Challenging +1.2 This is a standard two-sample confidence interval problem requiring calculation of sample means, pooled variance, and application of the t-distribution formula. While it involves multiple computational steps (8 marks) and requires stating assumptions, it follows a well-established procedure taught in statistics courses with no novel insight needed. The interpretation question is straightforward. Slightly above average difficulty due to the computational burden and need to recall the pooled variance method, but remains a textbook exercise.
An examination involved writing an essay. In order to compare the time taken to write the essay by students from two large colleges, a sample of \(12\) students from college A and a sample of \(8\) students from college B were randomly selected. The times, \(t_A\) and \(t_B\), taken for these students to write the essay were measured, correct to the nearest minute, and are summarised by
\(n_A = 12\), \(\Sigma t_A = 257\), \(\Sigma t_A^2 = 5629\), \(n_B = 8\), \(\Sigma t_B = 206\), \(\Sigma t_B^2 = 5359\).
Stating any required assumptions, calculate a \(95\%\) confidence interval for the difference in the population means. [8]
State, giving a reason, whether your confidence interval supports the statement that the population means, for the two colleges, are equal. [1]
Find difference in sample means to 2 dp, e.g.: x –x = 21⋅417 – 25⋅75 = –4⋅33 B1
A B
2 2
Estimate common population variance: s = (5629 – 257 /12
2
+ 5359 – 206 /8) / 18
= (124⋅9 + 54⋅5)/18
= 9⋅968 or 3⋅157 2 (3 sf) M1 A1
Use of correct tabular t value: t18, 0.975 = 2⋅101 (to 3 sf) B1
Find confidence interval for e.g. µ –µ : x –x ± ts √(1/12 + 1/8) M1
A B A B
Evaluate: –4⋅33 ± 3⋅03 or [–7⋅36, –1⋅31] A1*
State reason and conclusion (A.E.F.): Interval does not include zero
Answer
Marks
Guidance
(dep *A1 apart from rounding) so statement not supported B1
8
1
[9]
Page 6
Mark Scheme: Teachers’ version
Syllabus
GCE A LEVEL – May/June 2010
9231
22
9 (i)
(ii)
Answer
Marks
(iii)
Valid comment on scatter diagram (A.E.F.): Approx str. line with neg. gradient B1
EITHER: Find gradient b directly using r: b = r √(S / S ) M2
yy xx
2 2
= r √{(8245 – 240 /20) / (2125 – 200 /20)} A1
= – 0⋅992 √ (5365 / 125)
OR: Find S (to 3 sf): S = r √(S S ) = – 812⋅37 (M1 A1)
xy xy xx yy
Find b: b = S / S [= – 812⋅37/125] (M1)
xy xx
Evaluate b: [= – 6⋅499] = – 6⋅50 A1
Find equation of regression line: y = b(x – 10) + 12 M1
= 77⋅0 – 6⋅50 x A1
Find b′ using r 2 = bb′: [or S / S ] b′ = –0⋅992 2 /6⋅499 = –0⋅151 M1 A1
Answer
Marks
xy yy
1
6
Answer
Marks
2
[9]
Question 8:
8 | State assumptions (A.E.F.): Equal variances B1
Normal populations B1
Find difference in sample means to 2 dp, e.g.: x –x = 21⋅417 – 25⋅75 = –4⋅33 B1
A B
2 2
Estimate common population variance: s = (5629 – 257 /12
2
+ 5359 – 206 /8) / 18
= (124⋅9 + 54⋅5)/18
= 9⋅968 or 3⋅157 2 (3 sf) M1 A1
Use of correct tabular t value: t18, 0.975 = 2⋅101 (to 3 sf) B1
Find confidence interval for e.g. µ –µ : x –x ± ts √(1/12 + 1/8) M1
A B A B
Evaluate: –4⋅33 ± 3⋅03 or [–7⋅36, –1⋅31] A1*
State reason and conclusion (A.E.F.): Interval does not include zero
(dep *A1 apart from rounding) so statement not supported B1 | 8
1 | [9]
Page 6 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – May/June 2010 | 9231 | 22
9 (i)
(ii)
(iii) | Valid comment on scatter diagram (A.E.F.): Approx str. line with neg. gradient B1
EITHER: Find gradient b directly using r: b = r √(S / S ) M2
yy xx
2 2
= r √{(8245 – 240 /20) / (2125 – 200 /20)} A1
= – 0⋅992 √ (5365 / 125)
OR: Find S (to 3 sf): S = r √(S S ) = – 812⋅37 (M1 A1)
xy xy xx yy
Find b: b = S / S [= – 812⋅37/125] (M1)
xy xx
Evaluate b: [= – 6⋅499] = – 6⋅50 A1
Find equation of regression line: y = b(x – 10) + 12 M1
= 77⋅0 – 6⋅50 x A1
Find b′ using r 2 = bb′: [or S / S ] b′ = –0⋅992 2 /6⋅499 = –0⋅151 M1 A1
xy yy | 1
6
2 | [9]
An examination involved writing an essay. In order to compare the time taken to write the essay by students from two large colleges, a sample of $12$ students from college A and a sample of $8$ students from college B were randomly selected. The times, $t_A$ and $t_B$, taken for these students to write the essay were measured, correct to the nearest minute, and are summarised by
$n_A = 12$, $\Sigma t_A = 257$, $\Sigma t_A^2 = 5629$, $n_B = 8$, $\Sigma t_B = 206$, $\Sigma t_B^2 = 5359$.
Stating any required assumptions, calculate a $95\%$ confidence interval for the difference in the population means. [8]
State, giving a reason, whether your confidence interval supports the statement that the population means, for the two colleges, are equal. [1]
\hfill \mbox{\textit{CAIE FP2 2010 Q8 [9]}}