| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of Pearson’s product-moment correlation coefficient |
| Type | One-tailed test for positive correlation |
| Difficulty | Standard +0.3 This is a straightforward application of the PMCC formula with given summary statistics, followed by a standard one-tailed hypothesis test using tables. Part (c) requires recalling that scatter diagrams check for linearity assumptions. All steps are routine for Further Statistics students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | DR |
| Answer | Marks |
|---|---|
| = 0.6306 | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | Correct calculation seen. |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | H : = 0, H : > 0 |
| Answer | Marks |
|---|---|
| results | B1 |
| Answer | Marks |
|---|---|
| [5] | 3.3 |
| Answer | Marks |
|---|---|
| 2.2b | For both hypotheses. Allow hypotheses in words provided |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | The test may not be valid since the population might not |
| be bivariate Normal. | B1 | |
| [1] | 3.5b | B0 for ‘the data might not be bivariate Normal’ |
Question 5:
5 | (a) | DR
S = 2 1 6 3 5 − 12 5 6 5 7 2 4 [= 1182]
x y 0
S =17103− 1 5652 [= 1141.75]
xx 20
S =29286− 1 7242 [= 3077.2]
yy 20
S xy 1182
r= =
S S 1141.753077.2
xx yy
= 0.6306 | B1
B1
M1
A1
[4] | 1.1a
1.1
3.3
1.1 | Correct calculation seen.
S
For either S or y y with correct calculation seen
xx
For general form including sq. root
Allow 0.631 www
5 | (b) | H : = 0, H : > 0
0 1
where is the population pmcc between x and y
For n = 20, the 5% 1-tailed critical value is 0.3783
Since 0.6306 > 0.3783 so there is sufficient evidence to
reject H
0
There is sufficient evidence at the 5% level to suggest that
there is positive correlation between practice exam
results | B1
B1
B1
M1
A1
[5] | 3.3
2.5
3.4
1.1
2.2b | For both hypotheses. Allow hypotheses in words provided
these refer to the population correlation coefficient.
For defining in context. Must include population.
For correct critical value
For suitable comparison and consistent conclusion
regarding rejection of H . FT their r (0 < r < 1) and their
0
critical value.
For non-assertive conclusion in context that refers to H . Do
1
not allow ‘relationship’ for ‘correlation’. No FT incorrect
cv.
5 | (c) | The test may not be valid since the population might not
be bivariate Normal. | B1
[1] | 3.5b | B0 for ‘the data might not be bivariate Normal’5 Two practice GCSE examinations in mathematics are given to all of the students in a large year group. A teacher wants to check whether there is a positive relationship between the marks obtained by the students in the two examinations. She selects a random sample of 20 students. Summary data for the marks obtained in the first and second practice examinations, $x$ and $y$ respectively, are as follows.
$$\sum x = 565 \quad \sum y = 724 \quad \sum x ^ { 2 } = 17103 \quad \sum y ^ { 2 } = 29286 \quad \sum x y = 21635$$
The teacher decides to carry out a hypothesis test based on Pearson's product moment correlation coefficient.
\begin{enumerate}[label=(\alph*)]
\item In this question you must show detailed reasoning.
Calculate the value of Pearson's product moment correlation coefficient.
\item Carry out the test at the $5 \%$ significance level.
\item Given that the teacher did not draw a scatter diagram before carrying out the test, comment on the validity of the test.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2023 Q5 [10]}}