OCR S3 2007 June — Question 3 8 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2007
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypePaired sample t-test
DifficultyStandard +0.3 This is a straightforward paired t-test application with clearly defined hypotheses (one-tailed test for overestimation). Students must calculate summary statistics from given differences and apply the standard t-test procedure, which is routine for S3 level with no conceptual complications beyond stating the normality assumption.
Spec5.05c Hypothesis test: normal distribution for population mean

3 A nurse was asked to measure the blood pressure of 12 patients using an aneroid device. The nurse's readings were immediately checked using an accurate electronic device. The differences, \(x\), given by \(x =\) (aneroid reading - electronic reading), in appropriate units, are shown below. $$\begin{array} { c c c c c c c c c c c } - 1.3 & 4.7 & - 0.9 & 3.8 & - 1.5 & 4.0 & - 1.9 & 4.4 & - 0.8 & 5.5 & - 2.9 \end{array} 4.1$$ Stating any assumption you need to make, test, at the \(10 \%\) significance level, whether readings with an aneroid device, on average, overestimate patients' blood pressure.

AnswerMarks Guidance
Assumes differences form a random sample from a normal distribution. \(H_0: \mu = 0\), \(H_1: \mu > 0\)B1
\(\bar{x} = 17.2/12\); \(s^2 = 10.155\) AEFB1B1 Other letters if defined; or in words. Or \((12/11)(136.36/12-(17.2/12)^2)\)aef
EITHER: \(t = \frac{\bar{x}}{s/\sqrt{12}}\) (+ or -)M1 With 12 or 9.309/11
\(= 1.558\)A1 Must be positive. Accept 1.56
1.363 seen. \(1.558 > 1.363\), so reject \(H_0\) and accept that there is evidence that the readings from the aneroid device overestimate blood pressure on averageB1 Allow CV of 1.372 or 1.356 evidence. Explicit comparison of CV(not - with +) and conclusion in context.
OR: For critical region or critical value of \(\bar{x}\): \(1.363(\sqrt{s/12})\)M1B1 B1 for correct \(t\)
Giving \(1.25(3)\)A1
Compare \(1.43(3)\) with \(1.25(3)\)B1√ 8 marks total
Conclusion in context
Assumes differences form a random sample from a normal distribution. $H_0: \mu = 0$, $H_1: \mu > 0$ | B1 |

$\bar{x} = 17.2/12$; $s^2 = 10.155$ AEF | B1B1 | Other letters if defined; or in words. Or $(12/11)(136.36/12-(17.2/12)^2)$aef

EITHER: $t = \frac{\bar{x}}{s/\sqrt{12}}$ (+ or -) | M1 | With 12 or 9.309/11

$= 1.558$ | A1 | Must be positive. Accept 1.56

1.363 seen. $1.558 > 1.363$, so reject $H_0$ and accept that there is evidence that the readings from the aneroid device overestimate blood pressure on average | B1 | Allow CV of 1.372 or 1.356 evidence. Explicit comparison of CV(not - with +) and conclusion in context.

OR: For critical region or critical value of $\bar{x}$: $1.363(\sqrt{s/12})$ | M1B1 | B1 for correct $t$

Giving $1.25(3)$ | A1 |

Compare $1.43(3)$ with $1.25(3)$ | B1√ | 8 marks total |

Conclusion in context | |
3 A nurse was asked to measure the blood pressure of 12 patients using an aneroid device. The nurse's readings were immediately checked using an accurate electronic device. The differences, $x$, given by $x =$ (aneroid reading - electronic reading), in appropriate units, are shown below.

$$\begin{array} { c c c c c c c c c c c } 
- 1.3 & 4.7 & - 0.9 & 3.8 & - 1.5 & 4.0 & - 1.9 & 4.4 & - 0.8 & 5.5 & - 2.9
\end{array} 4.1$$

Stating any assumption you need to make, test, at the $10 \%$ significance level, whether readings with an aneroid device, on average, overestimate patients' blood pressure.

\hfill \mbox{\textit{OCR S3 2007 Q3 [8]}}