| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample z-test large samples |
| Difficulty | Challenging +1.2 This is a two-sample t-test with unequal variances requiring hypothesis setup (H₀: μ₁ - μ₂ = 5), calculation of the test statistic using Welch's approximation, and interpretation. While it involves multiple steps and careful reading of 'exceeds...by less than 5kg', the mechanics are standard Further Maths Statistics procedures with large sample sizes (n=75 each) making the test robust. The conceptual demand is moderate—recognizing when to use pooled vs unpooled variance and setting up the correct one-tailed test—but this is routine material for FM students. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Sample mean | Unbiased estimate of population variance | |
| Controlled grazing | 19.14 | 20.54 |
| Free grazing | 15.36 | 9.84 |
| Answer | Marks |
|---|---|
| 8 | State if assumption necessary with valid reason: No, since large samples (A.E.F.) B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Consistent conclusion (A.E.F.; A1 dep *A1, *B1): Does exceed by less than 5 kg. M1√ A1 | (1) | |
| (8) | [9] | |
| Page 7 | Mark Scheme: Teachers’ version | Syllabus |
| GCE A LEVEL – October/November 2009 | 9231 | 02 |
Question 8:
8 | State if assumption necessary with valid reason: No, since large samples (A.E.F.) B1
State another valid assumption: Samples must be random (A.E.F.) B1
State hypotheses: H0: µc – µf = 5, H1: µc – µf < 5 B1
Find z (A.E.F.): z = (19⋅14 – 15⋅36 – 5) /
√(20⋅54/75 + 9⋅84/75) M1 A1
= –1⋅22/√(2×15⋅19/75) = –1⋅92 *A1
Use correct tabular value of z: z 0.95 = 1⋅64[5] *B1
Consistent conclusion (A.E.F.; A1 dep *A1, *B1): Does exceed by less than 5 kg. M1√ A1 | (1)
(8) | [9]
Page 7 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – October/November 2009 | 9231 | 02150 sheep, chosen from a large flock of sheep, were divided into two groups of 75. Over a fixed period, one group had their grazing controlled and the other group grazed freely. The gains in weight, in kg, were recorded for each animal and the table below shows the sample means and the unbiased estimates of the population variances for the two samples.
\begin{tabular}{|c|c|c|}
\hline
& Sample mean & Unbiased estimate of population variance \\
\hline
Controlled grazing & 19.14 & 20.54 \\
Free grazing & 15.36 & 9.84 \\
\hline
\end{tabular}
It is required to test whether the population mean for sheep having their grazing controlled exceeds the population mean for sheep grazing freely by less than 5 kg. State, giving a reason, if it is necessary for the validity of the test to assume that the two population variances are equal. [1]
Stating any other assumption, carry out the test at the 5\% significance level. [8]
\hfill \mbox{\textit{CAIE FP2 2009 Q8 [9]}}