Home / IB Mathematics SL 4.11 Formulation of null and alternative hypotheses AI SL Paper 2 – Exam Style Questions

IB Mathematics SL 4.11 Formulation of null and alternative hypotheses AI SL Paper 2 - Exam Style Questions - New Syllabus

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Question

The stopping distances for Bike travelling at \(20\text{ km h}^{-1}\) are assumed to follow a normal distribution with mean \(6.76\text{ m}\) and standard deviation \(0.12\text{ m}\).
(a) Under this assumption, find, correct to four decimal places, the probability that a randomly chosen bicycle travelling at \(20\text{ km h}^{-1}\) manages to stop
(i) in less than \(6.5\text{ m}\).
(ii) in more than \(7\text{ m}\). [3]
1000 randomly selected Bike are tested and their stopping distances (at \(20\text{ km h}^{-1}\)) are measured.
(b) Find, correct to four significant figures, the expected number of Bike tested that stop between
(i) \(6.5\text{ m}\) and \(6.75\text{ m}\).
(ii) \(6.75\text{ m}\) and \(7\text{ m}\). [3]
The measured stopping distances of the 1000 Bike are given in the table.
Measured stopping distanceNumber of Bike (observed)
Less than \(6.5\text{ m}\)12
Between \(6.5\text{ m}\) and \(6.75\text{ m}\)428
Between \(6.75\text{ m}\) and \(7\text{ m}\)527
More than \(7\text{ m}\)33
It is decided to perform a \(\chi^2\) goodness-of-fit test at the 5% level of significance to decide whether the stopping distances can be modelled by \(N(6.76,\,0.12^2)\).
(c) State the null and alternative hypotheses. [2]
(d) Find the p-value for the test. [3]
(e) State the conclusion of the test. Give a reason for your answer. [2]
▶️Answer/Explanation
Markscheme (with detailed working)

Let \(X \sim N(\mu,\sigma^2)=N(6.76,\,0.12^2)\). For probabilities we standardize: \(Z=\frac{X-\mu}{\sigma}\).

(a)

(i) \(P(X<6.5)=P\!\left(Z<\dfrac{6.5-6.76}{0.12}\right)=P(Z<-2.1667\ldots)=\boxed{0.0151}\) (4 d.p.).
(ii) \(P(X>7)=P\!\left(Z>\dfrac{7-6.76}{0.12}\right)=P(Z>2.0000)=1-\Phi(2.0000)=\boxed{0.0228}\) (4 d.p.).
[3 marks]

(b) Multiply model probabilities by 1000 (the sample size) and round to 4 s.f.

(i) \(P(6.5\le X<6.75)=\Phi\!\left(\dfrac{6.75-6.76}{0.12}\right)-\Phi\!\left(\dfrac{6.5-6.76}{0.12}\right)=\Phi(-0.0833\ldots)-\Phi(-2.1667\ldots)\) \(=0.4668-0.0151=0.4517\Rightarrow \boxed{451.7}\) Bike.
(ii) \(P(6.75\le X<7)=\Phi\!\left(\dfrac{7-6.76}{0.12}\right)-\Phi\!\left(\dfrac{6.75-6.76}{0.12}\right)=\Phi(2.0000)-\Phi(-0.0833\ldots)\) \(=0.9773-0.4668=0.5105\Rightarrow \boxed{510.5}\) Bike.
[3 marks]

(c)

\(H_0:\) The stopping distances follow \(N(6.76,\,0.12^2)\).
\(H_1:\) The stopping distances do not follow \(N(6.76,\,0.12^2)\).

[2 marks]

(d) \(\chi^2\) goodness-of-fit (4 categories ⇒ degrees of freedom \(=4-1=3\), parameters specified).

Model probabilities and expected counts (from (a) and (b)):

  • \(<6.5\text{ m}\): \(p=0.0151\Rightarrow E=15.1\)
  • \(6.5\text{–}6.75\text{ m}\): \(p=0.4517\Rightarrow E=451.7\)
  • \(6.75\text{–}7\text{ m}\): \(p=0.5105\Rightarrow E=510.5\)
  • \(>7\text{ m}\): \(p=0.0228\Rightarrow E=22.8\)
Observed counts: \(O=(12,\,428,\,527,\,33)\).
\[ \chi^2=\sum\frac{(O-E)^2}{E} =\frac{(12-15.1)^2}{15.1} +\frac{(428-451.7)^2}{451.7} +\frac{(527-510.5)^2}{510.5} +\frac{(33-22.8)^2}{22.8} \] \[ \approx 0.6364+1.2435+0.5333+4.5632 =6.9764\ (\text{to 4 d.p.}) \] Using \(\chi^2_3\), the p-value is \(\boxed{0.0727}\) (7.27%).
[3 marks]

(e)

Since \(p=0.0727>0.05\), we fail to reject \(H_0\). There is insufficient evidence to conclude the distances are not \(N(6.76,\,0.12^2)\); the model is consistent with the data at the 5% level.

[2 marks]
Total Marks: 13
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