Samples of 2000 were taken from an encoding machine at a bank over a five-week period.Management is willing to use the historical average for the central line of the chart.
$\begin{array} { | c | c | } \hline \begin{array} { c } \text { Sample } \\\text { Number }\end{array} & \begin{array} { c } \text { Proportion } \\\text { Defective }\end{array} \\\hline 1 & 0.001 \\2 & 0.005 \\3 & 0.006 \\4 & 0.006 \\5 & 0.002 \\\hline\end{array}$
Which one of the following would be the limits for an attribute control chart with z = 2?

Question

Samples of 2000 were taken from an encoding machine at a bank over a five-week period.Management is willing to use the historical average for the central line of the chart.

\begin{array} { | c | c | } \hline \begin{array} { c } \text { Sample } \\\text { Number }\end{array} & \begin{array} { c } \text { Proportion } \\\text { Defective }\end{array} \\\hline 1 & 0.001 \\2 & 0.005 \\3 & 0.006 \\4 & 0.006 \\5 & 0.002 \\\hline\end{array}

Which one of the following would be the limits for an attribute control chart with z = 2?

Quizplus · Accepted Answer

To calculate the control limits for an attribute control chart with z = 2, we need to first calculate the proportion defective for each sample:

Sample 1: 0.001
Sample 2: 0.005
Sample 3: 0.006
Sample 4: 0.006
Sample 5: 0.002

Next, we need to calculate the average proportion defective:

Average proportion defective = (0.001 + 0.005 + 0.006 + 0.006 + 0.002)/5 = 0.004

Now we can calculate the standard deviation of the proportion defective:

σp = sqrt[p*(1-p)/n] = sqrt[0.004*(1-0.004)/2000] = 0.002

Finally, we can calculate the control limits:

UCL = p̄ + z*σp = 0.004 + 2*0.002 = 0.008
LCL = p̄ - z*σp = 0.004 - 2*0.002 = 0.000

Therefore, the limits for an attribute control chart with z = 2 are UCL < 0.010 and LCL > 0.001, which is answer choice C.

Samples of 2000 Were Taken from an Encoding Machine at a Bank