Higher Engineering Mathematics

618 STATISTICS AND PROBABILITY

(iii) Taking the difference between the time taken
for each order and 7 h gives:

− 214 h + 234 h + 812 h +4h + 114 h

−^12 h +2h + 134 h + 334 h − 312 h

+ 112 h + 212 h + 814 h +6h +1h

+^34 h −^14 h

(iv) These differences may now be ranked from 1

to 17, ignoring whether they are positive or

negative:

Rank 1 2 34 56

Difference −^14 −^12341114112

Rank 78 9101112

Difference 134 2 − 214 212 234 − (^312)
Rank 13 14 15 16 17
Difference (^33446814812)
(v) The Wilcoxon signed-rank statisticTis calcu-
lated asthe sum of the ranksof the negative
differences for a one-tailed test.
The sum of the ranks for the negative values is:
T= 1 + 2 + 9 + 12 = 24.
(vi) Table 63.4 gives the critical values ofTfor the
Wilcoxon signed-rank test. Forn=17 and a
significance levelα 1 =5%,T≤41.
Hence the conclusion is that sinceT=24 the result
is within the 5% critical region.There is therefore
strong evidence to supportH 1 , the alternative
hypothesis, that the median processing time is
greater than 7 hours.
Problem 9. The following data represents the
number of hours that a portable car vacuum
cleaner operates before recharging is required.
Operating
time (h) 1.4 2.3 0.8 1.4 1.8 1.5
1.9 1.4 2.1 1.1 1.6
Use the Wilcoxon signed-rank test to test the
hypothesis, at a 5% level of significance, that this
particular vacuum cleaner operates, on average,
1.7 hours before needing a recharge.
(This is the same as Problem 6 where the sign test
was used).
Using the procedure:
(i)H 0 :t= 1 .7handH 1 :t= 1 .7h.
(ii) Significance level,α 2 =5%(since this is a two-
tailed test).
(iii) Taking the difference between each operating
time and 1.7 h gives:
− 0 .3h + 0 .6h − 0 .9h − 0 .3h

• 0 .1h − 0 .2h + 0 .2h − 0 .3h


• 0 .4h − 0 .6h − 0 .1h
  (iv) These differences may now be ranked from 1
  to 11 (ignoring whether they are positive or
  negative).
  Some of the differences are equal to each
  other. For example, there are two 0.1’s (ignor-
  ing signs) that would occupy positions 1 and
  2 when ordered. We average these as far as
  rankings are concerned i.e. each is assigned a
  ranking of
  1 + 2
  2
  i.e. 1.5. Similarly the two 0.2
  values in positions 3 and 4 when ordered are
  each assigned rankings of
  3 + 4
  2
  i.e. 3.5, and
  the three 0.3 values in positions 5, 6, and 7 are
  each assigned a ranking of
  5 + 6 + 7
  3
  i.e. 6, and
  so on. The rankings are therefore:
  Rank 1.5 1.5 3.5 3.5
  Difference +0.1 −0.1 −0.2 +0.2
  Rank 66 68
  Difference −0.3 −0.3 −0.3 +0.4
  Rank 9.5 9.5 11
  Difference +0.6 −0.6 −0.9
  (v) There are 4 positive terms and 7 negative
  terms. Taking the smaller number, the four
  positive terms have rankings of 1.5, 3.5, 8
  and 9.5. Summing the positive ranks gives:
  T= 1. 5 + 3. 5 + 8 + 9. 5 = 22. 5.
  (vi) From Table 63.4, whenn=11 andα 2 =5%,
  T≤ 10.
  SinceT= 22 .5 falls in the acceptance region
  (i.e. in this case is greater than 10),the null