3 At a cinema there are three film sessions each Saturday, "early", "middle" and "late". The numbers of the audience, in different age groups, at the three showings on a randomly chosen Saturday are given in Table 1.
\begin{table}[h]
\end{table}
| Question | Solution | Marks | AOs | Guidance |
| 1 | (a) | | -0.954 BC | B2 [2] | 1.1 1.1 | SC: If B0, give B1 if two of 7.04, 29.0[4], -13.6[4] (or 35.2, 145[.2], -68.2) seen |
| 1 | (b) | | Points lie close to a straight line Line has negative gradient | B1 B1 [2] | 2.2b 1.1 | Must refer to line, not just "negative correlation" |
| 1 | (c) | | No, it will be the same as \(x \rightarrow a\) is a linear transformation | B1 [1] | 2.2a | OE. Either "same" with correct reason, or "disagree" with correct reason. Allow any clear valid technical term |
| 2 | (a) | | Neither | B1 [1] | 1.2 | |
| 2 | (b) | | \(q = 1.13 + 0.620 p\) | B1B1 B1 [3] | 1.1,1.1 1.1 | 0.62(0) correct; both numbers correct Fully correct answer including letters |
| 2 | (c) | (i) | 2.68 | B1ft [1] | 1.1 | awrt 2.68, ft on their (b) if letters correct |
| 2 | (c) | (ii) | 2.5 is within data range, and points (here) are close to line/well correlated | B1 B1 [2] | 2.2b 2.2b | At least one reason, allow "no because points not close to line" Full argument, two reasons needed |
| 2 | (d) | | | Not much data here/points scattered/ possible outliers | | So not very reliable |
| M1 A1 [2] | 2.3 1.1 | Reason for not very reliable (not "extrapolation") Full argument and conclusion, not too assertive (not wholly unreliable!) |
| 3 | (a) | | Expected frequency for Middle/25 to 60 is 4.4 which is < 5 so must combine cells | B1*ft depB1 [2] | 2.4 3.5b | Correctly obtain this \(F _ { E }\), ft on addition errors " < 5" explicit and correct deduction |
| 3 | (b) | | | Early | Middle | Late | | 29.4 | 23.1 | 31.5 | | 26.6 | 20.9 | 28.5 |
| Early | Middle | Late | | 0.9918 | 0.4160 | 2.2937 | | 1.0962 | 0.4598 | 2.5351 |
| B1 | 1.1 | | Both, allow 28.4 for 28.5 | | awrt 2.29, but allow 2.3 In range [2.53, 2.54] |
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| Question | Solution | Marks | AOs | Guidance |
| 3 | (c) | | | \(\mathrm { H } _ { 0 }\) : no association between session and age group. \(\mathrm { H } _ { 1 }\) : some association | | \(\Sigma X ^ { 2 } = 7.793\) | | \(v = 2 , \chi ^ { 2 } ( 2 ) _ { \text {crit } } = 5.991\) | | Reject \(\mathrm { H } _ { 0 }\). | | Significant evidence of association between session attended and age group. |
| | | | Both. Allow "independent" etc | | Correct value of \(X ^ { 2 }\), awrt 7.79 (allow even if wrong in (b)) | | Correct CV and comparison | | Correct first conclusion, FT on their TS only | | Contextualised, not too assertive |
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| 3 | (d) | | The two biggest contributions to \(\chi ^ { 2 }\) are both for the late session ... ... when the proportion of younger people is higher, and of older people is lower, than the null hypothesis would suggest. | | | | Refer to biggest contribution(s), FT on their answers to (b), needs "reject \(\mathrm { H } _ { 0 }\) " | | Full answer, referring to at least one cell (ignore comments on next highest cells) |
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| \multirow[t]{2}{*}{4} | \multirow{2}{*}{} | \multirow{2}{*}{OR:} | | \(\frac { { } ^ { 2 m } C _ { 2 } \times m } { { } ^ { 3 m } C _ { 3 } }\) | | \(= \frac { 2 m ( 2 m - 1 ) } { 2 } \times m \div \frac { 3 m ( 3 m - 1 ) ( 3 m - 2 ) } { 6 }\) | | \(= \frac { 2 m ( 2 m - 1 ) } { ( 3 m - 1 ) ( 3 m - 2 ) }\) \(\frac { 2 m ( 2 m - 1 ) } { ( 3 m - 1 ) ( 3 m - 2 ) } = \frac { 28 } { 55 }\) | | \(\Rightarrow 16 m ^ { 2 } - 71 m + 28 = 0\) | | \(m = 4\) BC | | Reject \(m = \frac { 7 } { 16 }\) as \(m\) is an integer |
| | | 3.1b | | 3.1b | | 2.1 | | 3.1a | | 2.1 | | 1.1 | | 3.2a |
| | Use \({ } ^ { 2 m } C _ { 2 }\) and \(m\) | | Divide by \({ } ^ { 3 m } C _ { 3 }\) | | Correct expression in terms of \(m\) (allow with \(m\) not cancelled yet) | | Equate to \(\frac { 28 } { 55 }\) \ | simplify to three-term quadratic | | Correct simplified quadratic, or (quadratic) \(\times m , = 0\), aef Solve to get both 4 and \(\frac { 7 } { 16 }\) | | Explicitly reject \(m = \frac { 7 } { 16 }\) |
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| | | \(\frac { 2 m ( 2 m - 1 ) \times m \times 3 ! } { 3 m ( 3 m - 1 ) ( 3 m - 2 ) \times 2 }\) then as above | | | | Multiplication method can get full marks, but if no 3 or 3 !, max | | M1M0A0 M1A0M0A0 |
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