June 2024 Paper 2 Q12
12 A survey conducted in 2021 showed that 10% of British adults were vegetarians.
A dietitian believes that the proportion of British adults who are vegetarians may have changed, so decides to conduct a hypothesis test at the 5% level of significance.
In a random sample of 112 adults, the dietitian finds that there are 19 vegetarians.
Carry out the hypothesis test to determine whether there is any evidence to support the dietitian’s belief. [7]
| Scheme | Marks | AO |
|---|---|---|
| H\(_0\): \(p = 0.1\) H\(_1\): \(p \neq 0.1\) | B1 | 1.1 |
| \(p\) is the probability that a (British) adult (selected at random) is a vegetarian | B1 | 2.5 |
| \(\mathrm{P}(X \geqslant k)\) found using B(112, 0.1), where \(k\) = 18, 19 or 20 | M1* | 3.3 |
| \([\mathrm{P}(X \geqslant 19) =]\ 0.015 - 0.015331\) | A1 | 1.1 |
| their 0.015 correctly compared with 0.025 or their 0.985 correctly compared with 0.975 | M1dep* | 3.4 |
| do not accept H\(_0\) or reject H\(_0\) or accept H\(_1\) or significant | A1FT | 1.1 |
| sufficient evidence at the 5% level to suggest that the probability that an adult is vegetarian is not 0.10 oe | A1 | 3.5a |
| [7] |
Notes
B1: allow equivalent in words; do not allow percentages
allow other variable only if correctly defined
B1: or \(p\) is the proportion of adults that are vegetarian
B1B1 if other symbol instead of \(p\) used if correctly defined
M1*: may be implied by
\((\mathrm{P}(X \geqslant 18)) = 0.0295 - 0.030\) or
\((\mathrm{P}(X \geqslant 19)) = 0.015 - 0.015331\) or
\((\mathrm{P}(X \geqslant 20)) = 0.0075 - 0.00754\)
NB M0 for \(\mathrm{P}(X = 19) = 0.00779\)
NB \(\mathrm{P}(X \leqslant 17) = 0.97049\ldots\), \(\mathrm{P}(X \leqslant 18) = 0.984669\ldots\) and \(\mathrm{P}(X \leqslant 19) = 0.99246\ldots\) imply M1
A1: or \([\mathrm{P}(X \leqslant 18) =]\ 0.984669 - 0.985\) or 0.98
A1FT: A0 if their \(0.015 \gt 0.025\) or their \(0.985 \lt 0.975\)
A1: 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
Alternatively, using critical region
| Scheme | Marks | AO |
|---|---|---|
| H\(_0\): \(p = 0.1\) H\(_1\): \(p \neq 0.1\) | B1 | |
| \(p\) is the probability that a (British) adult (selected at random) is a vegetarian | B1 | |
| critical region is \(X \leqslant k \cup X \geqslant l\ X \geqslant k\) found from calculation of probability; allow \(k\) = 4 or 5, \(l\) = 18, 19 or 20 | M1* | |
| [critical region is \(X]\ \geqslant 19\ \cup [X]\ \leqslant 4\) | A1 | |
| 19 correctly compared with their critical value | M1dep* | 3.4 |
| do not accept H\(_0\) or reject H\(_0\) or accept H\(_1\) or significant | A1FT | 1.1 |
| sufficient evidence at the 5% level to suggest that the probability that an adult is vegetarian is not 0.10 oe | A1 | 3.5a |
B1: allow equivalent in words; do not allow percentages
allow other variable only if correctly defined
B1: or \(p\) is the proportion of adults that are vegetarian
B1B1 if other symbol instead of \(p\) used if correctly defined
M1*: allow calculation of upper tail only for M1
A1: from \(\mathrm{P}(X \geqslant 19) = 0.015 - 0.015331\) and \(\mathrm{P}(X \leqslant 4) = 0.010\)
must see both tails for A1
A1FT: A0 if \(19 \lt\) their critical value
A1: 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
Alternatively, using Normal approximation
| Scheme | Marks | AO |
|---|---|---|
| H\(_0\): \(p = 0.1\) H\(_1\): \(p \neq 0.1\) | B1 | |
| \(p\) is the probability that a (British) adult (selected at random) is a vegetarian | B1 | |
| \(\mathrm{P}(X \geqslant 18.5)\) or \(\mathrm{P}(X \geqslant 19.5)\) found using N(11.2, 10.08) or \(\mathrm{P}(X \geqslant 19.5) = 0.00447(1)\) | M1* | |
| \([\mathrm{P}(X \geqslant 18.5) =]\ 0.0107 - 0.011\) | A1 | |
| their 0.0107 correctly compared with 0.025 or their 0.989 correctly compared with 0.975 | M1dep* | |
| do not accept H\(_0\) or reject H\(_0\) or accept H\(_1\) or significant | A1FT | |
| sufficient evidence at the 5% level to suggest that the probability that an adult is vegetarian is not 0.10 oe | A1 |
B1: allow equivalent in words; do not allow percentages
allow other variable only if correctly defined
B1: or \(p\) is the proportion of adults that are vegetarian
B1B1 if other symbol instead of \(p\) used if correctly defined
M1*: NB M0 for \(\mathrm{P}(X = 19) = 0.006145\) (from normPdf\((19, 11.2, \sqrt{10.08})\)) or \(\mathrm{P}(X = 19) = 0.00627\) (from using continuity correction, may see normCdf\((18.5, 19.5, 11.2, \sqrt{10.08})\))
NB \(\mathrm{P}(X \leqslant 19.5) = 0.99553\ldots\) and \(\mathrm{P}(X \leqslant 18.5) = 0.989255\ldots\) imply M1
A1FT: A0 if their \(0.0107 \gt 0.025\) or their \(0.989 \lt 0.975\)
A1: 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