Standard +0.3 This is a standard chi-squared test of independence question requiring students to identify that an expected frequency is below 5 and conclude cells must be combined. It's a routine application of a well-known rule from the specification with straightforward calculation, making it slightly easier than average for A-level Further Statistics.
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]
Question
Solution
Marks
AOs
Guidance
3
(c)
\(\mathrm { H } _ { 0 }\) : no association between session and age group. \(\mathrm { H } _ { 1 }\) : some association
Significant evidence of association between session attended and age group.
B1
B1
B1
M1ft
A1ft [5]
1.1
1.1
1.1
1.1
2.2b
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
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.
M1ft
A1ft
[2]
1.1
2.4
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)
\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
M1
M1
A1
M1
A1
M1
A1
[7]
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 \(m = \frac { 7 } { 16 }\)
\(\frac { 2 m ( 2 m - 1 ) \times m \times 3 ! } { 3 m ( 3 m - 1 ) ( 3 m - 2 ) \times 2 }\) then as above
Multiplication method can get full marks, but if no 3 or 3 !, max
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}
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
\multicolumn{3}{|c|}{Question} & Solution & Marks & AOs & Guidance \\
\hline
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 \\
\hline
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" \\
\hline
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 \\
\hline
2 & (a) & & Neither & B1 [1] & 1.2 & \\
\hline
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 \\
\hline
2 & (c) & (i) & 2.68 & B1ft [1] & 1.1 & awrt 2.68, ft on their (b) if letters correct \\
\hline
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 \\
\hline
2 & (d) & & \begin{tabular}{l}
Not much data here/points scattered/ possible outliers \\
So not very reliable \\
\end{tabular} & M1 A1 [2] & 2.3 1.1 & Reason for not very reliable (not "extrapolation") Full argument and conclusion, not too assertive (not wholly unreliable!) \\
\hline
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 \\
\hline
3 & (b) & & \begin{tabular}{ | c | c | c | }
\hline
Early & Middle & Late \\
\hline
29.4 & 23.1 & 31.5 \\
\hline
26.6 & 20.9 & 28.5 \\
\hline
\end{tabular}\begin{tabular}{ | c | c | c | }
\hline
Early & Middle & Late \\
\hline
0.9918 & 0.4160 & 2.2937 \\
\hline
1.0962 & 0.4598 & 2.5351 \\
\hline
\end{tabular} & B1 & 1.1 & \begin{tabular}{l}
Both, allow 28.4 for 28.5 \\
awrt 2.29, but allow 2.3 In range [2.53, 2.54] \\
\end{tabular} \\
\hline
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
\multicolumn{3}{|c|}{Question} & Solution & Marks & AOs & Guidance \\
\hline
3 & (c) & & \begin{tabular}{l}
$\mathrm { H } _ { 0 }$ : 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. \\
\end{tabular} & \begin{tabular}{l}
B1 \\
B1 \\
B1 \\
M1ft \\
A1ft [5] \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
1.1 \\
1.1 \\
1.1 \\
2.2b \\
\end{tabular} & \begin{tabular}{l}
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 \\
\end{tabular} \\
\hline
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. & \begin{tabular}{l}
M1ft \\
A1ft \\[0pt]
[2] \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
2.4 \\
\end{tabular} & \begin{tabular}{l}
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) \\
\end{tabular} \\
\hline
\multirow[t]{2}{*}{4} & \multirow{2}{*}{} & \multirow{2}{*}{OR:} & \begin{tabular}{l}
$\frac { { } ^ { 2 m } C _ { 2 } \times m } { { } ^ { 3 m } C _ { 3 } }$ \\
$= \frac { 2 m ( 2 m - 1 ) } { 2 } \times m \div \frac { 3 m ( 3 m - 1 ) ( 3 m - 2 ) } { 6 }$ \\
$= \frac { 2 m ( 2 m - 1 ) } { ( 3 m - 1 ) ( 3 m - 2 ) }$ \(\frac { 2 m ( 2 m - 1 ) } { ( 3 m - 1 ) ( 3 m - 2 ) } = \frac { 28 } { 55 }\) \\
$\Rightarrow 16 m ^ { 2 } - 71 m + 28 = 0$ \\
$m = 4$ BC \\
Reject $m = \frac { 7 } { 16 }$ as $m$ is an integer \\
\end{tabular} & \begin{tabular}{l}
M1 \\
M1 \\
A1 \\
M1 \\
A1 \\
M1 \\
A1 \\[0pt]
[7] \\
\end{tabular} & \begin{tabular}{l}
3.1b \\
3.1b \\
2.1 \\
3.1a \\
2.1 \\
1.1 \\
3.2a \\
\end{tabular} & \begin{tabular}{l}
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 $m = \frac { 7 } { 16 }$ \\
\end{tabular} \\
\hline
& & & $\frac { 2 m ( 2 m - 1 ) \times m \times 3 ! } { 3 m ( 3 m - 1 ) ( 3 m - 2 ) \times 2 }$ then as above & & & \begin{tabular}{l}
Multiplication method can get full marks, but if no 3 or 3 !, max \\
M1M0A0 M1A0M0A0 \\
\end{tabular} \\
\hline
\end{tabular}
\end{center}
\hfill \mbox{\textit{OCR FS1 AS 2021 Q3 [12]}}