| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Standard +0.3 This is a standard S2 hypothesis testing question with routine calculations: computing sample mean/variance from summations, performing a straightforward two-tail z-test, and finding a proportion using normal tables. The multi-part structure and interpretation question add slight complexity, but all techniques are textbook applications requiring no novel insight. |
| Spec | 2.05e Hypothesis test for normal mean: known variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\hat{\mu} = \bar{x} = 81\) | B1 | 81 only, can be implied |
| \(\dfrac{329800}{50} - 81^2\) \([= 35]\) | M1 | Correct formula for biased estimate, their "81", can be implied |
| \(\times \dfrac{50}{49}\); \(= 35.71\) | M1 | Multiply by 50/49. SC: single formula: M2, or M1 if wrong but divisor 49 anywhere [can be recovered if correctly done in part (ii)] |
| A1 | A.r.t. \(35.7\) – can't be recovered from part (ii). Can be implied | |
| \(1 - \Phi\!\left(\dfrac{90-81}{\sqrt{35.71}}\right) = 1 - \Phi(1.506) = 1 - 0.9339\) | M1 | Standardise with their \(\mu\) and \(\sigma\); allow \(\sigma^2\), cc but not \(\sqrt{50}\) |
| \(= \mathbf{6.61\%}\) or \(\mathbf{0.0661}\) | A1 | Answer a.r.t. 6.6% or 0.066 |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \mu = 80\) | B2 | Correct, B2. One error e.g. wrong or no symbol, \(>\), B1, but \(x\) or \(\bar{x}\) or \(t\) etc, or 81, B0. NB: If both hypotheses involve 81, *can't* get final M1. |
| \(H_1: \mu \neq 80\) | ||
| \(\alpha\): \(z = \dfrac{81 - 80}{\sqrt{35.71/50}} = 1.183\) \([or\; p = 0.1183]\) | M1 | Standardise with \(\sqrt{50}\); allow \(\sqrt{}\), sign or cc errors; allow from biased variance |
| A1 | \(z\), a.r.t. 1.18, or \(p\), a.r.t. 0.118. Allow \(-1.18\). | |
| \(< 1.645\) | B1 | Their \(z < 1.645\) or \(p > 0.05\), *not* if one-tail. Allow \(-1.18 > -1.645\). *Not* just 1.645 seen. |
| \(\beta\): CV \(80 + 1.645\sqrt{\dfrac{35.71}{50}} = 81.39\) | M1 | \(80 + zs/\sqrt{50}\); allow \(\sqrt{}\) or cc errors; ignore \(-\) (no marks for \(-\) alone) |
| B1 | \(z = 1.645\) used in this expression (not just seen), *not* from one-tail | |
| \(81 < 81.39\) | A1 | Compare CV with 81; allow 81.08 from one-tailed (\(z = 1.282\)) (but not on their \(\sigma\)). SC: \(81 - 1.645\sqrt{\frac{35.71}{50}}\): If \(H_0: \mu = 80\): (B2) M1B1A0M0A0. If \(H_0: \mu = 81\): (B0) M1B1A1 (79.61) M0A0. |
| Do not reject \(H_0\). | M1 | Correct first conclusion; needs \(\sqrt{50}\), correct comparison type; \(\mu\) and \(\bar{x}\) not consistently wrong way round (thus \(H_0: \mu = 81\) can get B0 M1A1A1 M0A0, max 3/7). In method \(\beta\), it needs to be clear that comparison involves \(\bar{x}\). |
| Insufficient evidence that the mean time is not 80 minutes. | A1ft | Contextualised (mention "time"), acknowledge uncertainty ("evidence that…"). *Not* "significant evidence that mean time is 80". FT on wrong \(z\)-value or wrong critical value if previous mark gained. SC: One-tailed: can get B1B0 M1A1B0 M1A1, max 5/7. No \(\sqrt{50}\): can get B2 M0 B1 M0, max 3/7. |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) Yes (single observation only) | B1 | No reason needed, but withhold if wrong reason seen. Allow "yes, no dist\(^n\) given" |
| (b) No, CLT applies to large sample | B1 | "No" *and* refer to central limit theorem or "large sample". {note for scoris zoning – (a) and (b) to be in single zone} |
| [2] |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\hat{\mu} = \bar{x} = 81$ | B1 | 81 only, can be implied |
| $\dfrac{329800}{50} - 81^2$ $[= 35]$ | M1 | Correct formula for biased estimate, their "81", can be implied |
| $\times \dfrac{50}{49}$; $= 35.71$ | M1 | Multiply by 50/49. SC: single formula: M2, or M1 if wrong but divisor 49 anywhere [can be recovered if correctly done in part (ii)] |
| | A1 | A.r.t. $35.7$ – can't be recovered from part (ii). Can be implied |
| $1 - \Phi\!\left(\dfrac{90-81}{\sqrt{35.71}}\right) = 1 - \Phi(1.506) = 1 - 0.9339$ | M1 | Standardise with their $\mu$ and $\sigma$; allow $\sigma^2$, cc but not $\sqrt{50}$ |
| $= \mathbf{6.61\%}$ or $\mathbf{0.0661}$ | A1 | Answer a.r.t. 6.6% or 0.066 |
| | **[6]** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu = 80$ | B2 | Correct, B2. One error e.g. wrong or no symbol, $>$, B1, but $x$ or $\bar{x}$ or $t$ etc, or 81, B0. NB: If both hypotheses involve 81, *can't* get final M1. |
| $H_1: \mu \neq 80$ | | |
| $\alpha$: $z = \dfrac{81 - 80}{\sqrt{35.71/50}} = 1.183$ $[or\; p = 0.1183]$ | M1 | Standardise with $\sqrt{50}$; allow $\sqrt{}$, sign or cc errors; allow from biased variance |
| | A1 | $z$, a.r.t. 1.18, or $p$, a.r.t. 0.118. Allow $-1.18$. |
| $< 1.645$ | B1 | Their $z < 1.645$ or $p > 0.05$, *not* if one-tail. Allow $-1.18 > -1.645$. *Not* just 1.645 seen. |
| $\beta$: CV $80 + 1.645\sqrt{\dfrac{35.71}{50}} = 81.39$ | M1 | $80 + zs/\sqrt{50}$; allow $\sqrt{}$ or cc errors; ignore $-$ (no marks for $-$ alone) |
| | B1 | $z = 1.645$ used in this expression (not just seen), *not* from one-tail |
| $81 < 81.39$ | A1 | Compare CV with 81; allow 81.08 from one-tailed ($z = 1.282$) (but not on their $\sigma$). SC: $81 - 1.645\sqrt{\frac{35.71}{50}}$: If $H_0: \mu = 80$: (B2) M1B1A0M0A0. If $H_0: \mu = 81$: (B0) M1B1A1 (79.61) M0A0. |
| Do not reject $H_0$. | M1 | Correct first conclusion; needs $\sqrt{50}$, correct comparison type; $\mu$ and $\bar{x}$ not consistently wrong way round (thus $H_0: \mu = 81$ can get B0 M1A1A1 M0A0, max 3/7). In method $\beta$, it needs to be clear that comparison involves $\bar{x}$. |
| Insufficient evidence that the mean time is not 80 minutes. | A1ft | Contextualised (mention "time"), acknowledge uncertainty ("evidence that…"). *Not* "significant evidence that mean time is 80". FT on wrong $z$-value or wrong critical value if previous mark gained. SC: One-tailed: can get B1B0 M1A1B0 M1A1, max 5/7. No $\sqrt{50}$: can get B2 M0 B1 M0, max 3/7. |
| | **[7]** | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| (a) Yes (single observation only) | B1 | No reason needed, but withhold if wrong reason seen. Allow "yes, no dist$^n$ given" |
| (b) No, CLT applies to large sample | B1 | "No" *and* refer to central limit theorem or "large sample". {note for scoris zoning – (a) and (b) to be in single zone} |
| | **[2]** | |
---7 An examination board is developing a new syllabus and wants to know if the question papers are the right length. A random sample of 50 candidates was given a pre-test on a dummy paper. The times, $t$ minutes, taken by these candidates to complete the paper can be summarised by\\
$n = 50$,\\
$\sum \boldsymbol { t } = \mathbf { 4 0 5 0 }$,\\
$\sum \boldsymbol { t } ^ { \mathbf { 2 } } \boldsymbol { = } \mathbf { 3 2 9 8 0 0 }$.\\
Assume that times are normally distributed.\\[0pt]
(i) Estimate the proportion of candidates that could not complete the paper within 90 minutes. [6]\\
(ii) Test, at the $10 \%$ significance level, whether the mean time for all candidates to complete this paper is $\mathbf { 8 0 }$ minutes. Use a two-tail test. [7]\\
(iii) Explain whether the assumption that times are normally distributed is necessary in answering
\begin{enumerate}[label=(\alph*)]
\item part (i),\\[0pt]
\item part (ii). [2]
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2014 Q7 [15]}}