IB Mathematics SL 4.11 Formulation of null and alternative hypotheses AI SL Paper 1- Exam Style Questions- New Syllabus
Question
| Age Group | Preferred Car Colour | Total | |||
|---|---|---|---|---|---|
| White | Black | Silver | Red | ||
| $18 \leq \text{Age} < 25$ | 12 | 7 | 4 | 17 | 40 |
| $25 \leq \text{Age} < 45$ | 15 | $b$ | 10 | 12 | 58 |
| $\text{Age} \geq 45$ | 12 | 18 | 16 | 6 | 52 |
| Total | 39 | 46 | 30 | 35 | 150 |
The hypotheses are:
$H_0$: Preferred car colour is independent of the driver’s age group.
$H_1$: Preferred car colour is not independent of the driver’s age group.
The critical value for this test is given as $16.81$.
Most-appropriate topic codes (IB Mathematics: Applications and Interpretation HL):
• SL 4.11: The $\chi^2$ test for independence: contingency tables, degrees of freedom, critical value — parts (b), (c)
▶️ Answer/Explanation
(a)
The total for the second row is 58. Therefore:
\(15 + b + 10 + 12 = 58\), solving gives \(b = 21\).
\( \boxed{21} \)
(b)
Use the \(\chi^2\) test function on a graphic display calculator. 1. Enter the observed frequencies (the data from the table) into a matrix. 2. Perform a \(\chi^2\) test for independence. The calculated test statistic \(\chi^2_{\text{calc}}\) is approximately 18.3.
\(\chi^2_{\text{calc}} = 18.3\) (to three significant figures) or 18.3 (e.g., from calculator output 18.3313…).
\( \boxed{\chi^2_{\text{calc}} = 18.3} \)
(c)
Reasoning for decision: Since the calculated test statistic \(\chi^2_{\text{calc}} = 18.3\) is greater than the given critical value of 16.81 (equivalently, the p-value is approximately 0.00546, which is less than the significance level of 0.01), we reject the null hypothesis \(H_0\) at the 1% significance level.
Conclusion: There is sufficient evidence to suggest that age and car colour preference are not independent (i.e., there is a significant association between age group and preferred car colour).
\( \boxed{\text{Reject } H_0 \text{; sufficient evidence that age and colour preference are not independent.}} \)