Higher Engineering Mathematics

CHI-SQUARE AND DISTRIBUTION-FREE TESTS 621

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(ii) Level of significanceα 1 =1%.

(iii) Let the sizes of the samples benPandnQ,

wherenP=8 andnQ=10. The Mann-Whitney

test compares every item in samplePin turn

with every item in sampleQ, a record being

kept of the number of times, say, that the item

fromPis greater thanQ, or vice-versa. In this

case there arenPnQ, i.e. (8)(10)=80 compar-

isons to be made. All the data is arranged into

ascending order whilst retaining their separate

identities—an easy way is to arrange a linear

scale as shown in Fig. 63.1, on page 624.

From Fig. 63.1, a list ofP’s andQ’s can be

ranked giving:

P P QP PQPQPPPQQQ

QQQQ

(iv) Write under each letterPthe number ofQ’s

that precede it in the sequence, giving:

P P QP P QPQPPPQ

00 11 2 333

QQQQQQ

(v) Add together these 8 numbers, denoting the

sum byU, i.e.

U= 0 + 0 + 1 + 1 + 2 + 3 + 3 + 3 = 13

(vi) The critical regions are of the formU≤critical

region.

From Table 63.5, for a sample size 8 and 10 at

significance levelα 1 =1% the critical regions

isU≤ 13.

The value ofUin our case, from (v), is 13 which

is significant at 1% significance level.

The Mann-Whitney test has therefore confirmed
thatthere is evidence that the non-British cars
have better reliability than the British cars in the
first 10 000 miles, i.e. the alternative hypothesis
applies.

Problem 12. Two machines, A and B, are used

to measure vibration in a particular rubber prod-

uct. The data given below are the vibrational

forces, in kilograms, of random samples from

each machine:

A 9.7 10.2 11.2 12.4 14.1 22.3

29.6 31.7 33.0 33.2 33.4 46.2

50.7 52.5 55.4

B 20.6 25.3 29.2 35.2 41.9 48.5

54.1 57.1 59.8 63.2 68.5

Use the Mann-Whitney test at a significance

level of 5% to determine if there is any evidence

of the two machines producing different results.

Using the procedure:

(i)H 0 : There is no difference in results from the

machines, on average.

H 1 : The results from the two machines are

different, on average.

(ii)α 2 =5%.

(iii) Arranging the data in order gives:

9.7 10.2 11.2 12.4 14.1 20.6 22.3

AAAAABA

25.3 29.2 29.6 31.7 33.0 33.2 33.4

BBAAAAA

35.2 41.9 46.2 48.5 50.7 52.5 54.1

BBABAAB

55.4 57.1 59.8 63.2 68.5

ABBBB

(iv) The number of B’s preceding the A’s in the

sequence is as follows:

AAAAABABB

00000 1

AAAAABBAB

33333 5

AABABBBB

66 7

(v) Adding the numbers from (iv) gives:

U= 0 + 0 + 0 + 0 + 0 + 1 + 3 + 3 + 3 + 3

+ 3 + 5 + 6 + 6 + 7 = 40

(vi) From Table 63.5, forn 1 =11 andn 2 =15, and

α 2 =5%,U≤ 44.

Since our value ofU from (v) is less than

44,H 0 is rejected andH 1 accepted, i.e.the

results from the two machines are different.