Challenging +1.2 This is a standard two-sample t-test with equal variances requiring calculation of sample statistics, pooled variance, and hypothesis testing. While it involves multiple computational steps (finding means, variances, pooled estimate, test statistic), it follows a completely routine procedure taught in Further Maths Statistics with no conceptual challenges or novel insights required. The 10-mark allocation reflects computational work rather than difficulty.
Two fish farmers \(X\) and \(Y\) produce a particular type of fish. Farmer \(X\) chooses a random sample of 8 of his fish and records the masses, \(x\) kg, as follows.
1.2 \quad 1.4 \quad 0.8 \quad 2.1 \quad 1.8 \quad 2.6 \quad 1.5 \quad 2.0
Farmer \(Y\) chooses a random sample of 10 of his fish and summarises the masses, \(y\) kg, as follows.
$$\Sigma y = 20.2 \qquad \Sigma y^2 = 44.6$$
You should assume that both distributions are normal with equal variances. Test at the 10% significance level whether the mean mass of fish produced by farmer \(X\) differs from the mean mass of fish produced by farmer \(Y\).
[10]
Find value of t (or can comparey –x = 0⋅345 with
0⋅509)
Answer
Marks
Guidance
t < 1⋅75 so mean masses are the same (AEF)
DB1 FT
Correct conclusion (FT on t, dep *B1)
SR: Ζ = (y –x) / √(s 2/8 + s 2/10)
X Y
Answer
Marks
Guidance
= 0⋅345 / √(0⋅078) = 1⋅20
(B1)
SR: Implicitly taking s 2, s 2 as unequal popln.
X Y
variances
(may also earn first B1 B1 M1)
Ζ < 1⋅645
Answer
Marks
Guidance
so mean masses are the same (AEF)
(B1FT)
Comparison with Ζ and conclusion (FT on Ζ)
0.95
(can earn at most 5/10)
Answer
Marks
Guidance
Total:
10
Question
Answer
Marks
Question 9:
9 | H : µ = µ , H : µ ≠ µ (AEF)
0 X Y 1 X Y | B1 | State hypotheses (B0 forx …)
x = 13⋅4/8 or 1⋅67[5], y = 2⋅02 (all to 3 s.f.) | B1 | Find sample means (values to 3 s.f. throughout)
s 2 = (24⋅7 – 13⋅42/8) / 7
X
= 451/1400 or 0⋅3221 or 0⋅56782 and
s 2 = (44⋅6 – 20⋅22/10) / 9
Y
= 949/2250 or 0⋅4218 or 0⋅64942 | M1 | Estimate or imply popln. variances
(allow biased here: 0⋅2819 or 0⋅53092)
(allow biased here: 0⋅3796 or 0⋅61612)
s2 = (7 s 2 + 9 s 2) / 16 (AEF)
X Y
or (24⋅7 – 13⋅42/8 + 44⋅6 – 20⋅22/10) / 16 | M1 A1 | Estimate (pooled) common variance
(note s 2 and s 2 not needed explicitly)
X Y
= 6051/16 000 or 0⋅3782 or 0⋅61502 | A1
t = 1⋅746
16, 0.95 | *B1 | State or use correct tabular t value
[–] t = (y –x) / s √(1/8 + 1/10) = 1⋅18 | M1 A1 | Find value of t (or can comparey –x = 0⋅345 with
0⋅509)
t < 1⋅75 so mean masses are the same (AEF) | DB1 FT | Correct conclusion (FT on t, dep *B1)
SR: Ζ = (y –x) / √(s 2/8 + s 2/10)
X Y
= 0⋅345 / √(0⋅078) = 1⋅20 | (B1) | SR: Implicitly taking s 2, s 2 as unequal popln.
X Y
variances
(may also earn first B1 B1 M1)
Ζ < 1⋅645
so mean masses are the same (AEF) | (B1FT) | Comparison with Ζ and conclusion (FT on Ζ)
0.95
(can earn at most 5/10)
Total: | 10
Question | Answer | Marks | Guidance
Two fish farmers $X$ and $Y$ produce a particular type of fish. Farmer $X$ chooses a random sample of 8 of his fish and records the masses, $x$ kg, as follows.
1.2 \quad 1.4 \quad 0.8 \quad 2.1 \quad 1.8 \quad 2.6 \quad 1.5 \quad 2.0
Farmer $Y$ chooses a random sample of 10 of his fish and summarises the masses, $y$ kg, as follows.
$$\Sigma y = 20.2 \qquad \Sigma y^2 = 44.6$$
You should assume that both distributions are normal with equal variances. Test at the 10% significance level whether the mean mass of fish produced by farmer $X$ differs from the mean mass of fish produced by farmer $Y$.
[10]
\hfill \mbox{\textit{CAIE FP2 2017 Q9 [10]}}