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

a
The stopping distances follow \( N(6.76, 0.12^2) \).
(i) For \( X < 6.5 \):
\( Z = \frac{6.5 – 6.76}{0.12} = \frac{-0.26}{0.12} \approx -2.1667 \)
\( P(Z < -2.1667) = \Phi(-2.1667) \approx 0.01513 \approx 0.0151 \)
(ii) For \( X > 7 \):
\( Z = \frac{7 – 6.76}{0.12} = \frac{0.24}{0.12} = 2.0 \)
\( P(Z > 2.0) = 1 – \Phi(2.0) \approx 1 – 0.97725 = 0.02275 \approx 0.0228 \)
Explanation:
Standardize the distances to z-scores and use standard normal tables to find probabilities, rounded to four decimal places.
Result:
(i) \( 0.0151 \)
(ii) \( 0.0228 \)

b
Expected number = \( 1000 \times P(a < X < b) \).
(i) For \( 6.5 < X < 6.75 \):
\( Z_1 = \frac{6.5 – 6.76}{0.12} = -2.1667 \), \( Z_2 = \frac{6.75 – 6.76}{0.12} = -0.0833 \)
\( P(-2.1667 < Z < -0.0833) = \Phi(-0.0833) – \Phi(-2.1667) \approx 0.4668 – 0.01513 = 0.45167 \)
\( 1000 \times 0.45167 = 451.67 \approx 451.7 \).
(ii) For \( 6.75 < X < 7 \):
\( Z_1 = -0.0833 \), \( Z_2 = 2.0 \)
\( P(-0.0833 < Z < 2.0) = \Phi(2.0) – \Phi(-0.0833) \approx 0.97725 – 0.4668 = 0.51045 \)
\( 1000 \times 0.51045 = 510.45 \approx 510.5 \).
Explanation:
Calculate probabilities for the intervals using z-scores and multiply by \( 1000 \), rounding to four significant figures.
Result:
(i) \( 451.7 \)
(ii) \( 510.5 \)

c
Hypotheses:
\[ H_0: \text{Stopping distances follow } N(6.76, 0.12^2) \]
\[ H_1: \text{Stopping distances do not follow } N(6.76, 0.12^2) \]
d
Observed frequencies from the table:
\[ \begin{array}{|c|c|} \hline \textbf{Measured stopping distance} & \textbf{Number of bicycles} \\ \hline \text{Less than } 6.5 \, \text{m} & 12 \\ \hline \text{Between } 6.5 \, \text{m and } 6.75 \, \text{m} & 428 \\ \hline \text{Between } 6.75 \, \text{m and } 7 \, \text{m} & 527 \\ \hline \text{More than } 7 \, \text{m} & 33 \\ \hline \end{array} \]
Compute each component \( \frac{(O_i – E_i)^2}{E_i} \):
– \( \frac{(12 – 15.1)^2}{15.1} \approx 0.647 \)
– \( \frac{(428 – 451.7)^2}{451.7} \approx 1.240 \)
– \( \frac{(527 – 510.5)^2}{510.5} \approx 0.537 \)
– \( \frac{(33 – 22.8)^2}{22.8} \approx 4.619 \)
Sum: \( \chi^2 \approx 0.647 + 1.240 + 0.537 + 4.619 = 7.043 \).
We have 4 classes and no parameters estimated from the data, so \( df = 4 – 1 = 3 \).
The \( p \)-value is the right-tail probability of a \( \chi^2 \) distribution with 3 degrees of freedom at the observed statistic:
\( p = \Pr(\chi^2_{(3)} \ge 7.043) = 1 – F_{\chi^2_{(3)}}(7.043) \).
You can evaluate this using a statistics package or calculator:
– Excel / Google Sheets: =CHISQ.DIST.RT(7.043, 3) → 0.072654… ≈ 0.0727
– TI-84 (or similar): use the chi-square upper tail function, e.g. χ²cdf(7.043,1E99,3) → 0.0727
– Python (SciPy): from scipy.stats import chi2; chi2.sf(7.043, 3) → 0.072654…
Rounded value reported: \( p = 0.0727 \) (≈ 7.27%).
For \( df = 3 \):
– \( \chi^2_{0.10} \approx 6.251 \) (right-tail 0.10)
– \( \chi^2_{0.05} \approx 7.815 \) (right-tail 0.05)
Since 7.043 lies between 6.251 and 7.815, the \( p \)-value lies between 0.10 and 0.05 — consistent with 0.0727.
e
Since \( 0.05 < 0.0727 \), there is insufficient evidence to reject \( H_0 \) (or “accept \( H_0 \)”).
Reason: The observed departures are not large enough at the 5% level (all expected counts ≥ 5, so the chi-squared approximation is valid).
Explanation:
State the hypotheses, compute the chi-squared statistic with detailed steps, determine the degrees of freedom, calculate the p-value using statistical tools, verify with percentiles, and conclude by comparing with the 5% significance level, noting the validity of the approximation.
Result:
(c) \( H_0 \): Stopping distances follow \( N(6.76, 0.12^2) \); \( H_1 \): Stopping distances do not follow \( N(6.76, 0.12^2) \)
(d) \( p = 0.0727 \)
(e) Do not reject \( H_0 \); the stopping distances can be modelled by \( N(6.76, 0.12^2) \) as \( 0.05 < 0.0727 \), with all expected counts ≥ 5 validating the chi-squared approximation.