Ten samples of 15 parts each were taken from an ongoing process to establish a p
ID: 383050 • Letter: T
Question
Ten samples of 15 parts each were taken from an ongoing process to establish a p-chart for control. The samples and the number of defectives in each are shown in the following table: SAMPLE n NUMBER OF DEFECTIVE ITEMS IN THE SAMPLE 1 15 1 2 15 0 3 15 0 4 15 0 5 15 2 6 15 0 7 15 3 8 15 1 9 15 0 10 15 3 a. Determine the 1formula221.mml, Sp, UCL and LCL for a p-chart of 95 percent confidence (1.96 standard deviations). (Leave no cells blank - be certain to enter "0" wherever required. Round your answers to 3 decimal places.) 1formula221.mml Sp UCL LCL b. What comments can you make about the process? Process is in statistical control Process is out of statistical control
Explanation / Answer
# of
p
Sample
n
Defectives
( # of defectives/ n)
1
15
1
0.067
2
15
0
0.000
3
15
0
0.000
4
15
0
0.000
5
15
2
0.133
6
15
0
0.000
7
15
3
0.200
8
15
1
0.067
9
15
0
0.000
10
15
3
0.200
Total
10
95% confidence interval --> Z-value corresponding =1.96
p =Total number of defects/(k*n)
Where,
K=number of samples=10
N=sample size=15
p =10/(10*15)=0.067
Standard deviation of process
Sp= sqrt [p(100- p)/n] = sqrt [0.067(1- 0.067)/15] =0.0645 0r 0.065
UCL = p + Z* Sp = 0.067 + 1.96(0.065 )=0.194
LCL = p + Z* Sp = 0.067 - 1.96( 0.065 ) = -0.060
LCL is adjusted to Zero since number of defectives cannot be negative
Comments:
With confidence interval at 95%, only one plot in 20 plots is allowed to be out of control. With ten samples, just one plot out of the limits shows the process is out of control
In the given process, two plots (samples 7 and 10) fall above UCL. Hence the process is not under statistical control
Also the p-chart shows wild swings up and down. Hence the process should be examined and corrected for poor performance.
# of
p
Sample
n
Defectives
( # of defectives/ n)
1
15
1
0.067
2
15
0
0.000
3
15
0
0.000
4
15
0
0.000
5
15
2
0.133
6
15
0
0.000
7
15
3
0.200
8
15
1
0.067
9
15
0
0.000
10
15
3
0.200
Total
10
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