Date | November 2018 | Marks available | 2 | Reference code | 18N.2.SL.TZ0.T_1 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Estimate | Question number | T_1 | Adapted from | N/A |
Question
The marks obtained by nine Mathematical Studies SL students in their projects (x) and their final IB examination scores (y) were recorded. These data were used to determine whether the project mark is a good predictor of the examination score. The results are shown in the table.
The equation of the regression line y on x is y = mx + c.
A tenth student, Jerome, obtained a project mark of 17.
Use your graphic display calculator to write down , the mean project mark.
Use your graphic display calculator to write down , the mean examination score.
Use your graphic display calculator to write down r , Pearson’s product–moment correlation coefficient.
Find the exact value of m and of c for these data.
Show that the point M (, ) lies on the regression line y on x.
Use the regression line y on x to estimate Jerome’s examination score.
Justify whether it is valid to use the regression line y on x to estimate Jerome’s examination score.
In his final IB examination Jerome scored 65.
Calculate the percentage error in Jerome’s estimated examination score.
Markscheme
14 (G1)
[1 mark]
54 (G1)
[1 mark]
0.5 (G2)
[2 marks]
m = 0.875, c = 41.75 (A1)(A1)
Note: Award (A1) for 0.875 seen. Award (A1) for 41.75 seen. If 41.75 is rounded to 41.8 do not award (A1).
[2 marks]
y = 0.875(14) + 41.75 (M1)
Note: Award (M1) for their correct substitution into their regression line. Follow through from parts (a)(i) and (b)(i).
= 54
and so the mean point lies on the regression line (A1)
(accept 54 is , the mean value of the y data)
Note: Do not award (A1) unless the conclusion is explicitly stated and the 54 seen. The (A1) can be awarded only if their conclusion is consistent with their equation and it lies on the line.
The use of 41.8 as their c value precludes awarding (A1).
OR
54 = 0.875(14) + 41.75 (M1)
54 = 54
Note: Award (M1) for their correct substitution into their regression line. Follow through from parts (a)(i) and (b)(i).
and so the mean point lies on the regression line (A1)
Note: Do not award (A1) unless the conclusion is explicitly stated. Follow through from part (a).
The use of 41.8 as their c value precludes the awarding of (A1).
[2 marks]
y = 0.875(17) + 41.75 (M1)
Note: Award (M1) for correct substitution into their regression line.
= 56.6 (56.625) (A1)(ft)(G2)
Note: Follow through from part (b)(i).
[2 marks]
the estimate is valid (A1)
since this is interpolation and the correlation coefficient is large enough (R1)
OR
the estimate is not valid (A1)
since the correlation coefficient is not large enough (R1)
Note: Do not award (A1)(R0). The (R1) may be awarded for reasoning based on strength of correlation, but do not accept “correlation coefficient is not strong enough” or “correlation is not large enough”.
Award (A0)(R0) for this method if no numerical answer to part (a)(iii) is seen.
[2 marks]
(M1)
Note: Award (M1) for correct substitution into percentage error formula. Follow through from part (c)(i).
= 12.9 (%)(12.9230…) (A1)(ft)(G2)
Note: Follow through from part (c)(i). Condone use of percentage symbol.
Award (G0) for an answer of −12.9 with no working.
[2 marks]