Table 6.4
The manager of Kewpie has recently implemented an attribute-based process control method. The accompanying table shows the results of five different samples of dolls from the five weeks of production after the process control method was implemented.
$$\begin{array} { | c | c | c | }
\hline \text { Week } & \begin{array} { c }
\text { Sample } \
\text { Size }
\end{array} & \begin{array} { c }
\text { Number } \
\text { Defective }
\end{array} \
\hline 1 & 100 & 3 \
2 & 100 & 20 \
3 & 100 & 23 \
4 & 100 & 12 \
5 & 100 & 13 \
\hline
\end{array}$$
-Use the information in Table 6.4.The manager of Kewpie has decided to redesign the process control method (both central line and control limits)based on just the data from weeks 1-5.What is the new UCL? (Assume z = 2)
A) less than or equal to 0.15
B) greater than 0.15 but less than or equal to 0.19
C) greater than 0.19 but less than or equal to 0.23
D) greater than 0.23

Question

Table 6.4
The manager of Kewpie has recently implemented an attribute-based process control method. The accompanying table shows the results of five different samples of dolls from the five weeks of production after the process control method was implemented.
 $$\begin{array} { | c | c | c | } 
\hline \text { Week } & \begin{array} { c } 
\text { Sample } \
\text { Size }
\end{array} & \begin{array} { c } 
\text { Number } \
\text { Defective }
\end{array} \
\hline 1 & 100 & 3 \
2 & 100 & 20 \
3 & 100 & 23 \
4 & 100 & 12 \
5 & 100 & 13 \
\hline
\end{array}$$
-Use the information in Table 6.4.The manager of Kewpie has decided to redesign the process control method (both central line and control limits)based on just the data from weeks 1-5.What is the new UCL? (Assume z = 2)
A) less than or equal to 0.15
B) greater than 0.15 but less than or equal to 0.19
C) greater than 0.19 but less than or equal to 0.23
D) greater than 0.23

Quizplus · Accepted Answer

To find the new UCL, we need to calculate the average number of defects per sample and the standard deviation of the number of defects per sample for weeks 1-5.

Average number of defects per sample:
(3+20+23+12+13)/5 = 14.2

Standard deviation of the number of defects per sample:
s = sqrt [ ( (3-14.2)^2 + (20-14.2)^2 + (23-14.2)^2 + (12-14.2)^2 + (13-14.2)^2 ) / 4 ] 
= 7.46

New UCL = 14.2 + (2 * 7.46) = 29.12

Therefore, the new UCL is greater than 0.19 but less than or equal to 0.23. The answer is C.

Table 64 the Manager of Kewpie Has Recently Implemented an Attribute-Based

Table 64
the Manager of Kewpie Has Recently Implemented an Attribute-Based