| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample t-test |
| Difficulty | Standard +0.3 This is a straightforward one-sample t-test with small sample size. Students must calculate sample mean and standard deviation from given data, then apply the standard t-test procedure. While it requires careful calculation and knowledge of the t-distribution, it's a routine application of a standard technique with no conceptual challenges or novel insights required. The additional part (b) asking for assumptions is standard bookwork. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{x} = \frac{56.56}{7} = 8.08\) | B1 | Accept unsimplified. |
| \(s^2 = \frac{1}{6}\left(457.275 - \frac{56.56^2}{7}\right) = 0.045033 \left[= \frac{1351}{30000}\right]\) | M1 | Accept unsimplified. |
| \(t = \frac{8.08 - 8.22}{\sqrt{\frac{s^2}{7}}} = -1.745\) | M1 A1 | \(1.74 - 1.75\) |
| Tabular value \(1.440\); \(1.745 > 1.440\), so reject \(H_0\) | M1 | Comparison with \(1.440\) and correct FT conclusion. |
| There is insufficient evidence to support the hypothesis that the (population) mean is \(8.22\) OR sufficient evidence to suggest mean is less than \(8.22\) OR sufficient evidence to suggest \(\mu < 8.22\) OE | A1 | Correct conclusion, in context, following correct work. Level of uncertainty in language is used. |
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Underlying distribution is normal / population is normal | B1 | \(X\) is normal distributed. |
| Total: 1 |
## Question 1:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = \frac{56.56}{7} = 8.08$ | **B1** | Accept unsimplified. |
| $s^2 = \frac{1}{6}\left(457.275 - \frac{56.56^2}{7}\right) = 0.045033 \left[= \frac{1351}{30000}\right]$ | **M1** | Accept unsimplified. |
| $t = \frac{8.08 - 8.22}{\sqrt{\frac{s^2}{7}}} = -1.745$ | **M1 A1** | $1.74 - 1.75$ |
| Tabular value $1.440$; $1.745 > 1.440$, so reject $H_0$ | **M1** | Comparison with $1.440$ and correct FT conclusion. |
| There is insufficient evidence to support the hypothesis that the (population) mean is $8.22$ OR sufficient evidence to suggest mean is less than $8.22$ OR sufficient evidence to suggest $\mu < 8.22$ OE | **A1** | Correct conclusion, in context, following correct work. Level of uncertainty in language is used. |
| | **Total: 6** | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Underlying distribution is normal / population is normal | **B1** | $X$ is normal distributed. |
| | **Total: 1** | |1 A random sample of 7 observations of a variable $X$ are as follows.
$$\begin{array} { l l l l l l l }
8.26 & 7.78 & 7.92 & 8.04 & 8.27 & 7.95 & 8.34
\end{array}$$
The population mean of $X$ is $\mu$.
\begin{enumerate}[label=(\alph*)]
\item Test, at the $10 \%$ significance level, the null hypothesis $\mu = 8.22$ against the alternative hypothesis $\mu < 8.22$.
\item State an assumption necessary for the test in part (a) to be valid.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q1 [7]}}