| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 2×2 contingency table |
| Difficulty | Standard +0.3 This is a straightforward chi-squared test of independence with a 2×2 contingency table, requiring standard calculation of expected frequencies, test statistic, and comparison to critical value. Parts (b) and (c) test basic understanding of independence and conditional probability, but require minimal calculation or conceptual depth beyond the routine application of the test. |
| Spec | 5.06a Chi-squared: contingency tables |
| \(L\) | \(L ^ { \prime }\) | |
| \(A\) | 14 | 9 |
| \(B\) | 26 | 51 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | H : no association between city and description |
| Answer | Marks |
|---|---|
| between city and description of handedness | B1 |
| Answer | Marks |
|---|---|
| [7] | 2.5 |
| Answer | Marks |
|---|---|
| 2.2b | Or equivalent. Ignore |
| Answer | Marks |
|---|---|
| too definite | Allow H : city and handed- |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | P(L | A) = P(L) |
| [1] | 3.3 | Or clearly equivalent statement |
| (c) | P(L A) = 14/100 | |
| P(A) = 23/100 so P(L | A) = 14/23 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 3.5c | 0.14 any prob or any prob 0.23 |
| Answer | Marks |
|---|---|
| 23 | E.g. 14/40 is M1A0 |
Question 6:
6 | (a) | H : no association between city and description
0
of handedness; H : some association …
1
Expected frequencies 9.2, 13.8, 30.8, 46.2
X 2 = 2.0097…+1.3398…+0.6003…+0.4002…
= 4.3501…
> 2.706
Reject H . Significant evidence of association
0
between city and description of handedness | B1
M1
M1
A1
A1
M1ft
A1ft
[7] | 2.5
1.1
3.3
3.4
1.1
1.1
2.2b | Or equivalent. Ignore
Evidence for correct method
Evidence for correct method
X 2, 4.35 or better
Compare their X2 with 2.706
Correct first conclusion, needs
correct method; contextualised, not
too definite | Allow H : city and handed-
0
ness are independent, etc
e.g. at least 2 correct values
e.g. at least 2 correct values
in same row or column
Allow 2.71
ft on wrong test statistic, not
on wrong critical value
SC No or wrong Yates (5.42): can get B1M1 M0A0; 5.42 > 2.71 SCB1; M1A1, total 5/7
(b) | P(L | A) = P(L) | B1
[1] | 3.3 | Or clearly equivalent statement
(c) | P(L A) = 14/100
P(A) = 23/100 so P(L | A) = 14/23 | M1
A1
[2] | 3.4
3.5c | 0.14 any prob or any prob 0.23
14 or awrt 0.609
23 | E.g. 14/40 is M1A06 Yusha investigates the proportion of left-handed people living in two cities, $A$ and $B$. He obtains data from random samples from the two cities. His results are shown in the table, in which $L$ denotes "left-handed".
\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
& $L$ & $L ^ { \prime }$ \\
\hline
$A$ & 14 & 9 \\
\hline
$B$ & 26 & 51 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Test at the 10\% significance level whether there is association between being left-handed and living in a particular city.
A person is chosen at random from one of the cities $A$ and $B$.\\
Let $A$ denote "the person lives in city $A$ ".
\item State the relationship between $\mathrm { P } ( L )$ and $P ( L \mid A )$ according to the model implied by the null hypothesis of your test.
\item Use the data in the table to suggest a value for $P ( L \mid A )$ given by an improved model.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2019 Q6 [10]}}