| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Find critical alpha or significance level |
| Difficulty | Standard +0.8 This is a two-sample t-test requiring calculation of sample means and variances from summary statistics, followed by a two-tailed test and then a more sophisticated one-tailed test with a non-zero hypothesized difference. Part (b) requires working backwards from the test statistic to find critical significance levels, which is conceptually more demanding than standard hypothesis testing. While the techniques are standard Further Statistics content, the multi-step calculations and reverse-engineering in part (b) elevate this above routine exercises. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s_x^2 = \frac{1}{39}\left(31400 - \frac{1120^2}{40}\right) = 1.0256\) and \(s_y^2 = \frac{1}{49}\left(37600 - \frac{1370^2}{50}\right) = 1.2653\) | M1 | \(\frac{40}{39}\) and \(\frac{62}{49}\) both required |
| \(s^2 = \frac{1.0256}{40} + \frac{1.2653}{50}\) | M1 | \(\frac{2434}{47775}\); condone \(s^2 = \frac{1.03}{40} + \frac{1.27}{50}\) or \(0.05115\) |
| \(s^2 = 0.050947\) | A1 | May be implied by \(2.66\) as \(z\) value |
| \(H_0: \mu_a - \mu_b = 0\) and \(H_1: \mu_a - \mu_b \neq 0\) | B1 | |
| \(z = \dfrac{\frac{1120}{40} - \frac{1370}{50}}{s} = \dfrac{0.6}{\sqrt{0.050947}}\) | M1 | |
| \(z = 2.66\) | A1 | |
| Critical value \(z = 2.326\); \(2.66 > 2.326\); Reject \(H_0\) | M1 | Compare their \(z\) value with \(2.326\) and correct follow-through conclusion |
| Sufficient evidence to support difference in population means oe | A1 | Correct conclusion in context; level of uncertainty in language is used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(z = \dfrac{0.6 - 0.25}{s}\) | M1 | Condone \(1.54\) |
| \(z = 1.55\) | A1 | |
| \(0.9392 - 0.9395\) or \(0.0605 - 0.0608\) | A1 | |
| \(\alpha \geqslant 6.05\) | A1 | Condone \(6.05 - 6.08\), accept \(>\) |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $s_x^2 = \frac{1}{39}\left(31400 - \frac{1120^2}{40}\right) = 1.0256$ and $s_y^2 = \frac{1}{49}\left(37600 - \frac{1370^2}{50}\right) = 1.2653$ | M1 | $\frac{40}{39}$ and $\frac{62}{49}$ both required |
| $s^2 = \frac{1.0256}{40} + \frac{1.2653}{50}$ | M1 | $\frac{2434}{47775}$; condone $s^2 = \frac{1.03}{40} + \frac{1.27}{50}$ or $0.05115$ |
| $s^2 = 0.050947$ | A1 | May be implied by $2.66$ as $z$ value |
| $H_0: \mu_a - \mu_b = 0$ and $H_1: \mu_a - \mu_b \neq 0$ | B1 | |
| $z = \dfrac{\frac{1120}{40} - \frac{1370}{50}}{s} = \dfrac{0.6}{\sqrt{0.050947}}$ | M1 | |
| $z = 2.66$ | A1 | |
| Critical value $z = 2.326$; $2.66 > 2.326$; Reject $H_0$ | M1 | Compare their $z$ value with $2.326$ and correct follow-through conclusion |
| Sufficient evidence to support difference in **population** means oe | A1 | Correct conclusion in context; level of uncertainty in language is used |
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## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $z = \dfrac{0.6 - 0.25}{s}$ | M1 | Condone $1.54$ |
| $z = 1.55$ | A1 | |
| $0.9392 - 0.9395$ or $0.0605 - 0.0608$ | A1 | |
| $\alpha \geqslant 6.05$ | A1 | Condone $6.05 - 6.08$, accept $>$ |6 A company has two machines, $A$ and $B$, which independently fill small bottles with a liquid. The volumes of liquid per bottle, in suitable units, filled by machines $A$ and $B$ are denoted by $x$ and $y$ respectively. A scientist at the company takes a random sample of 40 bottles filled by machine $A$ and a random sample of 50 bottles filled by machine $B$. The results are summarised as follows.
$$\sum x = 1120 \quad \sum x ^ { 2 } = 31400 \quad \sum y = 1370 \quad \sum y ^ { 2 } = 37600$$
The population means of the volumes of liquid in the bottles filled by machines $A$ and $B$ are denoted by $\mu _ { A }$ and $\mu _ { B }$.
\begin{enumerate}[label=(\alph*)]
\item Test at the $2 \%$ significance level whether there is any difference between $\mu _ { A }$ and $\mu _ { B }$.
\item Find the set of values of $\alpha$ for which there would be evidence at the $\alpha \%$ significance level that $\mu _ { \mathrm { A } } - \mu _ { \mathrm { B } }$ is greater than 0.25.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q6 [12]}}