Moderate -0.3 This is a standard AS-level binomial hypothesis test with clearly stated hypotheses (H₀: p=0.7, H₁: p>0.7), straightforward calculation of P(X≥18) under the null hypothesis, and comparison with 5% significance level. While it requires multiple steps (setup, calculation, conclusion), it follows a well-rehearsed procedure with no conceptual surprises, making it slightly easier than average for AS Paper 2.
Martin grows cucumbers from seed.
In the past, he has found that 70% of all seeds successfully germinate and grow into cucumber plants.
He decides to try out a new brand of seed.
The producer of this brand claims that these seeds are more likely to successfully germinate than other brands of seeds.
Martin sows 20 of this new brand of seed and 18 successfully germinate.
Carry out a hypothesis test at the 5% level of significance to investigate the producer's claim.
[7 marks]
Question 19:
19 | States both hypotheses correctly
for one-tailed test accept
‘population proportion’ = 0.7 or
70%, but NOT x = or 𝑥 = or µ = | AO2.5 | B1 | X = number of seeds which
germinate
H :p=0.7
0
H :p>0.7
1
Under null hypothesis XB(20,0.7)
𝑃(𝑋 ≥ 18)= 1−𝑃(𝑋 ≤ 17)
= 1−0.965
= 0.035
0.035 < 0.05
Reject H
0
There is sufficient evidence to
conclude that the new seeds are
more likely to germinate.
States model used PI can also be
implied by 0.0278 or 0.00684 | AO3.3 | M1
Calculates
𝑃(𝑋 ≥ 18) 𝑜𝑟 𝑃(𝑋 ≥ 19), (PI by
0.035(5) or 0.0076) but not
P(X = 18) or P(X = 19) | AO1.1a | M1
Obtains correct probability for
𝑃(𝑋 ≥ 18) | AO1.1b | A1
Evaluates binomial model by
comparing 0.035(5) or 0.0076 with
0.05 but not P(X = 18) or P(X =19) | AO3.5a | M1
Infers H rejected (FT 0.0076),
0
condone ‘accept H ’
1 | AO2.2b | A1F
Concludes correctly in context.
‘Sufficient evidence’ or equivalent
required. Only award for full
complete correct solution. | AO3.2a | R1
Total | 7
Q | Alternative Marking Instructions | AO | Marks | Typical Solution
19 | States both hypotheses correctly
for one-tailed test accept
‘population proportion’ = 0.7 or
70%, but NOT x = or 𝑥 = or µ = | AO2.5 | B1 | X = number of seeds which
germinate
H :p=0.7
0
H :p>0.7
1
Under null hypothesis XB(20,0.7)
𝑃(𝑋 ≥ 17)= 0.1071 > 0.05 and
𝑃(𝑋 ≥ 18)= 0.0355 < 0.05
Hence 𝑋 ≥ 18 is critical region
X = 18 is in critical region
Reject H
0
There is sufficient evidence to
conclude that the new seeds are
more likely to germinate.
States model used PI can also be
implied by 0.0716 or 0.0278 | AO3.3 | M1
Finds P(X ≥ 17) and P(X ≥ 18) but
not P(X=17) or P(X=18) accept
0.035 | AO1.1a | M1
Identifies correct critical region | AO1.1b | A1
Evaluates Binomial model by
comparing 𝑋 = 18 with critical
region (condone CR of X ≥ 17) | AO3.5a | M1
Infers H rejected, condone ‘accept
0
H ’ FT CR of X ≥ 17
1 | AO2.2b | A1F
Concludes correctly in context.
‘Sufficient evidence’ or equivalent
required. Only award for full
complete correct solution. | AO3.2a | R1
Total | 7
TOTAL | 80
Martin grows cucumbers from seed.
In the past, he has found that 70% of all seeds successfully germinate and grow into cucumber plants.
He decides to try out a new brand of seed.
The producer of this brand claims that these seeds are more likely to successfully germinate than other brands of seeds.
Martin sows 20 of this new brand of seed and 18 successfully germinate.
Carry out a hypothesis test at the 5% level of significance to investigate the producer's claim.
[7 marks]
\hfill \mbox{\textit{AQA AS Paper 2 2018 Q19 [7]}}