IB Mathematics AHL 4.14 Linear transformation AI HL Paper 3- Exam Style Questions- New Syllabus
Question
This question considers the analysis of several datasets of examination marks using a variety of standard procedures and also an unfamiliar statistical test.
A class of eight students sits two examinations, one in French and one in German.
The marks in these examinations are given in Table 1.
| Student | French mark | German mark |
|---|---|---|
| \(S_1\) | 42 | 39 |
| \(S_2\) | 65 | 66 |
| \(S_3\) | 82 | 71 |
| \(S_4\) | 50 | 53 |
| \(S_5\) | 48 | 32 |
| \(S_6\) | 73 | 59 |
| \(S_7\) | 34 | 40 |
| \(S_8\) | 59 | 56 |
The maximum mark in both examinations is the same.
You may assume that these data are a random sample from a bivariate normal distribution with mean \(\mu_F\) for the French examination, mean \(\mu_G\) for the German examination and Pearson’s product-moment correlation coefficient \(\rho\).
Before the examinations were sat, the Head of Languages, Liam, decided to investigate whether there would be significant evidence of a difference between \(\mu_F\) and \(\mu_G\). Liam plans to analyze the data using a two-tailed paired \(t\)-test with significance level 5%.
(a)
(i) Explain briefly why he chose to use a \(t\)-test and not a \(z\)-test. [1]
(ii) Explain briefly why he chose to use a two-tailed test and not a one-tailed test. [1]
(i) Explain briefly why he chose to use a \(t\)-test and not a \(z\)-test. [1]
(ii) Explain briefly why he chose to use a two-tailed test and not a one-tailed test. [1]
(b)
(i) State suitable hypotheses for the \(t\)-test. [1]
(ii) Find the p-value for this test. [2]
(iii) The p-value is a probability. State the event for which it gives the probability. [1]
(iv) State, giving a reason, what conclusion Liam should reach. [2]
(i) State suitable hypotheses for the \(t\)-test. [1]
(ii) Find the p-value for this test. [2]
(iii) The p-value is a probability. State the event for which it gives the probability. [1]
(iv) State, giving a reason, what conclusion Liam should reach. [2]
Liam believes that students who score well in one language examination tend to score well in the other language examination. He therefore decides to carry out a test at the 5% significance level to investigate whether there is a positive correlation between the French examination marks and the German examination marks.
(c)
(i) State appropriate hypotheses in terms of \(\rho\). [1]
(ii) Perform a suitable test and state the p-value. State, in context, the conclusion that Liam should reach, giving a reason. [4]
(i) State appropriate hypotheses in terms of \(\rho\). [1]
(ii) Perform a suitable test and state the p-value. State, in context, the conclusion that Liam should reach, giving a reason. [4]
There are actually two more students in this particular class, Paul and Sue. Paul sat the French examination, but he was unable to sit the German examination. Sue sat the German examination, but she was unable to sit the French examination.
(d)
(i) Paul’s mark in the French examination was 58. Use the data in Table 1 to predict the mark that Paul would have obtained in his German examination. [3]
(ii) Based on her mark in the German examination, Sue’s mark in the French examination was predicted to be 71. Find the mark she obtained in the German examination. [2]
(i) Paul’s mark in the French examination was 58. Use the data in Table 1 to predict the mark that Paul would have obtained in his German examination. [3]
(ii) Based on her mark in the German examination, Sue’s mark in the French examination was predicted to be 71. Find the mark she obtained in the German examination. [2]
Six students sit examinations in mathematics and history and their marks are shown in Table 2.
| Student | Mathematics mark (\(x\)) | History mark (\(y\)) |
|---|---|---|
| \(P_1\) | 53 | 41 |
| \(P_2\) | 76 | 70 |
| \(P_3\) | 50 | 67 |
| \(P_4\) | 65 | 47 |
| \(P_5\) | 61 | 66 |
| \(P_6\) | 84 | 50 |
Sophie is informed that the maximum mark in each subject is 100.
Sophie believes that the data might not be normally distributed, so she investigates what suitable tests are available which do not assume the data is normally distributed. She decides to use an unfamiliar test based on a statistic called Kendall’s \(\tau\).
Consider \(n\) bivariate observations \((x_i, y_i), i=1,2,\dots,n\), such that there are no equal \(x\)-values and no equal \(y\)-values. Any pair of distinct bivariate observations \((x_i, y_i)\) and \((x_j, y_j)\) is said to be concordant if \((x_i-x_j) \times (y_i-y_j) > 0\) and discordant if \((x_i-x_j) \times (y_i-y_j) < 0\). For \(n\) bivariate observations, there are \(\frac{n \times (n-1)}{2}\) distinct pairs.
Kendall’s \(\tau\) is defined as \[ \tau = \frac{2 \times (C-D)}{n \times (n-1)} \] where \(C\) and \(D\) denote respectively the number of concordant pairs and discordant pairs.
(e)
(i) Show that the value of Kendall’s \(\tau\) always lies in the interval \([-1, +1]\). [1]
(ii) For students \(P_1\) and \(P_2\), show that their pair is concordant. [1]
(iii) Show that the value of Kendall’s \(\tau\) for the mathematics and history data is 0.2. [4]
(i) Show that the value of Kendall’s \(\tau\) always lies in the interval \([-1, +1]\). [1]
(ii) For students \(P_1\) and \(P_2\), show that their pair is concordant. [1]
(iii) Show that the value of Kendall’s \(\tau\) for the mathematics and history data is 0.2. [4]
Sophie decided to use this statistic in a two-tailed test at the 10% significance level. The critical region for her test is \(|\tau| \geq 0.733\).
(f)
(i) State, in words, her null and alternative hypotheses. [1]
(ii) State the conclusion that Sophie should reach. Give a reason for your answer. [2]
(i) State, in words, her null and alternative hypotheses. [1]
(ii) State the conclusion that Sophie should reach. Give a reason for your answer. [2]
Sophie now finds that the history marks are actually out of 120. The history teacher advises Sophie to scale the history marks so that they are out of 100 and then redo the calculations for the value of \(\tau\).
(g) State, with a reason, whether you agree with the advice given by the history teacher. [2]
▶️ Answer/Explanation
Markscheme
(a)(i)
The population standard deviations are unknown, so a t-test is appropriate.
Use t-test due to unknown standard deviations A1
[1 mark]
The population standard deviations are unknown, so a t-test is appropriate.
Use t-test due to unknown standard deviations A1
[1 mark]
(a)(ii)
Liam is interested in any difference between means, not a specific direction, requiring a two-tailed test.
Two-tailed test for detecting any difference A1
[1 mark]
Liam is interested in any difference between means, not a specific direction, requiring a two-tailed test.
Two-tailed test for detecting any difference A1
[1 mark]
(b)(i)
\( H_0: \mu_D = 0 \), \( H_1: \mu_D \neq 0 \), where \(\mu_D\) is the mean difference (French − German).
\( H_0: \mu_D = 0 \), \( H_1: \mu_D \neq 0 \) A1
[1 mark]
\( H_0: \mu_D = 0 \), \( H_1: \mu_D \neq 0 \), where \(\mu_D\) is the mean difference (French − German).
\( H_0: \mu_D = 0 \), \( H_1: \mu_D \neq 0 \) A1
[1 mark]
(b)(ii)
Differences (French − German):
Mean difference: \( \frac{3 + (-1) + 11 + (-3) + 16 + 14 + (-6) + 3}{8} = 4.625 \).
Standard deviation: \( \sqrt{\frac{(3-4.625)^2 + (-1-4.625)^2 + \ldots + (3-4.625)^2}{7}} \approx 7.203 \).
t-statistic: \( \frac{4.625}{\frac{7.203}{\sqrt{8}}} \approx 1.817 \), df = 7.
p-value (two-tailed): \( \approx 0.153 \).
p-value \(\approx 0.153\) M1A1
[2 marks]
Differences (French − German):
| French mark | German mark | Difference |
|---|---|---|
| 42 | 39 | 3 |
| 65 | 66 | -1 |
| 82 | 71 | 11 |
| 50 | 53 | -3 |
| 48 | 32 | 16 |
| 73 | 59 | 14 |
| 34 | 40 | -6 |
| 59 | 56 | 3 |
Mean difference: \( \frac{3 + (-1) + 11 + (-3) + 16 + 14 + (-6) + 3}{8} = 4.625 \).
Standard deviation: \( \sqrt{\frac{(3-4.625)^2 + (-1-4.625)^2 + \ldots + (3-4.625)^2}{7}} \approx 7.203 \).
t-statistic: \( \frac{4.625}{\frac{7.203}{\sqrt{8}}} \approx 1.817 \), df = 7.
p-value (two-tailed): \( \approx 0.153 \).
p-value \(\approx 0.153\) M1A1
[2 marks]
(b)(iii)
The p-value is the probability of observing a mean difference at least as extreme as 4.625, assuming the null hypothesis is true.
Probability of observing the mean difference or larger A1
[1 mark]
The p-value is the probability of observing a mean difference at least as extreme as 4.625, assuming the null hypothesis is true.
Probability of observing the mean difference or larger A1
[1 mark]
(b)(iv)
Since \( 0.153 > 0.05 \), fail to reject \( H_0 \). There is insufficient evidence of a difference between population means.
No significant difference, as p-value > 0.05 R1A1
[2 marks]
Since \( 0.153 > 0.05 \), fail to reject \( H_0 \). There is insufficient evidence of a difference between population means.
No significant difference, as p-value > 0.05 R1A1
[2 marks]
(c)(i)
\( H_0: \rho = 0 \), \( H_1: \rho > 0 \).
\( H_0: \rho = 0 \), \( H_1: \rho > 0 \) A1
[1 mark]
\( H_0: \rho = 0 \), \( H_1: \rho > 0 \).
\( H_0: \rho = 0 \), \( H_1: \rho > 0 \) A1
[1 mark]
(c)(ii)
Pearson’s correlation: \( r \approx 0.849 \).
Test statistic: \( t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}} = \frac{0.849 \sqrt{6}}{\sqrt{1-0.849^2}} \approx 3.693 \), df = 6.
p-value (one-tailed): \( \approx 0.00286 \).
Since \( 0.00286 < 0.05 \), reject \( H_0 \). The French and German marks are positively correlated.
p-value \(\approx 0.00286\), positive correlation A2R1A1
[4 marks]
Pearson’s correlation: \( r \approx 0.849 \).
Test statistic: \( t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}} = \frac{0.849 \sqrt{6}}{\sqrt{1-0.849^2}} \approx 3.693 \), df = 6.
p-value (one-tailed): \( \approx 0.00286 \).
Since \( 0.00286 < 0.05 \), reject \( H_0 \). The French and German marks are positively correlated.
p-value \(\approx 0.00286\), positive correlation A2R1A1
[4 marks]
(d)(i)
Regression line (German on French): \( \text{German} = 10.2393 + 0.737495 \times \text{French} \).
For French = 58: \( \text{German} = 10.2393 + 0.737495 \times 58 \approx 10.2393 + 42.7747 \approx 53.014 \approx 53 \).
Paul’s German mark \(\approx 53\) A1M1A1
[3 marks]
Regression line (German on French): \( \text{German} = 10.2393 + 0.737495 \times \text{French} \).
For French = 58: \( \text{German} = 10.2393 + 0.737495 \times 58 \approx 10.2393 + 42.7747 \approx 53.014 \approx 53 \).
Paul’s German mark \(\approx 53\) A1M1A1
[3 marks]
(d)(ii)
Regression line (French on German): \( \text{French} = 4.04116 + 1.01122 \times \text{German} \).
For French = 71: \( 71 = 4.04116 + 1.01122 \times \text{German} \).
\( \text{German} = \frac{71 – 4.04116}{1.01122} \approx \frac{66.95884}{1.01122} \approx 66.216 \approx 66 \).
Sue’s German mark \(\approx 66\) A1A1
[2 marks]
Regression line (French on German): \( \text{French} = 4.04116 + 1.01122 \times \text{German} \).
For French = 71: \( 71 = 4.04116 + 1.01122 \times \text{German} \).
\( \text{German} = \frac{71 – 4.04116}{1.01122} \approx \frac{66.95884}{1.01122} \approx 66.216 \approx 66 \).
Sue’s German mark \(\approx 66\) A1A1
[2 marks]
(e)(i)
\( \tau = \frac{2 \times (C – D)}{n \times (n – 1)} \). If all pairs concordant, \( C = \frac{n \times (n – 1)}{2} \), \( D = 0 \), so \( \tau = 1 \). If all discordant, \( C = 0 \), \( D = \frac{n \times (n – 1)}{2} \), so \( \tau = -1 \). Thus, \( \tau \in [-1, +1] \).
\(\tau \in [-1, +1]\) A1
[1 mark]
\( \tau = \frac{2 \times (C – D)}{n \times (n – 1)} \). If all pairs concordant, \( C = \frac{n \times (n – 1)}{2} \), \( D = 0 \), so \( \tau = 1 \). If all discordant, \( C = 0 \), \( D = \frac{n \times (n – 1)}{2} \), so \( \tau = -1 \). Thus, \( \tau \in [-1, +1] \).
\(\tau \in [-1, +1]\) A1
[1 mark]
(e)(ii)
For \( P_1 (53, 41) \), \( P_2 (76, 70) \): \( (53 – 76) \times (41 – 70) = (-23) \times (-29) > 0 \), so concordant.
Pair \( P_1 \), \( P_2 \) is concordant A1
[1 mark]
For \( P_1 (53, 41) \), \( P_2 (76, 70) \): \( (53 – 76) \times (41 – 70) = (-23) \times (-29) > 0 \), so concordant.
Pair \( P_1 \), \( P_2 \) is concordant A1
[1 mark]
(e)(iii)
For \( n = 6 \), total pairs = \( \frac{6 \times 5}{2} = 15 \).
Concordant pairs (e.g., \( (x_i – x_j) \times (y_i – y_j) > 0 \)): \( (P_1, P_2), (P_1, P_4), (P_1, P_5), (P_2, P_4), (P_2, P_5), (P_2, P_6), (P_4, P_6), (P_5, P_6), (P_3, P_5) \), so \( C = 9 \).
Discordant pairs: \( (P_1, P_3), (P_1, P_6), (P_2, P_3), (P_3, P_4), (P_3, P_6), (P_4, P_5) \), so \( D = 6 \).
\( \tau = \frac{2 \times (9 – 6)}{6 \times 5} = \frac{2 \times 3}{30} = 0.2 \).
\(\tau = 0.2\) M1A1A1M1
[4 marks]
For \( n = 6 \), total pairs = \( \frac{6 \times 5}{2} = 15 \).
Concordant pairs (e.g., \( (x_i – x_j) \times (y_i – y_j) > 0 \)): \( (P_1, P_2), (P_1, P_4), (P_1, P_5), (P_2, P_4), (P_2, P_5), (P_2, P_6), (P_4, P_6), (P_5, P_6), (P_3, P_5) \), so \( C = 9 \).
Discordant pairs: \( (P_1, P_3), (P_1, P_6), (P_2, P_3), (P_3, P_4), (P_3, P_6), (P_4, P_5) \), so \( D = 6 \).
\( \tau = \frac{2 \times (9 – 6)}{6 \times 5} = \frac{2 \times 3}{30} = 0.2 \).
\(\tau = 0.2\) M1A1A1M1
[4 marks]
(f)(i)
Null hypothesis: There is no association between mathematics and history marks.
Alternative hypothesis: There is an association between mathematics and history marks.
No association vs. association A1
[1 mark]
Null hypothesis: There is no association between mathematics and history marks.
Alternative hypothesis: There is an association between mathematics and history marks.
No association vs. association A1
[1 mark]
(f)(ii)
Since \( |\tau| = 0.2 < 0.733 \), it is not in the critical region. Insufficient evidence to reject \( H_0 \).
No association, as \( |\tau| < 0.733 \) R1A1
[2 marks]
Since \( |\tau| = 0.2 < 0.733 \), it is not in the critical region. Insufficient evidence to reject \( H_0 \).
No association, as \( |\tau| < 0.733 \) R1A1
[2 marks]
(g)
Disagree. Rescaling history marks does not affect concordant or discordant pair counts, so \(\tau\) remains unchanged.
Disagree, scaling does not affect \(\tau\) A1R1
[2 marks]
Disagree. Rescaling history marks does not affect concordant or discordant pair counts, so \(\tau\) remains unchanged.
Disagree, scaling does not affect \(\tau\) A1R1
[2 marks]
Total Marks: 29