| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (upper tail, H₁: p > p₀) |
| Difficulty | Moderate -0.3 This is a straightforward hypothesis test for a proportion with standard setup (H₀: p=0.35, H₁: p>0.35) and clear parameters given. Part (a) requires recognizing that a single sample doesn't prove causation/evolution. Part (b) is routine: calculate test statistic using normal approximation, compare to critical value, and conclude. The question is slightly easier than average because it's a textbook one-tailed test with no complications, though it does require careful execution across multiple steps for full marks. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| 13 | (a) | it can’t be fully justified because |
| Answer | Marks | Guidance |
|---|---|---|
| population oe | B1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 13 | (b) | H : p = 0.35 |
| Answer | Marks |
|---|---|
| born without wisdom teeth is more than 0.35 | B1 |
| Answer | Marks |
|---|---|
| A1 | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | allow equivalent in words; |
Question 13:
13 | (a) | it can’t be fully justified because
eg different samples may lead to different
conclusions oe
eg the proportion could be 0.35 and 61/140
may have arisen by chance oe
eg the sample may not be representative oe
eg the researcher used a sample not a
population oe | B1 | 2.4 | do not allow
eg the sample is too small
eg the sample is too small to be representative
[1]
13 | (b) | H : p = 0.35
0
H : p > 0.35
1
p is the probability that a baby (selected at
random) is born without wisdom teeth
P(X ≥ k) found using B(140, 0.35), where
k = 60, 61 or 62
NB P(X ≥ 60) = 0.03272 – 0.033 or
P(X ≥ 62) = 0.01438 – 0.015
NB 0.967…, 0.978… and 0.985…imply M1
P(X ≥ 61) = 0.02197 – 0.022
their 0.022 correctly compared with 0.05
do not accept H or reject H or accept H or
0 0 1
significant
sufficient evidence at the 5% level to
suggest that the probability that a baby is
born without wisdom teeth is more than 0.35 | B1
B1
M1*
A1
M1dep*
A1FT
A1 | 1.1
2.5
3.3
1.1
3.4
1.1
1.1 | allow equivalent in words;
do not allow percentages
or p is the proportion of babies that are born without wisdom
teeth
B1B1 if other symbol instead of p used if correctly defined
or critical region is X ≥ k found from calculation of
probability; allow k = 58, 59 or 60
or critical region is X ≥ 59 from 0.0475…. or 0.048
or 61 correctly compared with their 59;
allow their 0.978 correctly compared with 0.95
A0 if their 0.022 > 0.05 or 61 < their 59
dependent on award of all other marks apart from second B1
do not allow eg conclude / prove / indicate or other assertive
statement instead of suggest
[7]Records from the 1950s showed that 35\% of human babies were born without wisdom teeth. It is believed that as part of the evolutionary process more babies are now born without wisdom teeth. In a random sample of 140 babies, collected in 2020, a researcher found that 61 were born without wisdom teeth.
The researcher made the following statement.
``This shows that the percentage of babies born without wisdom teeth has increased from 35\%.''
\begin{enumerate}[label=(\alph*)]
\item Explain whether this statement can be fully justified. [1]
\item Conduct a hypothesis test at the 5\% level to determine whether there is any evidence to suggest that more than 35\% of babies are now born without wisdom teeth. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2022 Q13 [8]}}