Page 4

Question Number

Scheme

3

Marks

⎯d = 1.41

Differences 2.1 -0.7 2.6 -1.7 3.3 1.6 1.7 1.2 1.6 2.4

M1 M1

H0 : µd = 0 H1 : µd > 0

B1

s=

t=

M1

40.65 − 10 ×1.412 = 1.5191… 9

1.41 = 2.935… ⎛ 1.519... ⎞ ⎜ ⎟ ⎝ 10 ⎠

awrt 2.94 /2.93

M1 A1

t9 (1%) = 2.821

B1

2.935.. > 2.821 Evidence to reject H0. There has been an increase in the mean weight of the mice.

B1ft (8)

2 sample test can score M0 M0 B1 for H0 : µA = µB H1 : µA < µB

M1

9 × 24.5 + 9 × 17.16 18

M0 A0 B1 2.552 B1 ft ie 4/8

S4 MS Jun 2008  

Edexcel GCE Mathematics Statistics 4 (6686) June 2008 Edexcel GCE Mark Scheme (Final) + + = 3 = 9 1 2 10 Var( θ 3 ) = 36 1 {9 σ 2 + 4 σ 2 +...

S4 MS Jun 2008  

Edexcel GCE Mathematics Statistics 4 (6686) June 2008 Edexcel GCE Mark Scheme (Final) + + = 3 = 9 1 2 10 Var( θ 3 ) = 36 1 {9 σ 2 + 4 σ 2 +...

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