| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Type I/II errors and power |
| Difficulty | Standard +0.3 This is a straightforward application of standard z-test procedures with known variance. Part (a) requires routine calculation of sample mean, standard deviation, test statistic, and comparison with critical value. Part (b) involves a standard Type II error calculation using normal distribution properties. All steps follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | Est (μ) = 23/50 = 0.46 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 49 50 | M1 | For an expression of the correct form for unbiased |
| Answer | Marks |
|---|---|
| 1225 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 50 | M1 | Standardising with their values. |
| = –1.268 or -1.267 or = –1.27 (3sf) | A1 | |
| −1.268 > –1.645 or 0.102 to 0.103 > 0.05 | M1 | For a valid comparison. |
| Answer | Marks | Guidance |
|---|---|---|
| concentration is less than 0.5. | A1 FT | In context, not definite. E.g., not ‘Mean concentration is |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 7(b) | cv−0.5 |
| Answer | Marks |
|---|---|
| 50 | M1 |
| cv = 0.448(1) or 0.448 (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 50 | M1 | |
| 1 – ɸ('1.524') | M1 | For area consistent with their working. |
| = 0.0638 to 0.0642 | A1 |
Question 7:
--- 7(a) ---
7(a) | Est (μ) = 23/50 = 0.46 | B1
Est (σ) = 50 13.02 −0.462 or Est (σ2) = 50 ( 13.02 −0.462 ) oe
49 50 49 50
1 (23.0)2
Or estimated unbiased variance = 13.02−
49 50 | M1 | For an expression of the correct form for unbiased
standard deviation or variance.
61
Est (σ) = 0.22315 or Est (σ2) = 0.0497959 = or 0.0498
1225 | A1
0.46−0.5
'0.22315'
50 | M1 | Standardising with their values.
= –1.268 or -1.267 or = –1.27 (3sf) | A1
−1.268 > –1.645 or 0.102 to 0.103 > 0.05 | M1 | For a valid comparison.
[Do not reject Ho] There is insufficient evidence [at 5% level] that the mean
concentration is less than 0.5. | A1 FT | In context, not definite. E.g., not ‘Mean concentration is
not less than 0.5’. No contradictions.
7
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | cv−0.5
=−1.645
'0.22315'
50 | M1
cv = 0.448(1) or 0.448 (3 sf) | A1
'0.448'− 0.4
[=1.521 to 1.524]
'0.22315'
50 | M1
1 – ɸ('1.524') | M1 | For area consistent with their working.
= 0.0638 to 0.0642 | A1
5A biologist wishes to test whether the mean concentration $\mu$, in suitable units, of a certain pollutant in a river is below the permitted level of 0.5. She measures the concentration, $x$, of the pollutant at 50 randomly chosen locations in the river. The results are summarised below.
$n = 50 \quad \Sigma x = 23.0 \quad \Sigma x^2 = 13.02$
\begin{enumerate}[label=(\alph*)]
\item Carry out a test at the 5% significance level of the null hypothesis $\mu = 0.5$ against the alternative hypothesis $\mu < 0.5$. [7]
\end{enumerate}
Later, a similar test is carried out at the 5% significance level using another sample of size 50 and the same hypotheses as before. You should assume that the standard deviation is unchanged.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Given that, in fact, the value of $\mu$ is 0.4, find the probability of a Type II error. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q7 [12]}}