OCR S3 2014 June — Question 3 7 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2014
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeSingle sample t-test
DifficultyStandard +0.3 This is a straightforward one-sample t-test application with clear hypotheses (μ < 11.8), small sample requiring calculation of sample mean and standard deviation, and standard critical value comparison. While it requires proper test structure and calculation accuracy, it follows a standard template with no conceptual surprises, making it slightly easier than average for S3 level.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean

3 An athlete finds that her times for running 100 m are normally distributed. Before a period of intensive training, her mean time is 11.8 s . After the period of intensive training, five randomly selected times, in seconds, are as follows. $$\begin{array} { l l l l l } 11.70 & 11.65 & 11.80 & 11.75 & 11.60 \end{array}$$ Carry out a suitable test, at the \(5 \%\) significance level, to investigate whether times after the training are less, on average, than times before the training.

Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: \mu = 11.8\), \(H_1: \mu < 11.8\)B1 NOT e.g. \(\mu_0, \mu_1\)
\(\bar{x} = 11.7\)B1 or \(\bar{d} = \pm 0.1\)
\(\hat{\sigma}^2 = 0.00625\)B1 Allow \(\frac{1}{160}\)
\(TS = \frac{11.7 - 11.8}{\sqrt{\frac{0.00625}{5}}}\)M1 Allow reversed if consistent. And for following marks.
\(= -2.828\)A1 Allow \(-2.83\)
\(TS < -2.132\), reject \(H_0\)M1 ft TS. OR \(p < 0.05\), reject \(H_0\). Must be \(t\), not \(z\).
There is sufficient evidence that the intensive training has improved the athlete's performanceA1 ft TS. Contextualised, not over-assertive. \(p = 0.0237\) and conclusion.
[7]
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 11.8$, $H_1: \mu < 11.8$ | B1 | NOT e.g. $\mu_0, \mu_1$ |
| $\bar{x} = 11.7$ | B1 | or $\bar{d} = \pm 0.1$ |
| $\hat{\sigma}^2 = 0.00625$ | B1 | Allow $\frac{1}{160}$ |
| $TS = \frac{11.7 - 11.8}{\sqrt{\frac{0.00625}{5}}}$ | M1 | Allow reversed if consistent. And for following marks. |
| $= -2.828$ | A1 | Allow $-2.83$ |
| $TS < -2.132$, reject $H_0$ | M1 | ft TS. OR $p < 0.05$, reject $H_0$. Must be $t$, not $z$. |
| There is sufficient evidence that the intensive training has improved the athlete's performance | A1 | ft TS. Contextualised, not over-assertive. $p = 0.0237$ and conclusion. |
| **[7]** | | |

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3 An athlete finds that her times for running 100 m are normally distributed. Before a period of intensive training, her mean time is 11.8 s . After the period of intensive training, five randomly selected times, in seconds, are as follows.

$$\begin{array} { l l l l l } 
11.70 & 11.65 & 11.80 & 11.75 & 11.60
\end{array}$$

Carry out a suitable test, at the $5 \%$ significance level, to investigate whether times after the training are less, on average, than times before the training.

\hfill \mbox{\textit{OCR S3 2014 Q3 [7]}}