OCR MEI S4 2011 June — Question 4 24 marks

Exam BoardOCR MEI
ModuleS4 (Statistics 4)
Year2011
SessionJune
Marks24
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNon-parametric tests
TypeExperimental design types
DifficultyStandard +0.3 This question tests standard experimental design concepts (part a) and routine one-way ANOVA with unequal replicates (part b). While it requires understanding of blocking principles and ANOVA calculations, these are textbook procedures at S4 level with no novel problem-solving or complex reasoning required. The calculations are straightforward given the summary statistics.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance
4
  1. Provide an example of an experimental situation where there is one factor of primary interest and where a suitable experimental design would be
    1. randomised blocks,
    2. a Latin square. In each case, explain carefully why the design is suitable and why the other design would not be appropriate.
  2. An industrial experiment to compare four treatments for increasing the tensile strength of steel is carried out according to a completely randomised design. For various reasons, it is not possible to use the same number of replicates for each treatment. The increases, in a suitable unit of tensile strength, are as follows.
    Treatment
    A
    Treatment
    B
    Treatment
    C
    Treatment
    D
    10.121.19.222.6
    21.220.38.817.4
    11.616.015.223.1
    13.615.019.2
    [The sum of these data items is 256.8 and the sum of their squares is 4471.92 .] Construct the usual one-way analysis of variance table. Carry out the appropriate test, using a \(5 \%\) significance level. RECOGNISING ACHIEVEMENT
Question 4 (4769 June 2011):
Part (a)(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Description of situation where randomised blocks would be suitable: one extraneous factor (e.g. stream down one side of a field)E2 Each E2 available as E2, E1, E0
Explanation of why RB is suitable (design allows the extraneous factor to be "taken out" separately)E2
Explanation of why LS is not appropriate (only one extraneous factor; LS unnecessarily complicated; not enough degrees of freedom for sensible estimate of experimental error)E2
Part (a)(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Description of situation where Latin square would be suitable: two extraneous factors, all with same number of levels (e.g. streams down two sides of a field)E2
Explanation of why LS is suitable (design allows extraneous factors to be "taken out" separately)E2
Explanation of why RB is not appropriate (RB cannot cope with two extraneous factors)E2
[12]
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Totals: 56.5, 57.4, 60.6, 82.3 from samples of sizes 4, 3, 5, 4
Grand total 256.8; CF \(= 256.8^2/16 = 4121.64\)
Total SS \(= 4471.92 - \text{CF} = 350.28\)M1 attempt to form three sums of squares
Between treatments SS \(= \frac{56.5^2}{4} + \frac{57.4^2}{3} + \frac{60.6^2}{5} + \frac{82.3^2}{4} - \text{CF}\)M1 correct method for any two
\(= 4324.1103 - \text{CF} = 202.47\)A1 if each calculated SS is correct
Residual SS (by subtraction) \(= 350.28 - 202.47 = 147.81\)
SourceSS df
Between treatments202.47 3 [B1]
Residual147.81 12 [B1]
Total350.28 15
Answer/WorkingMarks Guidance
Refer MS ratio to \(F_{3,12}\); upper 5% point is 3.49M1, A1 No FT if wrong
SignificantE1
Seems the effects of the treatments are not all the sameE1
[12] 
# Question 4 (4769 June 2011):

## Part (a)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Description of situation where randomised blocks would be suitable: one extraneous factor (e.g. stream down one side of a field) | E2 | Each E2 available as E2, E1, E0 |
| Explanation of why RB is suitable (design allows the extraneous factor to be "taken out" separately) | E2 | |
| Explanation of why LS is not appropriate (only one extraneous factor; LS unnecessarily complicated; not enough degrees of freedom for sensible estimate of experimental error) | E2 | |

## Part (a)(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Description of situation where Latin square would be suitable: two extraneous factors, all with same number of levels (e.g. streams down two sides of a field) | E2 | |
| Explanation of why LS is suitable (design allows extraneous factors to be "taken out" separately) | E2 | |
| Explanation of why RB is not appropriate (RB cannot cope with two extraneous factors) | E2 | |
| **[12]** | | |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Totals: 56.5, 57.4, 60.6, 82.3 from samples of sizes 4, 3, 5, 4 | | |
| Grand total 256.8; CF $= 256.8^2/16 = 4121.64$ | | |
| Total SS $= 4471.92 - \text{CF} = 350.28$ | M1 | attempt to form three sums of squares |
| Between treatments SS $= \frac{56.5^2}{4} + \frac{57.4^2}{3} + \frac{60.6^2}{5} + \frac{82.3^2}{4} - \text{CF}$ | M1 | correct method for any two |
| $= 4324.1103 - \text{CF} = 202.47$ | A1 | if each calculated SS is correct |
| Residual SS (by subtraction) $= 350.28 - 202.47 = 147.81$ | | |

| Source | SS | df | MS | MS ratio |
|---|---|---|---|---|
| Between treatments | 202.47 | 3 [B1] | 67.49 [M1] | 5.47(92) [A1 cao] |
| Residual | 147.81 | 12 [B1] | 12.3175 [M1] | |
| Total | 350.28 | 15 | | |

| Answer/Working | Marks | Guidance |
|---|---|---|
| Refer MS ratio to $F_{3,12}$; upper 5% point is 3.49 | M1, A1 | No FT if wrong |
| Significant | E1 | |
| Seems the effects of the treatments are not all the same | E1 | |
| **[12]** | | |
4
\begin{enumerate}[label=(\alph*)]
\item Provide an example of an experimental situation where there is one factor of primary interest and where a suitable experimental design would be
\begin{enumerate}[label=(\roman*)]
\item randomised blocks,
\item a Latin square.

In each case, explain carefully why the design is suitable and why the other design would not be appropriate.
\end{enumerate}\item An industrial experiment to compare four treatments for increasing the tensile strength of steel is carried out according to a completely randomised design. For various reasons, it is not possible to use the same number of replicates for each treatment. The increases, in a suitable unit of tensile strength, are as follows.

\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
\begin{tabular}{ c }
Treatment \\
A \\
\end{tabular} & \begin{tabular}{ c }
Treatment \\
B \\
\end{tabular} & \begin{tabular}{ c }
Treatment \\
C \\
\end{tabular} & \begin{tabular}{ c }
Treatment \\
D \\
\end{tabular} \\
\hline
10.1 & 21.1 & 9.2 & 22.6 \\
21.2 & 20.3 & 8.8 & 17.4 \\
11.6 & 16.0 & 15.2 & 23.1 \\
13.6 &  & 15.0 & 19.2 \\
\hline
\end{tabular}
\end{center}

[The sum of these data items is 256.8 and the sum of their squares is 4471.92 .]

Construct the usual one-way analysis of variance table. Carry out the appropriate test, using a $5 \%$ significance level.

RECOGNISING ACHIEVEMENT
\end{enumerate}

\hfill \mbox{\textit{OCR MEI S4 2011 Q4 [24]}}
This paper (4 questions)
View full paper
Q1 24 Q2 24 Q3 24 Q4 24