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HG2MPS Solutions to Worksheet 1

A1 (i) X =1

n

Xi =

63.7

10= 6.37, Y =

1

n

Yi =

604.7

10= 60.47.

(ii) S2X =1

n 1

X2i (

Xi)2

n

=

1

9

411.85 (63.7)

2

10

= 0.6757,

S2Y =1

n 1

Y2i (

Yi)2

n

=

1

9

36614 (6047)

2

10

= 5.310.

(iii) SX =

0.6757 = 0.8220SY =

5.310 = 2.304

(iv) The heavy is more variable than the light filament.

A2 (a) Ordered data is

98 98 99 99 99 99 100 100 100 100100 100 101 101 101 101 102 102 102 103

(i) Sample median = M = average of 10th and 11th obs = 12{100 + 100} = 100.

(ii) Lower quartile = QL =n + 1

4obs = 51

4obs

= 14

of way from 5th to 6th obs

= 99 (as 5th obs = 99 and 6th obs = 99).

(iii) Upper quartile = QU =3(n + 1)

4obs =

63

4obs = 153

4obs

= 34

way from 15th to 16th obs= 101 (as 15th obs is 101 and 16th obs is 101).

(iv) Interquartile range = QU QL = 101 99 = 2.(v) QU M = 101 100 = 1 and M QL = 100 99 = 1.

This suggests the underlying population is symmetric.

(b)

cumulative relative cumulativeClass Interval mid-point frequency frequency frequency97.5 98.5 98 2 2 0.198.5 99.5 99 4 6 0.3

99.5 100.5 100 6 12 0.6100.5 101.5 101 4 16 0.8101.5 102.5 102 3 19 0.95102.5 103.5 103 1 20 1.0

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(c)

xj fj fjxj fjx2

j

98 2 196 1920899 4 396 39204

100 6 600 60000101 4 404 40804102 3 304 31212103 1 103 10609

fj = 20

fjxj = 2005

fjx2

j = 201037

x =1

n

k

j=1

fjxj =1

20{2005

}= 100.25 ohms

s2 =1

n 1

k

j=1

fjx2

j nx2

=1

19{201037 20(100.25)2}

= 1.8816

Hence sample mean = 100.25 ohms and sample variance = 1.8816.

A3 (a) Let zj = axj + b, for j = 1, 2, . . . , n. Then

z =1

n

nj=1

zj =1

n

nj=1

(axj + b) = a

1

n

nj=1

xj

+

nb

n= ax + b

and

s2Z =1

n

1

n

j=1

(zj z)2 = 1n

1

n

j=1 axj+ b (ax+ b)

2

=1

n 1n

j=1

a2(xj x)2

= a2

1

n 1n

j=1

(xj x)2

= a2s2x.

(Note s2Z does not depend on b.)

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(b) Transform the data by subtracting 100 from the xis and working with the codeddata zj = xj

100 i.e. a = 1 and b =

100.

zj fj fjzj fjz2

j

2 2 4 81 4 4 40 6 0 01 4 4 42 3 6 123 1 3 9

fjzj = 5

fjz2 = 37

So z =1

n

fjzj =5

20 = 0.25 and

s2Z =1

n{

fjz2

j nz2} =1

19{37 20(0.25)2} = 1.8816.

Now return to uncoded values: x = z+ 100 = 100.25, s2X = a2s2Z = 1.8816.

A4 (a) X =1

n

Xi =

1

15 109800 = 7320

Y =1

n Yi =1

15 107122500 = 9890

(b) S2X =1

15 1

849865000 (109800)215

= (1815.2)2

S2Y =1

15 1

1470077500 (148350)2

15

= (454.8)2

(c) SXY =1

15 1

1085657500 (109800 148350)15

= 18893

(d) Sample correlation coefficient

r =SXY

SXSY

=18893

1815.2 454.8=

0.023

(e) Sample correlation coefficient is close to zero. Hence fitting a straight line istotally unreasonable as can be seen from a plot of the data.

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A5 (a) X =1

n Xi =

1

6 16.0 = 2.6667

Y = 1n

Yi = 1

6 0.231 = 0.0385

(b) S2X =1

6 1

48.5 (16.0)2

6

= 1.1667 = (1.080)2

S2Y =1

6 1

0.0097050 (0.231)2

6= (0.0127)3

(c) SXY =1

6 1

0.68450 (16.0 0.231)6

= 0.0137

(d) r =SXY

SXSY=

0.0137

1.080 0.0127= 0.9987 indicating a very good fit.

(e) Least squares line of Y on X

y y = SXYS2X

(x x)

y 0.0385 = 0.0137(1.080)2

(x 2.6667)

which becomes

y = 0.0385

0.0117

2.6667 + 0.0117x

y = 0.0073 + 0.0117x

(f) At x = 2.5, y = 0.0073, 0.0117 2.5 = 0.0366 reliable as inside the range of thedata.

At x = 5.0, y = 0.0073 + 0.0117 5 = 0.0658 unreliable as outside of the rangeof the data.

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A6 (a) Model is R = aebt and taking natural logs we get ln R = ln a bt which may bewritten in the form

y = cx + d

where y = ln R, d = ln a, x = t, c = b.(b) y y = SXY

S2X(x x)

Substituting sample values we get

y 0.391 = 0.335(3.742)2

(x 7).

Thus

y = 0.024x + 0.56 = c = 0.024, d = 0.56.

(c) Here a = ed = e0.56 = 1.75, b = c = 0.024.(d) Model is R = 1.75e0.024t.

When t = 24, R = 1.75e0.02424 = 0.98.

(e) IfR = 1 what is the value of t?

Solve

1 = 1.75e0.024t = t = ln 1.750.024

= 23.3 hours.

(f) r =SXY

SXSY=

0.3353.742 0.103 = 0.869.

High negative value but not below 0.9. Reasonable fit but not a good fit. Firstdata point is perhaps a little suspect.

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