Introduction to Probability and Statistics for Engineers and Scientists

13.6Other Control Charts for Detecting Changes in the Population Mean 569

the repair shop plots a standard exponentially weighted moving-average (EWMA) control
chart with each data value being the average of 4 successive times, and with a weighting
factor ofα=.25. If the present value of the chart is 60 and the following are the next 16
subgroup averages, what can we conclude?

48, 52, 70, 62, 57, 81, 56, 59, 77, 82, 78, 80, 74, 82, 68, 84

SOLUTION Starting withW 0 =60, the successive values ofW 1 ,...,W 16 can be obtained
from the formula

Wt=.25Xt+.75Wt− 1

This gives

W 1 =(.25)(48)+(.75)(60)= 57

W 2 =(.25)(52)+(.75)(57)=55.75

W 3 =(.25)(70)+(.75)(55.75)=59.31

W 4 =(.25)(62)+(.75)(59.31)=59.98

W 5 =(.25)(57)+(.75)(59.98)=59.24

W 6 =(.25)(81)+(.75)(59.24)=64.68

and so on, with the following being the values ofW 7 throughW 16 :

62.50, 61.61, 65.48, 69.60, 71.70, 73.78, 73.83, 75.87, 73.90, 76.43

Since

3

√

.25

1.75

24

√

4

=13.61

the control limits of the standard EWMA control chart with weighting factorα=.25 are

LCL= 62 −13.61=48.39

UCL= 62 +13.61=75.61

Thus, the EWMA control chart would have declared the system out of control after
determiningW 14 (and also afterW 16 ). On the other hand, since a subgroup standard
deviation isσ/

√
n=12, it is interesting that no data value differed fromμ=62 by even
as much as 2 subgroup standard deviations, and so the standardX-control chart would
not have declared the system out of control. ■