Chapter 14
Statistical Process Control
14-2 Control Charts for Variation and Mean
1. No. Control charts are constructed entirely from the observed data, and they are totally
independent of any product specifications. The upper and lower control limits of control
charts, for example, are based on the actual behavior of the process under study, and not the
desired behavior. It is possible for the process of filling 12 oz cans of Coke to be well within
statistical control by consistently filling the cans with, say, 11.5 oz.
3. Random variation is the natural fluctuation due to chance alone that is inherent in every
process which does not function with perfect precision. Assignable variation is deviance from
perfect precision that can be attributed to an identifiable source – e.g., machinery that needs
adjustment.
5. a. Within statistical control.
b. Not applicable; none of the three criteria applies.
c. No; the variability is too large. Even though the process may be within control statistically,
it is not within control practically. For the amounts of cola in 12 oz cans to vary from 10 oz
to 13 oz is clearly unacceptable.
7. a. Not within statistical control.
b. Criterion 3 applies; there are 8 consecutive points below the centerline. For these data,
the criterion applies two times – as there are also 8 consecutive points above the
centerline.
c. No; there appears to have been a shift in the process after observation 8, as all the values
before then are below the centerline and all the observations after that are above the
centerline. It is usually important to identify the cause for such a shift.
Exercises 9-12 refer to the following table.
decade 00
1880 290.7
1890 293.2
1900 295.6
1910 299.4
1920 301.4
1930 305.9
1940 307.4
1950 311.3
1960 316.9
1970 325.7
1980 338.7
1990 354.2
2000 369.5
01
291.0
293.5
295.3
299.6
301.6
306.1
307.6
311.7
317.6
326.3
340.0
355.6
371.0
02
291.2
293.8
295.1
299.9
302.3
306.2
307.7
312.2
318.5
327.5
341.1
356.4
373.1
03
291.4
294.1
294.8
300.1
302.9
306.3
307.9
312.7
319.0
329.6
342.8
357.1
375.6
04
291.6
294.3
295.9
300.3
303.6
306.5
308.4
313.2
319.5
330.3
344.4
358.9
377.4
05
291.9
294.6
296.9
300.5
304.2
306.6
308.9
313.7
320.1
331.2
345.9
360.9
379.6
06
292.1
294.9
297.5
300.7
304.9
306.8
309.3
314.3
321.3
332.2
347.1
362.6
379.6
07
292.3
295.2
298.1
300.9
305.5
306.9
309.8
314.8
322.1
333.9
349.0
363.8
381.2
08
292.6
295.5
298.6
301.1
305.6
307.1
310.3
315.3
323.1
335.5
351.4
366.6
382.8
09
292.9
295.8
299.2
301.2
305.8
307.3
310.8
316.0
324.6
336.9
352.9
368.3
384.4
x
291.77
294.49
296.70
300.37
303.78
306.57
308.81
313.52
320.27
330.91
345.33
360.44
377.42
R
2.2
2.6
4.4
1.8
4.4
1.4
3.4
4.7
7.7
11.2
14.2
14.1
14.9
314
CHAPTER 14 Statistical Process Control
9. There are k=13 samples (decades) of n=10 observations (years) each.
x = (Σx)/k = 4150.38/13 = 319.26
R = (ΣR)/k = 87.0/13 = 6.692
For the R chart: UCL = D4 R = 1.777(6.692) = 11.89
LCL = D3 R = 0.223(6.692) = 1.49
For the x chart: UCL = x + A2 R = 319.26 + 0.308(6.692) = 319.26 + 2.06 = 321.32
LCL = x – A2 R = 319.26 – 0.308(6.692) = 319.26 – 2.06 = 317.20
11. The R chart is given below. The process variation is not within statistical control because
there is an upward trend, there are 8 consecutive points below the centerline, and there are
points above the upper control limit.
R CHART
16
14
R (CO2 ppm)
12
UCL = 11.89
10
8
_
R = 6.69
6
4
2
LCL = 1.49
0
0
2
4
6
8
10
12
14
decade
13. There are k=8 samples (years) of n=6 observations (bi-monthly readings) each.
x = (Σx)/k = 22253.67/8 = 2781.71
year
x
R = (ΣR)/k = 13835/8 = 1729.375
1
3077.67
For the R chart: UCL = D4 R = 2.004(1729.375) = 3465.67
2
2816.83
LCL = D3 R = 0.000(1729.375) = 0
3
2640.33
For the x chart: UCL = x + A2 R = 2781.71 + 0.483(1729.375)
4
2779.00
= 2781.71 + 835.29 = 3617.00
5
2809.33
6
2706.50
LCL = x – A2 R = 2781.71 – 0.483(1729.375)
7
2771.67
= 2781.71 – 835.29 = 1946.42
8
2652.33
R
1345
2175
1302
1020
3214
1311
2342
1306
Control Charts for Variation and Mean SECTION 14-2
315
15. The x chart is given below. The process mean is within statistical control. None of the
three criteria for being out of statistical control apply.
X-BAR CHART
3800
UCL = 3617.00
3600
3400
x-bar (kWh)
3200
3000
=
x = 2781.71
2800
2600
2400
2200
2000
LCL = 1946.42
0
1
2
3
4
5
6
7
8
year
17. There are k=8 samples (weeks) of n=5 observations (weekdays) each.
x = (Σx)/k = 989.30/8 = 123.66
R = (ΣR)/k = 3.0/8 = 0.375
For the R chart: UCL = D4 R = 2.114(0.375) = 0.793
LCL = D3 R = 0.000(0.375) = 0
For the x chart: UCL = x + A2 R = 123.66 + 0.577(0.375)
= 123.66 + 0.22 = 123.88
LCL = x – A2 R = 123.66 – 0.577(0.375)
= 123.66 – 0.22 = 123.45
week
1
2
3
4
5
6
7
8
x
123.66
123.44
123.68
123.94
123.68
123.36
123.74
123.80
R
0.6
0.3
0.4
0.4
0.6
0.1
0.4
0.2
19. The run chart is given below. There appears to be a pattern of long range cycles which suggest
that the process may not be within statistical control. Several consecutive day of lower voltage
are followed by several consecutive days of higher voltage, followed by several consecutive
days of lower voltage, followed by several consecutive days of higher voltage.
RUN CHART
124.2
power (volts)
124.0
123.8
=
x = 123.66
123.6
123.4
123.2
0
10
20
day
30
40
316
CHAPTER 14 Statistical Process Control
21. There are k=13 samples of size n=10 each.
s = Σs/k = 1.56029/13 = 0.1200
UCL = B4 s = 1.716(0.1200) = 0.2060
LCL = B3 s = 0.284(0.1200) = 0.0341
This is very similar to the R chart given in the text, and both charts indicate that the process
variation is within statistical control. For relatively small n’s in the sub-samples (n=10 in this
example), there will be a very high correlation between the values of R and s – and so the two
charts will be almost identical, but with different labels on the vertical axis.
s CHART
UCL = 0.2060
s (degrees C)
0.20
0.15
_
s = 0.1200
0.10
0.05
LCL = 0.0341
0
2
4
6
8
10
12
14
decade
1
2
3
4
5
6
7
8
9
10
11
12
13
s
0.164415
0.121033
0.133870
0.157113
0.079889
0.081131
0.077431
0.110156
0.138948
0.124740
0.114232
0.159109
0.098229
x
13.819
13.696
13.741
13.788
13.906
14.016
14.052
13.983
13.938
14.014
14.264
14.396
14.636
14-3 Control Charts for Attributes
1. No. Control charts determine whether a process is within statistical control relative to the
values it is actually producing, not relative to the values it is supposed to be producing.
NOTE: The text defines p as the pooled estimate of the proportion of successes among all items
sampled (i.e., the weighted mean of the k values for p) and gives
p = (Σx)/(Σn)
= (total number of “successes” among all items samples)/(total number of items sampled).
When there are k samples of size n each, this reduces to p = (Σp)/k. Since each of k values for p
must be calculated to make the chart, and since this is consistent with the notation in the last
section that x = (Σx)/k and R = (ΣR)/k, this manual uses p = (Σp)/k as the primary formula.
3. There are k=5 samples of size n=100 each.
p = (Σp)/k = 0.10/5 = 0.02
p q/n =
(0.02)(0.98)/100 = 0.014
UCL = p + 3 p q/n = 0.02 + 3(0.014) = 0.02 + 0.042 = 0.062
LCL = p – 3 p q/n = 0.02 – 3(0.014) = 0.02 – 0.042 = -0.022, adjusted to 0.00
Yes. The lower control limit was adjusted from -0.022 to 0.00 because negative values are not
possible for proportions. Since the process is not within statistical control when data fall
outside the control limits (i.e., not on the control limits), a process with the above control limits
cannot fail to be within statistical control by having a sample proportion that is too low.
Control Charts for Attributes SECTION 14-3
317
5. The process is within statistical control; none of the three criteria applies. In this case,
however, there are warning signs that suggest continuing monitoring of the process is in order
– two of the data points are dangerously close to the upper control limit, and the final n=7
consecutive data points below the centerline (along with the general pattern of the data) are
dangerously close to indicating the presence of long term cycles.
7. The process is not within statistical control – and in this case, all three criteria apply. The data
end with an upward trend. There is a data point above the upper control limit. There are at
least 8 consecutive data points above the centerline.
9. There are k=20 samples of n=10,000 each.
p = (Σp)/k = (0.0020 + 0.0014 + 0.0022 +…+ 0.0020)/20 = 0.0400/20 = 0.00200
p q/n =
(0.0020)(0.9980)/10000 = 0.0004468
UCL = p + 3 p q/n = 0.00200 + 3(0.0004468) = 0.00200 + 0.00134 = 0.00334
LCL = p – 3 p q/n = 0.00200 – 3(0.0004468) = 0.00200 – 0.00134 = 0.00066
The p chart is given below. The process is within statistical control. None of the three criteria
for being out of statistical control applies.
p CHART
0.0035
UCL = 0.00334
0.0030
p
0.0025
_
p = 0.00200
0.0020
0.0015
0.0010
LCL = 0.00066
0.0005
0
5
10
15
20
sample
11. There are k=15 samples of n=1000 each.
p = (Σp)/k = (0.601 + 0.625 + 0.619 +…+ 0.667)/15 = 9.523/15 = 0.6349
p q/n =
(0.6349)(0.3651)/1000 = 0.015225
UCL = p + 3 p q/n = 0.6349 + 3(0.015225) = 0.6349 + 0.0457 = 0.6805
LCL = p – 3 p q/n = 0.6349 – 3(0.015225) = 0.6349 – 0.0457 = 0.5892
318
CHAPTER 14 Statistical Process Control
The p chart is given below. The process is within statistical control. None of the three criteria
for being out of statistical control applies. Whether or not the enrollments are high enough
depends on the desired values – but control charts determine whether a process is within
statistical control relative to the values it is actually producing, not relative to the values
desired.
p CHART
UCL = 0.6805
0.68
0.66
0.64
p
_
p = 0.6349
0.62
0.60
LCL = 0.5892
0
2
4
6
8
10
12
14
16
sample
13. There are k=23 samples of n=1000 each.
p = (Σp)/k = (0.585 + 0.454 + 0.577 +…+ 0.514)/23 = 10.199/23 = 0.4434
p q/n =
(0.4434)(0.5566)/1000 = 0.0157099
UCL = p + 3 p q/n = 0.4434 + 3(0.0157099) = 0.4434 + 0.0471 = 0.4906
LCL = p – 3 p q/n = 0.4434 – 3(0.0157099) = 0.4434 – 0.0471 = 0.3963
The p chart is given below. The process is not within statistical control because there is an
p CHART
0.60
0.55
p
0.50
UCL = 0.4906
0.45
_
p = 0.4434
0.40
LCL = 0.3963
0.35
0.30
0
5
10
15
20
25
sample
alternating trend, there are points above the upper control limit, and there are points below the
lower control limit. The alternating trend is caused by the fact that the odd-numbered samples
represent presidential election years and the even-numbered samples represent non-presidential
election years. In addition, there may be a long-term cycle: it appears that voting in both types
of elections declined, reached a low point, and is beginning to increase again.
Control Charts for Attributes SECTION 14-3
319
15. There are k=20 samples of size n=10000 each. The example calculated p = 0.001.
The np chart is literally the p chart multiplied by n. It is based on x=np instead of on p.
n∙ p = 10000(0.001) = 10
[OR n∙ p = n∙(Σp)/k = (Σnp)/k = (Σx)/k = (15+12+14+…+7)/20 = 200/20 = 10]
n∙ p q/n =
n p q = 10000(0.001)(0.999) = 3.1607
UCL = n∙ p + 3 n p q = 10.00 + 3(3.1607) = 10.0 + 9.48 = 19.48
LCL = n∙ p – 3 n p q = 10.00 – 3(3.1607) = 10.0 – 0.48 = 0.52
The np control chart that follows is identical to the p control chart in the text except for the
labeling on the vertical axis. The values on the vertical axis of the np chart are n=10000 times
the values on the p chart – e.g., the np UCL is 10000(0.001948) = 19.48, the np center line is
10000(0.001) = 10, and the np LCL is 10000(0.000052) = 0.52.
np CHART
20
UCL = 19.48
x = np
15
_
n p = 10.00
10
5
LCL = 0.52
0
0
5
10
15
20
sample
Statistical Literacy and Critical Thinking
1. Statistical process control is the monitoring of data over time in order to detect trends or
problems that indicate the production process is out of statistical control and/or problems that
lead to the production of a product that does not meet prescribed specifications.
2. It is important to monitor the manufacturing process over time because conditions affecting the
product may occur over time – i.e., the hiring of new employees, the wearing out of the
machinery, etc. A possible adverse consequence of not monitoring the manufacturing process
would be the financial and/or public relations loss in dealing with products not meeting the
proper specifications leaving the plant and entering the market. In addition, the product could
become so far out of specification that only a temporary shutdown (and subsequent loss of
revenue) could remedy the problem.
3. In general, a process can go out of control because of a change in mean or a change in
variation. The fact that the mean is within statistical control does not indicate that the variation
is within statistical control, and vice-versa. For this reason, one should monitor a process with
both an x chart and an R chart.
4. No. Control charts determine whether a process is within statistical control relative to the
values it is actually producing, not relative to the values it is supposed to be producing.
320
CHAPTER 14 Statistical Process Control
Chapter Quick Quiz
1. Process data are values in chronological sequence that measure a characteristic of a
controllable activity.
2. Random variation is the natural fluctuation due to chance alone that is inherent in every
process which does not function with perfect precision. Assignable variation is deviance from
perfect precision that can be attributed to an identifiable source – e.g., machinery that needs
adjustment.
3. The test states a process is out of statistical control if any of the following three criteria applies.
(1) There is a data pattern that is obviously not random.
(2) There is a data value above the upper control limit or below the lower control limit.
(3) There are 8 consecutive data points either all above or all below the centerline.
4. An R chart monitors the variation of a process, while an x chart monitors the mean of a
process.
5. No. The process variation is not within statistical control because the R chart indicates a data
point that is above the upper control limit.
6. The R chart indicates that R = 21.2. In this context, R is the mean of the k=20 values for R
obtained from the 20 subsamples of size n.
7. No. The process mean is not within statistical control because the X-bar chart indicates that
there are at least 8 consecutive data points below the centerline.
8. The X-bar chart indicates that x = 6.45. In this context, x is the mean of the k=20 values for
x obtained from the 20 subsamples of size n.
9. A p chart is a control chart to monitor the proportion of items having a particular attribute over
time. It is used to determine whether the corresponding process is within statistical control.
10. False. A process is out of control when it appears to be under the influence of some factor or
force that causes other than the usual expected random deviation. That factor or force could be
causing a smaller proportion of defects, or it could be acting in some way (e.g., increasing the
variation) that does not alter the overall proportion of defects.
Review Exercises
321
Review Exercises
1. The run chart is given below. There is not a pattern suggesting that the process is not within
statistical control.
RUN CHART
290
280
axial load (lbs)
270
260
=
x = 256.95
250
240
230
220
210
200
0
5
10
15
20
item number
2. There are k=3 samples of n=7 observations each. For the R chart,
R = (ΣR)/k = (78+77+31)/3 = 62.0
UCL = D4 R = 1.924(62.0) = 119.3
LCL = D3 R = 0.076(62.0) = 4.7
3. The process variation is within statistical control because none of the three criteria for being
out of statistical control applies.
4. There are k=3 samples of n=7 observations each. For the x chart,
x = (Σx)/k = (252.71+247.86+270.29)/3 = 256.95
R = (ΣR)/k = (78+77+31)/3 = 62.0
UCL = x + A2 R = 256.95 + 0.419(62.0) = 256.95 + 25.98 = 282.93
LCL = x – A2 R = 256.95 – 0.419(62.0) = 256.95 – 25.98 = 230.97
5. The process mean is within statistical control because none of the three criteria for being out of
statistical control applies.
6. There are k=15 samples of n=1,000,000 each.
p = (Σp)/k = (0.000085 + 0.000087 + 0.000094 +…+ 0.000056)/15
= 0.001153/15 = 0.0000769
p q/n = (0.0000769)(0.9999231)/1000000 = 0.000008767
UCL = p + 3 p q/n = 0.0000769 + 3(0.000008767) = 0.0000769 + 0.0000263 = 0.0001032
LCL = p – 3 p q/n = 0.0000769 – 3(0.000008767) = 0.0000769 – 0.0000263 = 0.0000506
322
CHAPTER 14 Statistical Process Control
The p chart is given below. The process is not within statistical control because there is a
downward trend, there are 8 consecutive points above the centerline. The control chart
suggests that the downward trend in the proportion of homicides may have ended and is
leveling off.
p CHART
0.00011
UCL = 0.0001032
0.00010
p
0.00009
0.00008
_
p = 0.0000769
0.00007
0.00006
LCL = 0.0000506
0.00005
0
2
4
6
8
10
12
14
16
year number
7. There are k=15 samples of n=100 each.
p = (Σp)/k = (0.05 + 0.04 + 0.03 +…+ 0.01)/15 = 0.81/15 = 0.054
p q/n =
(0.054)(0.946)/100 = 0.0226
UCL = p + 3 p q/n = 0.054 + 3(0.0226) = 0.054 + 0.068 = 0.122
LCL = p – 3 p q/n = 0.054 – 3(0.0226) = 0.054 – 0.068 = -0.014, truncated at 0.000
The p chart is given below. The process is not within statistical control because there is a
pattern of increasing variation.
p CHART
UCL = 0.122
0.12
0.10
p
0.08
_
p = 0.054
0.06
0.04
0.02
0.00
LCL = 0.000
0
2
4
6
8
10
day number
12
14
16
Cumulative Review Exercises
323
Cumulative Review Exercises
1. The following summary information applies to all parts of the exercise.
Let CO2 = x and temperature = y.
n = 10
Σx = 3774.2
x = (Σx)/n = 3774.2/10 = 377.42
Σy = 146.36
y = (Σy)/n = 146.36/10 = 14.636
Σxy = 55242.261
n(Σxy) – (Σx)(Σy) = 30.698
Σx2 = 1424685.94
n(Σx2) – (Σx)2 = 2273.76
Σy2 = 2142.2118
n(Σy2) – (Σy)2 = 0.8684
r = [n(Σxy) - (Σx)(Σy)]/[ n(Σx 2 ) - (Σx) 2 n(Σy 2 ) - (y) 2 ]
= 30.698/[ 2273.76 0.8684 ]
= 0.6908
a. Ho: ρ = 0
H1: ρ ≠ 0
α = 0.05 and df = 8
C.V. t = ±tα/2 = ±t0.025 = ±2.306 [or r = ±0.632]
calculations:
tr = (r – μr)/sr
= (0.691 – 0)/ [1 - (0.691) 2 ]/8
0.025
0.025
= 0.691/0.2556
-0.632
0
0.632
r
= 2.703
-2.306
0
2.306
t
P-value = 2∙tcdf(2.703,99,8) = 0.0270
conclusion:
Reject Ho; there is sufficient evidence to conclude that ρ ≠ 0 (in fact, that ρ > 0). Yes;
there is sufficient evidence to support the claim of a linear correlation between carbon
dioxide concentration and temperature.
8
b. No. Linear correlation measures association, not causation. And if there does happen to be
an cause-and-effect relationship between the two variables, there is nothing in the
mathematics that would allow a person to say which variable is the cause and which variable
is the effect.
c. b1 = [n(Σxy) – (Σx)(Σy)]/[n(Σx2) – (Σx)2] = 30.698/2273.76 = 0.0135
bo = y – b1 x = 14.636 – 0.0135(377.42) = 9.540
ŷ = bo + b1x = 9.540 + 0.0135x
d. ŷ 290.7 = 9.540 + 0.0135(290.7) = 13.46 °C
The predicted temperature does not appear to be very close to the actual temperature of
13.88 °C. But a carbon dioxide concentration of 290.7 is well outside the range of x values
used to determine the regression line. Since 290.7 is (290.7 – 377.42)/5.02633 = -17.253
standard deviations from the mean of the x scores used to determine the line, the regression
line should probably not be used for that prediction.
324
CHAPTER 14 Statistical Process Control
2. There are k=10 samples (weeks) of n=200 belts each.
p = (Σp)/k = (0.010 + 0.015 + 0.005 +…+ 0.085)/10 = 0.335/10 = 0.0335
p q/n =
(0.0335)(0.9665)/200 = 0.01272
UCL = p + 3 p q/n = 0.0335 + 3(0.01272) = 0.0335 + 0.0382 = 0.0717
LCL = p – 3 p q/n = 0.0335 – 3(0.01272) = 0.0335 – 0.0382 = -0.0047, truncated at 0.0000
The p chart is given below. The process is not within statistical control because there is a
pattern of increasing variability and there are points above the upper control limit.
p CHART
0.09
0.08
UCL = 0.0717
0.07
0.06
p
0.05
0.04
_
p = 0.0335
0.03
0.02
0.01
0.00
LCL = 0.0000
0
2
4
6
8
10
week number
3. α = 0.05 and zα/2 = z0.025 = 1.96; p̂ = x/n = 67/2000 = 0.0335
ˆˆ
pˆ ± z /2 pq/n
0.0335 ± 1.96 (0.0335)(0.9665)/2000
0.0335  0.0079
0.0256 < p < 0.0414
We have 95% confidence that the interval from 0.0256 to 0.0414 contains the true proportion
of defective seat belts produced by this process.
4. original claim: p > 0.03
p̂ = x/n = 67/2000 = 0.0335
Ho: p = 0.03
H1: p > 0.03
α = 0.05
C.V. z = zα = z0.05 = 1.645
calculations:
z p̂ = ( p̂ – μ p̂ )/σ p̂
0.05
= (0.0335 – 0.03)/ (0.03)(0.97) / 2000
^
p
0.03
= 0.0035/0.003814
z
0
1.645
= 0.92
P-value = P(z>0.92) = 1 – 0.8212 = 0.1788
conclusion:
Do not reject Ho; there is not sufficient evidence to support the claim that p > 0.03. There
is not sufficient evidence to support the claim that the proportion of defects is greater than
3%.
Cumulative Review Exercises
325
5. a. P(all 8 above) = P(A1 and A2 and A3 and…and A8)
= P(A1)∙P(A2)∙P(A3)∙…∙P(A8)
= (1/2)(1/2)(1/2)…(1/2)
= (1/2)8 = 1/256 or 0.0039
b. P(all 8 below) = P(B1 and B2 and B3 and…and B8)
= P(B1)∙P(B2)∙P(B3)∙…∙P(B8)
= (1/2)(1/2)(1/2)…(1/2)
= (1/2)8 = 1/256 or 0.0039
c. P(all above OR all below) = P(all above) + P(all below) – P(all above AND all below)
= 1/256 + 1/256 – 0
= 2/256 = 1/128 or 0.0078
6. normal distribution: μ = 6.0, σ = 1.0
a. P(x>7.0) = P(z>1.00)
= 1 – 0.8413
= 0.1587 or 15.87%
Yes; if the helmets were made to fit only males with head breadths less than 7.0 inches, too
many men would be excluded.
b. For the smallest 5%,
For the largest 5%,
A = 0.0500 and z = -1.645
A = 0.9500 and z = 1.645
x = μ + zσ
x = μ + zσ
= 6.0 + (-1.645)(1)
= 6.0 + (1.645)(1)
= 6.0 – 1.645
= 6.0 + 1.645
= 4.355 inches
= 7.645 inches
The helmets will men with head breadths from 4.355 inches to 7.645 inches.
7. No. Since the sample is a voluntary response sample, which is not necessarily representative
of the general population, it is not appropriate to use these results to make any conclusion
about the general population
8. No; the values do not appear to come from a normal distribution. The histogram indicates a
distribution that is positively skewed and far from bell-shaped, and the normal probability plot
is not close to a straight line.
Histogram of Coin Values
Normal Probability Plot of Coin Values
18
2.0
16
1.5
14
1.0
z score
Frequency
12
10
8
0.5
0.0
6
-0.5
4
-1.0
2
-1.5
0
0
120
240
value (cents)
360
480
0
100
200
300
value (cents)
400
500
326
CHAPTER 14 Statistical Process Control
9. Listed in order, the values are as follows.
0 0 0 0 0 0 0 0 0 0 0 0 5 10 20 23 35 35 40 43 62 73 90 100 130 178 185 200 250 500
The n=30 have Σx = 1979 and Σx2 = 469535.
a. x = (Σx)/n = 1979/30 = 66.0 cents
b. x = (20+23)/2 = 21.5 cents
c. s2 = [n(Σx2) – (Σx)2]/[n(n-1)]
= [30(469535) – (1979)2]/[30(29)] = 10169609/870 = 11689.2
s = 11689.2 = 108.1 cents
10. original claim: p > 0.50
p̂ = x/n = 18/30 = 0.60
Ho: p = 0.50
H1: p > 0.50
α = 0.05
C.V. z = zα = z0.05 = 1.645
calculations:
0.05
z p̂ = ( p̂ – μ p̂ )/σ p̂
^
p
0.50
= (0.60 – 0.50)/ (0.50)(0.50) / 30
z
0
1.645
= 0.10/0.09128
= 1.10
P-value = P(z>1.10) = 1 – 0.8643 = 0.1357
conclusion:
Do not reject Ho; there is not sufficient evidence to support the claim that p > 0.50. There
is not sufficient evidence to support the claim that most students have change in their
possession.
FINAL NOTE: Congratulations! You have completed statistics – the course that everybody likes to
hate. I trust that this manual has helped to make the course a little more understandable – and that
you leave the course with an appreciation of broad principles, and mot merely memories of
manipulating formulas. I wish you well in your continued studies, and that you achieve your full
potential wherever your journey of life may lead.