Higher Engineering Mathematics

624 STATISTICS AND PROBABILITY

41875 1410 12 5

20 21 25 28 34

0 10 20 30

1196 16 18

SAMPLE P

SAMPLE Q

Figure 63.1

Now try the following exercise.

Exercise 230 Further problems on the

Mann-Whitney test

1. The tar content of two brands of cigarettes (in
   mg) was measured as follows:

BrandP 22.6 4.1 3.9 0.7 3.2

BrandQ 3.4 6.2 3.5 4.7 6.3

BrandP 6.1 1.7 2.3 5.6 2.0

BrandQ 5.5 3.8 2.1

Use the Mann-Whitney test at a 0.05 level of

significance to determine if the tar contents

of the two brands are equal.

⎡

⎢

⎢

⎣

H 0 :TA=TB,H 1 :TA=TB,

U= 30 .From Table 63.5,

U≤17, hence acceptH 0 ,

i.e. there is no difference

between the brands

⎤

⎥

⎥

⎦

2. A component is manufactured by two pro-
   cesses. Some components from each process
   are selected at random and tested for breaking
   strength to determine if there is a difference
   between the processes. The results are:

ProcessA 9.7 10.5 10.1 11.6 9.8

ProcessB 11.3 8.6 9.6 10.2 10.9

ProcessA 8.9 11.2 12.0 9.2

ProcessB 9.4 10.8

At a level of significance of 10%, use the

Mann-Whitney test to determine if there

is a difference between the mean breaking

strengths of the components manufactured by

the two processes.

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

H 0 :B.S.A=B.S.B,

H 1 :B.S.A=B.S.B,

α 2 =10%,U= 28 .From

Table 63.5,U≤15, hence

acceptH 0 , i.e. there is no

difference between the

processes

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

3. An experiment, designed to compare two pre-
   ventive methods against corrosion gave the
   following results for the maximum depths of
   pits (in mm) in metal strands:

Method

A 143 106 135 147 139 132 153 140

Method

B 98 105 137 94 112 103

Use the Mann-Whitney test, at a level of

significance of 0.05, to determine whether

the two tests are equally effective.

⎡ ⎢ ⎢ ⎢ ⎢ ⎣

H 0 :A=B,H 1 :A=B,

α 2 =5%,U= 4 .From

Table 63.5,U≤8, hence

null hypothesis is rejected,

i.e. the two methods are

not equally effective

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

4. Repeat Problem 3 of Exercise 228, page 616
   using the Mann-Whitney test.
   ⎡ ⎢ ⎢ ⎢ ⎢ ⎣
   H 0 : meanA=meanB,
   H 1 : meanA=meanB,
   α 2 =5%,U= 90
   From Table 63.5,U≤99,
   henceH 0 is rejected
   andH 1 accepted

⎤ ⎥ ⎥ ⎥ ⎥ ⎦