CAIE FP2 2014 November — Question 6 5 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2014
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeFind critical alpha or significance level
DifficultyChallenging +1.2 This is a standard two-sample t-test with given summary statistics requiring calculation of the test statistic, finding the critical value from tables, and working backwards to find the significance level range. While it involves multiple steps and understanding of hypothesis testing mechanics, the procedure is routine for Further Maths students with no novel insight required beyond careful arithmetic and table lookup.
Spec5.05c Hypothesis test: normal distribution for population mean

6 A random sample of 50 observations of a random variable \(X\) and a random sample of 60 observations of a random variable \(Y\) are taken. The results for the sample means, \(\bar { x }\) and \(\bar { y }\), and the unbiased estimates for the population variances, \(s _ { x } ^ { 2 }\) and \(s _ { y } ^ { 2 }\), respectively, are as follows. $$\bar { x } = 25.4 \quad \bar { y } = 23.6 \quad s _ { x } ^ { 2 } = 23.2 \quad s _ { y } ^ { 2 } = 27.8$$ A test, at the \(\alpha \%\) significance level, of the null hypothesis that the population means of \(X\) and \(Y\) are equal against the alternative hypothesis that they are not equal is carried out. Given that the null hypothesis is not rejected, find the set of possible values of \(\alpha\).

Question 6:
(Combined sample method)
AnswerMarks Guidance
AnswerMarks Guidance
\(s^2 = \frac{s_X^2}{50} + \frac{s_Y^2}{60} = \frac{1391}{1500}\) or \(0.9273\) or \(0.9630^2\)M1 Estimate population variance for combined sample
\(z = \frac{1.8}{s} = 1.869\)M1 A1 Calculate value of \(z\) to 2 d.p., either sign
\(\Phi(z) = 0.9692\) [or 96.92%]; \(\alpha \leq\) (or \(<\)) \(6.2\) (allow 6.1)M1 A1 Find \(\Phi(z)\); M1 A0 for \(\alpha \leq 3.1\) or \(\alpha > 93.8\)
(Equal variance method — S.R.)
AnswerMarks Guidance
AnswerMarks Guidance
Assuming equal population variances(B1) Explicit assumption
\(s^2 = \frac{(49s_X^2 + 59s_Y^2)}{108} = \frac{2777}{108}\) or \(25.71\) or \(5.071^2\) Find pooled estimate of common variance
\(z = \frac{1.8}{s\sqrt{50^{-1}+60^{-1}}} = 1.854\)M1 A1
\(\Phi(z) = 0.9681\) [or 96.81%]; \(\alpha \leq\) (or \(<\)) \(6.4\)M1 A1 M1 A0 for \(\alpha \leq 3.2\) or \(\alpha > 93.6\)
# Question 6:

## (Combined sample method)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s^2 = \frac{s_X^2}{50} + \frac{s_Y^2}{60} = \frac{1391}{1500}$ or $0.9273$ or $0.9630^2$ | M1 | Estimate population variance for combined sample |
| $z = \frac{1.8}{s} = 1.869$ | M1 A1 | Calculate value of $z$ to 2 d.p., either sign |
| $\Phi(z) = 0.9692$ [or 96.92%]; $\alpha \leq$ (or $<$) $6.2$ (allow 6.1) | M1 A1 | Find $\Phi(z)$; M1 A0 for $\alpha \leq 3.1$ or $\alpha > 93.8$ |

## (Equal variance method — S.R.)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Assuming equal population variances | (B1) | Explicit assumption |
| $s^2 = \frac{(49s_X^2 + 59s_Y^2)}{108} = \frac{2777}{108}$ or $25.71$ or $5.071^2$ | | Find pooled estimate of common variance |
| $z = \frac{1.8}{s\sqrt{50^{-1}+60^{-1}}} = 1.854$ | M1 A1 | |
| $\Phi(z) = 0.9681$ [or 96.81%]; $\alpha \leq$ (or $<$) $6.4$ | M1 A1 | M1 A0 for $\alpha \leq 3.2$ or $\alpha > 93.6$ |

---
6 A random sample of 50 observations of a random variable $X$ and a random sample of 60 observations of a random variable $Y$ are taken. The results for the sample means, $\bar { x }$ and $\bar { y }$, and the unbiased estimates for the population variances, $s _ { x } ^ { 2 }$ and $s _ { y } ^ { 2 }$, respectively, are as follows.

$$\bar { x } = 25.4 \quad \bar { y } = 23.6 \quad s _ { x } ^ { 2 } = 23.2 \quad s _ { y } ^ { 2 } = 27.8$$

A test, at the $\alpha \%$ significance level, of the null hypothesis that the population means of $X$ and $Y$ are equal against the alternative hypothesis that they are not equal is carried out. Given that the null hypothesis is not rejected, find the set of possible values of $\alpha$.

\hfill \mbox{\textit{CAIE FP2 2014 Q6 [5]}}