Question 6:
6 | (a) | Because the scatter diagram does not appear to be
elliptical (due to the possible outlier)
so the (underlying) distribution is (probably) not bivariate
Normal. | E1
E1
[2] | 3.5a
2.4 | For not elliptical.
Alternatively, ‘the points appear to be funnel-shaped’
For full answer (dependent on first mark)
“data is not bivariate Normal” is E0
“Normal bivariate” is E0
6 | (b) | Rank saw 7 6 3 2 1 8 5 4
Rank chop 7 4 6 1 5 8 3 2
𝑑2 0 4 9 1 16 0 4 4
23
Spearman’s rank coefficient or 0.5476
42 | M1
depM1
A1
[3] | 1.1
1.1
1.1 | For ranking saw or chop correctly
For ranking saw and chop consistently (i.e. same way
round) with at most one adjacent pair of ranks transposed.
If rankings not seen they may be inferred from correct
𝑑 or 𝑑2 values or ∑𝑑2 = 38 seen
Accept 0.55 or better.
SC B1 only if 0.5476 stated with no working.
6 | (c) | H : There is no association between sawing and chopping
0
(times) in the population
H : There is positive association between sawing and
1
chopping (times) in the population
Critical value 0.6429
0.5476 < 0.6429 (so do not reject H /accept H )
0 0
There is insufficient evidence to suggest that there is
positive association between sawing and chopping
times (in the population) | B1 B1
B1
M1
A1FT
[5] | 3.3
1.2
3.4
1.1
2.2b | For first B1 need to see one correct hypothesis with context
(need not have population at this point).
For B1B1 need to see two correct hypotheses with each of
context and population in at least one of the hypotheses.
n = 8, 5%, 1-tailed
For comparison provided 0 < 𝑟 < 1
𝑠
FT their r and sensible critical value from correct table
s
A0 for “(Sufficient evidence to suggest) no association”.
Needs to include context.
Rank saw | 7 | 6 | 3 | 2 | 1 | 8 | 5 | 4
Rank chop | 7 | 4 | 6 | 1 | 5 | 8 | 3 | 2
𝑑2 | 0 | 4 | 9 | 1 | 16 | 0 | 4 | 4