OCR MEI Further Statistics A AS 2022 June — Question 3 10 marks

Exam BoardOCR MEI
ModuleFurther Statistics A AS (Further Statistics A AS)
Year2022
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of Spearman’s rank correlation coefficien
TypeHypothesis test for association
DifficultyStandard +0.3 This is a straightforward application of Spearman's rank correlation test with clear data provided. Students must identify non-linearity from a scatter diagram (routine interpretation), calculate ranks and apply the standard formula, then perform a hypothesis test using critical value tables. All steps are standard textbook procedures with no novel problem-solving required, though it does require careful calculation with 10 data points.
Spec5.08e Spearman rank correlation5.08f Hypothesis test: Spearman rank

3 A biology student is doing an experiment in which plants are inoculated with a particular microorganism in an attempt to help them grow. She is investigating whether there is any association between the percentage of roots which have been colonised by the microorganism and the dry weight of the plant shoots. After the plants have grown for a few weeks, the student takes a random sample of 10 plants and measures the percentage of roots which have been colonised by the microorganism and the dry weight of the plant shoots. The spreadsheet output shows the data, together with a scatter diagram to illustrate the data. \includegraphics[max width=\textwidth, alt={}, center]{8f1e0c68-a334-4657-823e-386ab0994c02-3_722_1648_635_244}
  1. The student decides that a test based on Pearson's product moment correlation coefficient may not be valid. Explain why she comes to this conclusion.
  2. Calculate the value of Spearman's rank correlation coefficient.
  3. Carry out a test based on this coefficient, at the \(5 \%\) significance level, to investigate whether there is any association between percentage colonisation and shoot dry weight.

Question 3:
AnswerMarks Guidance
3(a) Because the scatter diagram does not appear to be
elliptical (but instead a curve) so the distribution is
AnswerMarks
probably not bivariate Normal.E1*
E1dep*
AnswerMarks
[2]3.5b
3.5bFor not elliptical
For full answer (dependent on first mark)
If no E marks awarded SC B1 for the association might not
be linear.
AnswerMarks Guidance
3(b) Rank PC 1 3 2 4 6 5 8 10 7 9
Rank DW 1 2 3 4 5 6 7 8 9 10
( =151)
Spearman’s rank coefficient = 0.9152
AnswerMarks Guidance
165Rank PC 1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1For ranking DW
For ranking PC. Allow both ranks reversed
BC Allow 0.915, 0.92
AnswerMarks Guidance
Rank DW1 2
3(c) H : There is no association between percentage
0
colonisation and shoot dry weight in the population
H : There is some association between percentage
1
colonisation and shoot dry weight in the population
(For n = 10, 2-tailed 5% critical value is) 0.6485
0.9152 > 0.6485
(Reject H ) There is evidence to suggest that there is
0
association between percentage colonisation and shoot
AnswerMarks
dry weight in the population.B1
B1
B1
M1
A1
AnswerMarks
[5]2.5
1.2
3.4
1.1
AnswerMarks
2.2bFor hypotheses in context
For population seen in either of the hypotheses
Allow 0.649
For correct comparison of their r with their critical value
s
AnswerMarks Guidance
providedr < 1
s
For correct non-assertive conclusion in context.
FT their r but A0 if wrong critical value used.
s
Question 3:
3 | (a) | Because the scatter diagram does not appear to be
elliptical (but instead a curve) so the distribution is
probably not bivariate Normal. | E1*
E1dep*
[2] | 3.5b
3.5b | For not elliptical
For full answer (dependent on first mark)
If no E marks awarded SC B1 for the association might not
be linear.
3 | (b) | Rank PC 1 3 2 4 6 5 8 10 7 9
Rank DW 1 2 3 4 5 6 7 8 9 10
( =151)
Spearman’s rank coefficient = 0.9152
165 | Rank PC | 1 | 3 | 2 | 4 | 6 | 5 | 8 | 10 | 7 | 9 | M1
M1
A1
[3] | 1.1
1.1
1.1 | For ranking DW
For ranking PC. Allow both ranks reversed
BC Allow 0.915, 0.92
Rank DW | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
3 | (c) | H : There is no association between percentage
0
colonisation and shoot dry weight in the population
H : There is some association between percentage
1
colonisation and shoot dry weight in the population
(For n = 10, 2-tailed 5% critical value is) 0.6485
0.9152 > 0.6485
(Reject H ) There is evidence to suggest that there is
0
association between percentage colonisation and shoot
dry weight in the population. | B1
B1
B1
M1
A1
[5] | 2.5
1.2
3.4
1.1
2.2b | For hypotheses in context
For population seen in either of the hypotheses
Allow 0.649
For correct comparison of their r with their critical value
s
provided |r| < 1
s
For correct non-assertive conclusion in context.
FT their r but A0 if wrong critical value used.
s
3 A biology student is doing an experiment in which plants are inoculated with a particular microorganism in an attempt to help them grow. She is investigating whether there is any association between the percentage of roots which have been colonised by the microorganism and the dry weight of the plant shoots. After the plants have grown for a few weeks, the student takes a random sample of 10 plants and measures the percentage of roots which have been colonised by the microorganism and the dry weight of the plant shoots.

The spreadsheet output shows the data, together with a scatter diagram to illustrate the data.\\
\includegraphics[max width=\textwidth, alt={}, center]{8f1e0c68-a334-4657-823e-386ab0994c02-3_722_1648_635_244}
\begin{enumerate}[label=(\alph*)]
\item The student decides that a test based on Pearson's product moment correlation coefficient may not be valid.

Explain why she comes to this conclusion.
\item Calculate the value of Spearman's rank correlation coefficient.
\item Carry out a test based on this coefficient, at the $5 \%$ significance level, to investigate whether there is any association between percentage colonisation and shoot dry weight.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2022 Q3 [10]}}