Statistical Process Control Charts
In an automatic transmission a part called a parking pawl is used to hold the vehicle in park. The size of the pawl depends on the model of the vehicle. Max Manufacturing Inc. makes two different thicknesses of pawls- a thin pawl, which has a nominal dimension of .2950 inch and a thick pawl which has a nominal dimension of .6850 inch. Both parts are stamped on the stamp machine. Max Manufacturing runs just-in-time operations and uses short –run control charts to monitor critical part dimensions. Pawl thickness for both parts is measured and recorded on the same short-run control chart. Create and interpret a short-run control chart for the following data:
0.2946 0.2947 .6850 .6853
0.2951 .2951 .6851 .6849
0.2957 .2949 .6852 .6852
0.2951 .2947 .6847 .6848
.2945 .2950 .6851
.2951 .2951 .6852
.2950 .2952 .6853
.2952 .2949 .6850
.2952 .2944 .6848
.2950 .2951 .6849
.2947 .2948 .6847
.2945 .2954 .6849
2 Answers | Add Yours
Attached is a graph of the control chart for the differences `d_i`
The observed differences from nominal values are in black for pawl 1 and in red for pawl 2.
In a short-run control chart we can transform observations to some comparable measure. This tells us about the overall quality of manufacturing when there are short runs of production on different manufactured parts where the short runs aren't long enough ie don't provide enough information to analyse the individual manufacturing processes separately.
In this example here there are two parking pawls being made, with nominal thickness 0.2950" and 0.6850". These are the given benchmarks to compare gathered measurements to.
In addition to this we require an estimate of standard deviation. Here we can calculate this from all the data and assume the standard deviation was known to be fixed at this value beforehand. In practice we would estimate the standard deviation from pilot data along with the nominal thickness values.
To estimate the (shared) standard deviation, calculate the differences `d` between the observations and their nominal values thus:
d = 0.2946-0.2950, 0.2947-0.2950, 0.6850-0.6850, 0.6853-0.6850 ,.... etc =
-0.0004, -0.0003, 0.0000, 0.0003, ... etc
The standard deviation is estimated using
`hat(sigma)_d = sqrt(sum_(i=1)^n (d_i-bar(d))^2/(n-1)) = sqrt(sum_(i=1)^40 (d_i - bar(d))^2/39)`
You should find with this data that `hat(sigma)_d = 0.000266 ` to 3 sf (` ` and `bar(d) = 0.0000225` : NB in actual pilot data we would necessarily have `bar(d)=0` ).
We then set up the control chart to be on the differences `d_i` where the nominal mean `mu_d = 0` and the standard deviation `sigma_d = 2.66 times 10^(-4)`
(assumed known and equal for the two types of pawl).
A standard Shewhart chart for Normal data signals an alarm/change in process if `d_i` falls outside of the limits `mu_d pm 3sigma_d` (a six sigma chart on the process mean).
Plot the data `d_i` and the limits `mu_d pm 3sigma_d = pm 7.99 times 10^(-4) ` on a chart, with the center line at the nominal mean `mu_d = 0`. You should find that none of the observations fall outside of the limits and therefore the overall manufacturing process for making both pawls remains 'in control'.
Ok I hink I understand..sort of. But can you help me a litle more? I want to be sure I understand it. So I take the sample number and subtract the nominal from it.
Ex. The first group will be the four vertical numbers starting with .2916 and ending with .2951. I subtract, add the differences and divide by 4. I will do this on each group and when I am done I will have 10 groups of numbers and I will use those to figure the centeline from.
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