1
Probability:
\(E(X) = 5\), OR \(2 \times P(X \leq 2)\)
AnswerMarks Guidance
\(P(X \leq 2) = 0.0913\), \(= 2 \times 0.0913 = 0.1826\)A1 each value
\(P(X \geq 8) = 1 - 0.8982 = 0.1018\)A1
Not significant (either \(x = 2\) not in C.R. or prob \(> 10\%\))M1
Insufficient evidence of a change in proportion of gold beadsA1✓ (7)
## Question 3:

**Hypotheses:**
$H_0: p = \frac{1}{4}$, $H_1: p \neq \frac{1}{4}$ | B1; B1 |

$X$ = no. of gold beads in sample of 20. Under $H_0$, $X \sim B(20, \frac{1}{4})$

**Critical Region:**
$P(X \leq 1) = 0.0243$, $P(X \leq 8) = 0.9591$

C.R.: $X \leq 1$ or $X \geq 9$ | M1 |

**Probability:**
$E(X) = 5$, OR $2 \times P(X \leq 2)$

$P(X \leq 2) = 0.0913$, $= 2 \times 0.0913 = 0.1826$ | A1 each value |

$P(X \geq 8) = 1 - 0.8982 = 0.1018$ | A1 |

Not significant (either $x = 2$ not in C.R. or prob $> 10\%$) | M1 |

Insufficient evidence of a change in proportion of gold beads | A1✓ | **(7)**

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3. In a sack containing a large number of beads $\frac { 1 } { 4 }$ are coloured gold and the remainder are of different colours. A group of children use some of the beads in a craft lesson and do not replace them. Afterwards the teacher wishes to know whether or not the proportion of gold beads left in the sack has changed. He selects a random sample of 20 beads and finds that 2 of them are coloured gold.

Stating your hypotheses clearly test, at the $10 \%$ level of significance, whether or not there is evidence that the proportion of gold beads has changed.\\

\hfill \mbox{\textit{Edexcel S2 2001 Q3 [7]}}