| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Standard +0.3 This is a straightforward application of the two-sample t-test with all summary statistics provided. Students must calculate pooled variance, compute the test statistic using a standard formula, compare to critical values, and state standard assumptions. While it requires careful arithmetic and knowledge of the procedure, it involves no conceptual challenges or novel problem-solving—just methodical execution of a learned technique with clearly labeled data. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Sample size | \(\bar { x }\) | \(\Sigma ( x - \bar { x } ) ^ { 2 }\) | |
| Experimental subjects | 18 | 32.16 | 1923.56 |
| Control subjects | 13 | 38.21 | 1147.58 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Anxiety scores; have normal distributions; common variance; independent samples | B2 | Context + 2 valid points B2; Context + 1VP no context +2VP B1; Not in words |
| \(H_0: \mu_E = \mu_C\), \(H_1: \mu_E < \mu_C\) | B1 | |
| \(s^2 = (1923.56 + 1147.58)/29\ (= 105.9)\) | B1 | Allow 1 error; eg \(s^2 = 1923.56/(17\text{ or }18)\) |
| \((t) = (32.16 - 38.21)/\sqrt{[105.9(18^{-1}+13^{-1})]}\) | M1 | |
| \(= -1.615\) | A1 | All correct \(+\) |
| \(t_{\text{crit}} = -1.699\) | B1 | \(47.5/(12\text{ or }13)\); Or \(+\); accept art \(\pm 1.70\) |
| Compare \(-1.615\) with \(-1.699\) and do not reject \(H_0\) | M1 | Or \(+\), \(+\). M0 if \(t\) not \(\pm1.699, \pm2.045\) |
| There is insufficient evidence at the 5% significance level to show that anxiety is reduced by listening to relaxation tapes | A1 ft 10 | In context, not over-assertive. OR Find CV or CR: B2B1B1; C= or \(\geq st\), \(t = \pm1.699\) or \(\pm2.015\); M1A1; \(t = \pm1.699\) B1; \(G = 6.11(2)\) A1; \(6.112 > 6.05\) and reject \(H_0\) etcM1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Sample sizes are too small (to appeal to CLT) | B1 1 |
# Question 6:
## Part (i):
| Working | Marks | Guidance |
|---------|-------|----------|
| Anxiety scores; have normal distributions; common variance; independent samples | B2 | Context + 2 valid points B2; Context + 1VP no context +2VP B1; Not in words |
| $H_0: \mu_E = \mu_C$, $H_1: \mu_E < \mu_C$ | B1 | |
| $s^2 = (1923.56 + 1147.58)/29\ (= 105.9)$ | B1 | Allow 1 error; eg $s^2 = 1923.56/(17\text{ or }18)$ |
| $(t) = (32.16 - 38.21)/\sqrt{[105.9(18^{-1}+13^{-1})]}$ | M1 | |
| $= -1.615$ | A1 | All correct $+$ |
| $t_{\text{crit}} = -1.699$ | B1 | $47.5/(12\text{ or }13)$; Or $+$; accept art $\pm 1.70$ |
| Compare $-1.615$ with $-1.699$ and do not reject $H_0$ | M1 | Or $+$, $+$. M0 if $t$ not $\pm1.699, \pm2.045$ |
| There is insufficient evidence at the 5% significance level to show that anxiety is reduced by listening to relaxation tapes | A1 ft **10** | In context, not over-assertive. OR Find CV or CR: B2B1B1; C= or $\geq st$, $t = \pm1.699$ or $\pm2.015$; M1A1; $t = \pm1.699$ B1; $G = 6.11(2)$ A1; $6.112 > 6.05$ and reject $H_0$ etcM1A1 |
## Part (ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| Sample sizes are too small (to appeal to CLT) | B1 **1** | |
---6 It has been suggested that people who suffer anxiety when they are about to undergo surgery can have their anxiety reduced by listening to relaxation tapes. A study was carried out on 18 experimental subjects who listened to relaxation tapes, and 13 control subjects who listened to neutral tapes. After listening to the tapes, the subjects were given a test which produced an anxiety score, $X$. Higher scores indicated higher anxiety. The results are summarised in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& Sample size & $\bar { x }$ & $\Sigma ( x - \bar { x } ) ^ { 2 }$ \\
\hline
Experimental subjects & 18 & 32.16 & 1923.56 \\
\hline
Control subjects & 13 & 38.21 & 1147.58 \\
\hline
\end{tabular}
\end{center}
(i) Use a two-sample $t$-test, at the $5 \%$ significance level, to test whether anxiety is reduced by listening to relaxation tapes. State two necessary assumptions for the validity of your test.\\
(ii) State why a test using a normal distribution would not have been appropriate.
\hfill \mbox{\textit{OCR S3 2010 Q6 [11]}}