Edexcel S2 2010 June — Question 6 15 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2010
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeTwo-tailed test critical region
DifficultyModerate -0.3 This is a standard S2 hypothesis testing question covering routine binomial test procedures. Parts (a)-(d) involve textbook applications of critical regions and two-tailed tests, while part (e) is a one-tailed test. All steps follow established algorithms with no novel problem-solving required, though the multi-part structure and cumulative probability calculations place it slightly below average difficulty for A-level.
Spec2.04b Binomial distribution: as model B(n,p)2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail

A company claims that a quarter of the bolts sent to them are faulty. To test this claim the number of faulty bolts in a random sample of 50 is recorded.
  1. Give two reasons why a binomial distribution may be a suitable model for the number of faulty bolts in the sample. [2]
  2. Using a 5\% significance level, find the critical region for a two-tailed test of the hypothesis that the probability of a bolt being faulty is \(\frac{1}{4}\). The probability of rejection in either tail should be as close as possible to 0.025 [3]
  3. Find the actual significance level of this test. [2]
In the sample of 50 the actual number of faulty bolts was 8.
  1. Comment on the company's claim in the light of this value. Justify your answer. [2]
The machine making the bolts was reset and another sample of 50 bolts was taken. Only 5 were found to be faulty.
  1. Test at the 1\% level of significance whether or not the probability of a faulty bolt has decreased. State your hypotheses clearly. [6]

AnswerMarks Guidance
(a) 2 outcomes/faulty or not faulty/success or failB1; B1 (2)
A constant probability
Independence
Fixed number of trials (fixed \(n\))
(b) \(X \sim B(50, 0.25)\)
\(P(X < 6) = 0.0194\)
\(P(X \leq 7) = 0.0453\)
\(P(X \geq 18) = 0.0551\)
AnswerMarks Guidance
\(P(X \geq 19) = 0.0287\)M1; A1 A1 (3)
CR \(X \leq 6\) and \(X \geq 19\)
(c) \(0.0194 + 0.0287 = 0.0481\)M1A1 (2)
(d) 8(It) is not in the Critical region or 8(It) is not significant or 0.0916 > 0.025; There is evidence that the probability of a faulty bolt is 0.25 or the company's claim is correct.M1; A1ft (2)
(e) \(H_0: p = 0.25\) \(H_1: p < 0.25\)B1B1; M1A1 (6)
Notes
(a) B1 B1 one mark for each of any of the four statements. Give first B1 if only one correct statement given. No context needed.
(b) M1 for writing or using \(B(50, 0.25)\) also may be implied by both CR being correct. Condone use of P in critical region for the method mark. A1 (\(X \leq 6\)) o.e. [0.6] DO NOT accept \(P(X < 6)\); A1 (\(X \geq 19\)) o.e. [19,50] DO NOT accept \(P(X > 19)\)
(c) M1 Adding two probabilities for two tails. Both probabilities must be less than 0.5; A1 awrt 0.0481
(d) M1 one of the given statements followed through from their CR. A1 contextual comment followed through from their CR. NB A correct contextual comment alone followed through from their CR, will get M1 A1
(e) B1 for \(H_0\) must use \(p\) or \(\pi\) (pi); B1 for \(H_1\) must use \(p\) or \(\pi\) (pi); M1 for finding or writing \(P(X \leq 5)\) or attempting to find a critical region or a correct critical region; A1 awrt 0.007/CR \(X \leq 5\); 0.007 < 0.01, 5 is in the critical region, reject \(H_0\), significant. There is evidence that the probability of faulty bolts has decreased. NB A correct contextual statement alone followed through from their prob and \(H_1\) get M1 A1
**(a)** 2 outcomes/faulty or not faulty/success or fail | B1; B1 | (2) | B1 B1 one mark for each of any of the four statements. Give first B1 if only one correct statement given. No context needed.

A constant probability | | |

Independence | | |

Fixed number of trials (fixed $n$) | | |

**(b)** $X \sim B(50, 0.25)$

$P(X < 6) = 0.0194$

$P(X \leq 7) = 0.0453$

$P(X \geq 18) = 0.0551$

$P(X \geq 19) = 0.0287$ | M1; A1 A1 | (3) | M1 for writing or using $B(50, 0.25)$ also may be implied by both CR being correct. Condone use of P in critical region for the method mark. **A1** ($X \leq 6$) o.e. [0.6] DO NOT accept $P(X < 6)$; **A1** ($X \geq 19$) o.e. [19,50] DO NOT accept $P(X > 19)$

CR $X \leq 6$ and $X \geq 19$ | | |

**(c)** $0.0194 + 0.0287 = 0.0481$ | M1A1 | (2) | M1 Adding two probabilities for two tails. Both probabilities must be less than 0.5; A1 awrt 0.0481

**(d)** 8(It) is not in the Critical region or 8(It) is not significant or 0.0916 > 0.025; There is evidence that the probability of a faulty bolt is 0.25 or the company's claim is correct. | M1; A1ft | (2) | M1 one of the given statements followed through from their CR. **A1** contextual comment followed through from their CR. **NB** A correct contextual comment alone followed through from their CR, will get M1 A1

**(e)** $H_0: p = 0.25$ $H_1: p < 0.25$ | B1B1; M1A1 | (6) | B1 for $H_0$ must use $p$ or $\pi$ (pi); B1 for $H_1$ must use $p$ or $\pi$ (pi); M1 for finding or writing $P(X \leq 5)$ or attempting to find a critical region or a correct critical region; **A1** awrt 0.007/CR $X \leq 5$; 0.007 < 0.01, 5 is in the critical region, reject $H_0$, significant. There is evidence that the probability of faulty bolts has decreased. | M1; A1ft | 6) |

**Notes**

**(a)** B1 B1 one mark for each of any of the four statements. Give first B1 if only one correct statement given. No context needed.

**(b)** M1 for writing or using $B(50, 0.25)$ also may be implied by both CR being correct. Condone use of P in critical region for the method mark. **A1** ($X \leq 6$) o.e. [0.6] DO NOT accept $P(X < 6)$; **A1** ($X \geq 19$) o.e. [19,50] DO NOT accept $P(X > 19)$

**(c)** M1 Adding two probabilities for two tails. Both probabilities must be less than 0.5; A1 awrt 0.0481

**(d)** M1 one of the given statements followed through from their CR. **A1** contextual comment followed through from their CR. **NB** A correct contextual comment alone followed through from their CR, will get M1 A1

**(e)** B1 for $H_0$ must use $p$ or $\pi$ (pi); B1 for $H_1$ must use $p$ or $\pi$ (pi); M1 for finding or writing $P(X \leq 5)$ or attempting to find a critical region or a correct critical region; A1 awrt 0.007/CR $X \leq 5$; 0.007 < 0.01, 5 is in the critical region, reject $H_0$, significant. There is evidence that the probability of faulty bolts has decreased. **NB** A correct contextual statement alone followed through from their prob and $H_1$ get M1 A1

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A company claims that a quarter of the bolts sent to them are faulty. To test this claim the number of faulty bolts in a random sample of 50 is recorded.

\begin{enumerate}[label=(\alph*)]
\item Give two reasons why a binomial distribution may be a suitable model for the number of faulty bolts in the sample. [2]
\item Using a 5\% significance level, find the critical region for a two-tailed test of the hypothesis that the probability of a bolt being faulty is $\frac{1}{4}$. The probability of rejection in either tail should be as close as possible to 0.025 [3]
\item Find the actual significance level of this test. [2]
\end{enumerate}

In the sample of 50 the actual number of faulty bolts was 8.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Comment on the company's claim in the light of this value. Justify your answer. [2]
\end{enumerate}

The machine making the bolts was reset and another sample of 50 bolts was taken. Only 5 were found to be faulty.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Test at the 1\% level of significance whether or not the probability of a faulty bolt has decreased. State your hypotheses clearly. [6]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2010 Q6 [15]}}