Question 3:
3 | (a) | H :  = 10.75, H :   10.75, where  is the
0 1
population mean reading age in the district
Or: H : The mean reading age is 10.75, H : The
0 1
mean reading age is not 10.75
x = 1 1 .1 6 ( 2 5 )
ˆ 2  = 3.78(3386) or ˆ = 1.945 | B2
B1
B2 | 1.1
2.5
1.1
1.1
1.1 | Allow μ defined in terms of population or context.
SC: One error, e.g.  not defined (at all), or one-tailed, B1, but
x or 11.16(25) used in hypotheses: B0
Stated or implied, allow 11.2 or better
Allow B1 for 3.736 or 3.74… or 1.93(28…). Can be implied.
B1B0 can be implied by p = 0.0281 or z = 1.91.
Second B1 can be allowed if multiplier 80/79 seen anywhere.
(1 1 .1 6 2 5 − 1 0 .7 5 )
z =
3 .7 8 3 3 8 6 / 8 0
p = 0.0289 or p = 0.05785 (if 0.05 used)
or z = 1.897
z < 1.96 or p > 0.025 or 0.05 as appropriate | M1
A1
B1 | 3.3
3.4
1.1 | Evidence of standardisation using 80, e.g. p = 0.0281, allow
square root errors only if algebraic evidence seen, allow signs
reversed. Not N(11.1625, …) stated (but can get last M1A1)
Awrt 0.029. Or p = 0.03075 from t , or 0.0615 if compared with
79
0.05. (Biased estimate gives p = 0.0281, z = 1.91: M1A0
here.)
If continuity correction used, can get M1A0 and final M1A1.
0.025 or 0.05 or 1.96 used for like-with-like comparison
or: | 10.75 + z√(3.783/80) | M1 | Recognisable z and 80 used, ignore LH CV (= 10.323), allow √
errors, their s (warning: ˆ 1 .9 4 5  = is very close to 1.96)
z = 1.96 | B1 | Stated or implied, allow from wrong tail
11.16 < 11.176 | 11.16 < 11.176 | A1ft | A1ft | FT on their z, e.g. 11.105 from z = 1.645 (conclusion reverses)
SC: CV 11.16 – 1.96√(3.783/80) < 10.75, M0B1A0, M1A1
Do not reject H . There is insufficient evidence
0
that the mean reading age is not 10.75 in this
district | M1ft
A1ft
[10] | 1.1
2.2b | FT on their z, needs 80 used, needs like-with-like comparison,
allow from comparing ½p with 0.05
Contextualised, not over-assertive, needs double negative oe
(b) | The central limit theorem allows assumption that
the distribution of the sample mean is
(approximately) normal. | B1
[1] | 1.2 | Must refer to sample mean [not “it”, or just “sample”] or
X or x, or “distribution of means” oe.
Their hypotheses | Comments | Mark
H :  = 10.75, H :   10.75, where μ is the population mean reading age
0 1 | Correct | B2
H :  = 10.75, H:   10.75, where μ is the mean reading age
0 i | OK: μ defined in context | B2
H :  = 10.75, H:   10.75, where μ is the population mean
0 i | OK: μ defined using “population” | B2
H :  = 10.75, H :   10.75, where μ is the mean
0 1 | μ not defined in context or “population” | B1
H :  = 10.75, H :  > 10.75
0 1 | Two errors: one-tailed, μ not defined (at all) | B0
H : The mean reading age is 10.75, H : The mean reading age is not 10.75
0 1 | OK | B2
H : The average reading age is 10.75. H : The average reading age is not 10.75
0 1 | Not “mean” | B1
H : There is no change in the mean reading age, H : there is change
0 1 | As in MS | B1
H : There is no evidence that the mean reading age is not 10.75, etc
0 | “evidence” does not belong in hypotheses | B1
Their calculation | Comments | Mark | Can get last M1A1?
z = 1.90 or p = 0.0289 | Correct | Full marks | Yes
p = 0.0578(5) > 0.05 | Correct | Full marks if compared with 0.05 | Yes
z = 1.91 or p = 0.0281 | Biased σ used | (Preceding B1 not B2) then M1A0 | Yes
p = 0.0563 > 0.05 | Biased σ used | (Preceding B1 not B2) then M1A0 if
compared with 0.05 | Yes
z = 0.2134 or 0.212, or
p = 0.4155 or 0.416 | Divisor 80 omitted | M0A0 | No
Other values of z or p | M0A0 unless they show evidence of using σ/√80 or σ2/√80
or σ2/80 or σ/80, in which case they can get M1A0 | No unless evidence as in previous
column, in which case yes
CV 10.75 – 1.96×0.217
11.16 > 10.32 | Wrong tail of
10.75  z√(s2/80) | M0 B1 A0 | No
10.75 + 1.645×0.217
11.16 > 11.11 | 1.645 wrong | M1 B0 A1 (ft on 1.645) | Yes
(Conclusion reversed: Reject H , etc.)
0
11.1625 – 1.96×0.217
10.737 < 10.75 | Centred on 11.1625
and not 10.75 | M0B1A0 | Yes
Their comparison | Comments | Mark | Can get last M1A1?
1.91 < 1.96 | Correct comparison, wrong z | B1 | Yes
0.0579 > 0.05 | Correct | B1 | Yes
0.0281 > 0.025 | Correct comparison, wrong p | B1 | Yes
0.0289 < 0.05 | Wrong | B0 | Yes, conclusion reverses
0.9711 < 0.975 | Correct | B1 | Yes
0.0289 < 0.975 | Wrong tail | B0 | No
0.0579 < 1.96 | Not like-with-like | B0 | No
Their conclusion | Comments | Mark
Accept H . There is insufficient evidence that reading age is not 10.75
0 | Correct, allow “accept H ”.
0 | M1A1
There is insufficient evidence to reject H . The reading age is 10.75
0 | BOD | M1A1
Accept H . There is insufficient evidence that reading age has changed
0 | BOD | M1A1
There is insufficient evidence that the average is not 10.75 | No context. Condone omission of “do not reject H ”
0 | M1A0
Do not reject H . There is significant evidence that the mean age is 10.75
0 | Wrong | M1A0
Do not reject H . The mean reading age is 10.75 in this district
0 | Too assertive | M1A0
Question | Answer | Marks | AO | Guidance