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Magazine Chapter 14: Q8CRE (page 654)
Defective Child Restraint Systems The Tracolyte Manufacturing Company produces plastic frames used for child booster seats in cars. During each week of production, 120 frames are selected and tested for conformance to all regulations by the Department of Transportation. Frames are considered defective if they do not meet all requirements. Listed below are the numbers of defective frames among the 120 that are tested each week. Use a control chart for p to verify that the process is within statistical control. If it is not in control, explain why it is not.
3 2 4 6 5 9 7 10 12 15
Short Answer
The following control chart is constructed:
It can be observed that the process is within statistical control.
Step by step solution
Given information
Data are given on the number of defective frames in 10 samples of plastic frames.
The sample size of each of the 10 samples is equal to 120.
Important values of p-chart
Let\(\bar p\)be the estimated proportion of defective frames in all the samples.
It is computed as follows.
\(\begin{aligned}{c}\bar p = \frac{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{defectives}}\;{\rm{from}}\;{\rm{all}}\;{\rm{samples}}\;{\rm{combined}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{samples}}}}\\ = \frac{{3 + 2 + ..... + 15}}{{10\left( {120} \right)}}\\ = \frac{{73}}{{1200}}\\ = 0.060833\end{aligned}\).
The value of\(\bar q\)is computed as shown below.
\(\begin{aligned}{c}\bar q = 1 - 0.060833\\ = 0.939167\end{aligned}\).
The value of the lower control limit (LCL) is computed below.
\[\begin{aligned}{c}LCL = \bar p - 3\sqrt {\frac{{\bar p\bar q}}{n}} \\ = 0.060833 - 3\sqrt {\frac{{\left( {0.060833} \right)\left( {0.939167} \right)}}{{120}}} \\ = - 0.00463\\ \approx 0\end{aligned}\].
The value of the upper control limit (UCL) is computed below.
\[\begin{aligned}{c}UCL = \bar p + 3\sqrt {\frac{{\bar p\bar q}}{n}} \\ = 0.060833 + 3\sqrt {\frac{{\left( {0.060833} \right)\left( {0.939167} \right)}}{{120}}} \\ = 0.126293\end{aligned}\].
Computation of the fraction defective
The sample fraction defective for the ithsample can be computed as follows.
\[{p_i} = \frac{{{d_i}}}{{120}}\].
Here,
\[{p_i}\]isthe sample fraction defective for the ithsample, and
\[{d_i}\]isthe number of defective frames in the ithsample.
The computation of the fraction defective for the ith batch is given as follows.
S.No. | Defectives (d) | Sample fraction defective (p) |
1 | 3 | 0.025 |
2 | 2 | 0.017 |
3 | 4 | 0.033 |
4 | 6 | 0.050 |
5 | 5 | 0.042 |
6 | 9 | 0.075 |
7 | 7 | 0.058 |
8 | 10 | 0.083 |
9 | 12 | 0.100 |
10 | 15 | 0.125 |
Construction of the p-chart
Follow the given steps to construct the p-chart:
- Mark the values 1, 2, ...,10 on the horizontal axis and label the axis as 鈥楽ample鈥�.
- Mark the values 0, 0.02, 0.04, 鈥︹��, 0.14 on the vertical axis and label the axis as 鈥楶roportion Defective鈥�.
- Plot a straight line corresponding to the value 0.060833 on the vertical axis and label the line (on the left side) as 鈥榎(CL = 0.060833\)鈥�.
- Plot a horizontal line corresponding to the value 0 on the vertical axis and label the line as 鈥楲CL=0鈥�.
- Similarly, plot a horizontal line corresponding to the value 0.126293 on the vertical axis and label the line as 鈥楿CL=0.126293鈥�.
- Mark the given10sample points (fraction defective of the ith lot) on the graph and join the dots using straight lines.
The following p-chart is plotted:
Analysis of the p-chart
It can be observed that the plotted chart shows no signs of being out of control.
Thus, it can be concluded that the process is not within statistical control.
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