Solved Problems

Solved Problem 15-1 The manufacturer of precision parts for drill presses produces round shafts for use in the construction of drill presses. The average diameter of a shaft is 0.56 inch. The inspection samples contain six shafts each. The average range of these samples is 0.006 inch. Determine the upper and lower control chart limits.

Solution

The mean factor $A_{2}$ $A_{2}$ from Table 15.2, where the sample size is six, is seen to be 0.483. With this factor, you can obtain the upper and lower control limits:

$\begin{array}{l} {UCL}_{\bar{x}} & = & 0.56 + (0.483) (0.006) \\ = & 0.56 + 0.0029 = 0.5629 \\ {LCL}_{\bar{x}} & = & 0.56 - 0.0029 \\ = & 0.5571 \end{array}$ $\begin{array}{l} {UCL}_{\bar{x}} & = & 0.56 + (0.483) (0.006) \\ = & 0.56 + 0.0029 = 0.5629 \\ {LCL}_{\bar{x}} & = & 0.56 - 0.0029 \\ = & 0.5571 \end{array}$
Solved Problem 15-2 Nocaf Drinks, Inc., a producer of decaffeinated coffee, bottles Nocaf. Each bottle should have a net weight of 4 ounces. The machine that fills the bottles with coffee is new, and the operations manager wants to make sure that it is properly adjusted. The operations manager takes a sample of $n = 8$ $n = 8$ bottles and records the average and range in ounces for each sample. The data for several samples are given in the following table. Note that every sample consists of 8 bottles.

Sample Sample Range Sample Average

A 0.41 4.00

B 0.55 4.16

C 0.44 3.99

D 0.48 4.00

E 0.56 4.17

F 0.62 3.93

G 0.54 3.98

H 0.44 4.01

Is the machine properly adjusted and in control?

Solution

We first find that $\bar{\bar{x}} = 4.03 and \bar{R} = 0.51.$ $\bar{\bar{x}} = 4.03 and \bar{R} = 0.51.$ Then, using Table 15.2, we find

$\begin{array}{l} {UCL}_{\bar{x}} & = & \bar{\bar{x}} + A_{2} \bar{R} = 4.03 + (0.373) (0.51) = 4.22 \\ {LCL}_{\bar{x}} & = & \bar{\bar{x}} - A_{2} \bar{R} = 4.03 - (0.373) (0.51) = 3.84 \\ {UCL}_{R} & = & D_{4} \bar{R} = (1.864) (0.51) = 0.95 \\ {LCL}_{R} & = & D_{3} \bar{R} = (0.136) (0.51) = 0.07 \end{array}$ $\begin{array}{l} {UCL}_{\bar{x}} & = & \bar{\bar{x}} + A_{2} \bar{R} = 4.03 + (0.373) (0.51) = 4.22 \\ {LCL}_{\bar{x}} & = & \bar{\bar{x}} - A_{2} \bar{R} = 4.03 - (0.373) (0.51) = 3.84 \\ {UCL}_{R} & = & D_{4} \bar{R} = (1.864) (0.51) = 0.95 \\ {LCL}_{R} & = & D_{3} \bar{R} = (0.136) (0.51) = 0.07 \end{array}$

It appears that the process average and range are both in control.
Solved Problem 15-3 Crabill Electronics, Inc., makes resistors, and among the last 100 resistors inspected, the percentage defective has been 0.05. Determine the upper and lower limits for this process for 99.7% confidence.

Solution

$\begin{array}{l} {UCL}_{p} & = & \bar{p} + 3 \sqrt{\frac{\bar{p} (1 - \bar{p})}{n}} = 0.05 + 3 \sqrt{\frac{(0.05) (1 - 0.05)}{100}} \\ = & 0.05 + 3 (0.0218) = 0.1154 \\ {LCL}_{p} & = & \bar{p} - 3 \sqrt{\frac{\bar{p} (1 - \bar{p})}{n}} = 0.05 - 3 (0.0218) \\ = & 0.05 - 0.0654 = 0 (since percent defective cannot be negative) \end{array}$ $\begin{array}{l} {UCL}_{p} & = & \bar{p} + 3 \sqrt{\frac{\bar{p} (1 - \bar{p})}{n}} = 0.05 + 3 \sqrt{\frac{(0.05) (1 - 0.05)}{100}} \\ = & 0.05 + 3 (0.0218) = 0.1154 \\ {LCL}_{p} & = & \bar{p} - 3 \sqrt{\frac{\bar{p} (1 - \bar{p})}{n}} = 0.05 - 3 (0.0218) \\ = & 0.05 - 0.0654 = 0 (since percent defective cannot be negative) \end{array}$

Sample	Sample Range	Sample Average
A	0.41	4.00
B	0.55	4.16
C	0.44	3.99
D	0.48	4.00
E	0.56	4.17
F	0.62	3.93
G	0.54	3.98
H	0.44	4.01

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