CAIE FP2 2013 June — Question 9 14 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeTwo-sample t-test equal variance
DifficultyChallenging +1.2 This is a standard two-part hypothesis testing question requiring a one-sample t-test followed by a two-sample t-test. While it involves multiple steps (calculating sample means/variances, finding test statistics, comparing to critical values), these are routine procedures for Further Maths students with no novel insight required. The question clearly signposts what to do and the calculations are straightforward from given summary statistics.
Spec5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution

9 A gardener \(P\) claims that a new type of fruit tree produces a higher annual mass of fruit than the type that he has previously grown. The old type of tree produced 5.2 kg of fruit per tree, on average. A random sample of 10 trees of the new type is chosen. The masses, \(x \mathrm {~kg}\), of fruit produced are summarised as follows. $$\Sigma x = 61.0 \quad \Sigma x ^ { 2 } = 384.0$$ Test, at the \(5 \%\) significance level, whether gardener \(P\) 's claim is justified, assuming a normal distribution. Another gardener \(Q\) has his own type of fruit tree. The masses, \(y \mathrm {~kg}\), of fruit produced by a random sample of 10 trees grown by gardener \(Q\) are summarised as follows. $$\Sigma y = 70.0 \quad \Sigma y ^ { 2 } = 500.6$$ Test, at the \(5 \%\) significance level, whether the mean mass of fruit produced by gardener \(Q\) 's trees is greater than the mean mass of fruit produced by gardener \(P\) 's trees. You may assume that both distributions are normal and you should state any additional assumption.

Question 9:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\bar{x} = 6.1\), \(s_x^2 = (384 - 61^2/10)/9 = 1.322\) or \(119/90\) or \(1.15^2\)M1 Find sample mean and estimate population variance (allow biased: \(1.19\) or \(1.091^2\))
\(H_0\): \(\mu = 5.2\), \(H_1\): \(\mu > 5.2\)B1 State hypotheses
\(t = (\bar{x} - 5.2)/(s_x/\sqrt{10}) = 2.47[5]\)M1 A1 Calculate value of \(t\) (to 3 s.f.)
\(t_{9,\,0.95} = 1.83[3]\)*B1 State or use correct tabular \(t\) value (or compare \(x = 6.1\) with \(5.2 + 0.667 = 5.87\))
\(t > 1.83\) so mean is greaterB1\(\sqrt{}\) Correct conclusion (A.E.F., \(\sqrt{}\) on \(t\), dep *B1)
\(H_0\): \(\mu_P = \mu_Q\), \(H_1\): \(\mu_P \leq \mu_Q\)B1 State hypotheses
Distributions have equal variancesB1 State assumption
\(y = 7.0\), \(\big[s_y^2 = (500.6 - 70^2/10)/9 = 1.178\) or \(53/45\) or \(1.085^2\big]\) Find sample mean and estimate variance for \(Y\)
\(s^2 = (9\times 1.322 + 9\times 1.178)/18\) or \((384 - 61^2/10 + 500.6 - 70^2/10)/18 = 1.25\) or \(1.18^2\)M1 A1 Estimate pooled common variance (EITHER)
\(s^2 = (1.322 + 1.178)/10 = 0.25\)(M1 A1) OR equivalently estimate common variance
\(t = (\bar{y} - \bar{x})/s = 1.8\)(M1 A1) Calculate value of \(t\) (to 3 s.f.)
\(t_{18,\,0.95} = 1.73[4]\)*B1 State or use correct tabular \(t\) value (or compare \(y - x = 0.9\) with \(0.867\))
\(t > 1.73\) so \(Q\)'s mean is greaterB1\(\sqrt{}\) Correct conclusion (A.E.F., \(\sqrt{}\) on \(t\), dep *B1)
Total: 14 marks
## Question 9:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = 6.1$, $s_x^2 = (384 - 61^2/10)/9 = 1.322$ or $119/90$ or $1.15^2$ | M1 | Find sample mean and estimate population variance (allow biased: $1.19$ or $1.091^2$) |
| $H_0$: $\mu = 5.2$, $H_1$: $\mu > 5.2$ | B1 | State hypotheses |
| $t = (\bar{x} - 5.2)/(s_x/\sqrt{10}) = 2.47[5]$ | M1 A1 | Calculate value of $t$ (to 3 s.f.) |
| $t_{9,\,0.95} = 1.83[3]$ | *B1 | State or use correct tabular $t$ value (or compare $x = 6.1$ with $5.2 + 0.667 = 5.87$) |
| $t > 1.83$ so mean is greater | B1$\sqrt{}$ | Correct conclusion (A.E.F., $\sqrt{}$ on $t$, dep *B1) |
| $H_0$: $\mu_P = \mu_Q$, $H_1$: $\mu_P \leq \mu_Q$ | B1 | State hypotheses |
| Distributions have equal variances | B1 | State assumption |
| $y = 7.0$, $\big[s_y^2 = (500.6 - 70^2/10)/9 = 1.178$ or $53/45$ or $1.085^2\big]$ | | Find sample mean and estimate variance for $Y$ |
| $s^2 = (9\times 1.322 + 9\times 1.178)/18$ or $(384 - 61^2/10 + 500.6 - 70^2/10)/18 = 1.25$ or $1.18^2$ | M1 A1 | Estimate pooled common variance (EITHER) |
| $s^2 = (1.322 + 1.178)/10 = 0.25$ | (M1 A1) | OR equivalently estimate common variance |
| $t = (\bar{y} - \bar{x})/s = 1.8$ | (M1 A1) | Calculate value of $t$ (to 3 s.f.) |
| $t_{18,\,0.95} = 1.73[4]$ | *B1 | State or use correct tabular $t$ value (or compare $y - x = 0.9$ with $0.867$) |
| $t > 1.73$ so $Q$'s mean is greater | B1$\sqrt{}$ | Correct conclusion (A.E.F., $\sqrt{}$ on $t$, dep *B1) |

**Total: 14 marks**
9 A gardener $P$ claims that a new type of fruit tree produces a higher annual mass of fruit than the type that he has previously grown. The old type of tree produced 5.2 kg of fruit per tree, on average. A random sample of 10 trees of the new type is chosen. The masses, $x \mathrm {~kg}$, of fruit produced are summarised as follows.

$$\Sigma x = 61.0 \quad \Sigma x ^ { 2 } = 384.0$$

Test, at the $5 \%$ significance level, whether gardener $P$ 's claim is justified, assuming a normal distribution.

Another gardener $Q$ has his own type of fruit tree. The masses, $y \mathrm {~kg}$, of fruit produced by a random sample of 10 trees grown by gardener $Q$ are summarised as follows.

$$\Sigma y = 70.0 \quad \Sigma y ^ { 2 } = 500.6$$

Test, at the $5 \%$ significance level, whether the mean mass of fruit produced by gardener $Q$ 's trees is greater than the mean mass of fruit produced by gardener $P$ 's trees. You may assume that both distributions are normal and you should state any additional assumption.

\hfill \mbox{\textit{CAIE FP2 2013 Q9 [14]}}