CHAPTER 18 Statistical Process Controls



Assignment six = 20 points:

1. In many educational institutions, the Athletic Department does a better job of teaching teamwork than do the other units. What skills can AET faculty learn about teaching teamwork by examining the analogy of a successful athletic team?

One thing that the Athletic department does is, it gives the team members a common goal and inspiration the achievement of winning the championship. The AD works the team together in drills even if the members do not like each other or have anything in common.

2. The tensile strength of a fiber used in manufacturing cloth is of interest to the purchaser. Previous experience indicates that the standard deviation of tensile strength is 2 psi. A random sample of eight fiber specimens is selected, and the average tensile strength is found to be 127 psi.

• Test the hypothesis that the mean tensile strength equals 125 psi versus the alternative that the mean exceeds 125 psi. Use α=0.05.

a) What is the P-value for this test?

b) Discuss why a one-sided alternative was chosen in part (a).

c) Construct a 95% lower confidence interval on the mean tensile strength.

A. H0 = 125 σ = 2 psi ẋ = 127 n =8

Ha > 125

Z0 =(ẋ-µ)/σ*√n = 127-125/2*√8=2.828

Za = .05 = 1.645

p-value = 1- Φ(Z0) = 1- Φ(2.828) = 1- 0.9977 =.00230

B. One sided was chosen because we are looking for a greater than value and not an “not equal value”

C. ẋ- Za(σ√n) ≤µ

127-1.645(2/√8)≤ µ

125.8 ≤ µ

3. (A machine is used to fill containers with a liquid product. Fill volume can be assumed to be normally distributed. A random sample of 10 containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.

a) Suppose that the manufacturer wants to be sure that the mean net content exceeds 12 oz. What conclusions can be drawn from the data (use α=0.01)?

ẋ =12.015 σ = .0303 n= 10 µ = 12

H0 = 12

Ha ≠ 12

a.

Test of mu = 12 vs not = 12

The assumed standard deviation = 0.03

Variable N Mean StDev SE Mean 99% CI Z P

C1 10 12.0150 0.0303 0.0095 (11.9906, 12.0394) 1.58 0.114

b.

b) Construct a 95% two-sided confidence interval on the mean fill volume.

Test of mu = 12 vs not = 12

The assumed standard deviation = 0.03

Variable N Mean StDev SE Mean 95% CI Z P

C1 10 12.0150 0.0303 0.0095 (11.9964, 12.0336) 1.58 0.114

4. Using the data bellow (Table 7.1 at your text) answer the following question (Using Minitab).

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a. Calculate mean, mode, median, standard deviation and variance for the data in Table 7.1.

Descriptive Statistics: Width

Variable N N* Mean SE Mean StDev Variance Minimum Maximum

Width 95 0 3.9948 0.000821 0.00800 0.000064 3.9680 4.0210

Descriptive Statistics: Gauge

Variable N N* Mean SE Mean StDev Variance Minimum Maximum

Gauge 95 0 0.24895 0.000432 0.00421 0.000018 0.23900 0.27300

b. Draw the Histogram for width and analyze the graph.

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c. Using correlation coefficient formula or Minitab, find the correlation coefficient between width and gauge. What does this mean?

Correlations: Width, Gauge

Pearson correlation of Width and Gauge = -0.125

P-Value = 0.227

5. Determine the trial central line and control limits for a p chart using the following data, which are for the payment of dental insurance claims. Plot the values on graph paper and determine if the process is stable. If there are any out -of-control points, assume an assignable cause and determine the revised central line and control limits.

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Trial

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Revised

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6. Determine the trial limits and revised control limits for a u chart using the data in the table for the surface finish of rolls of white paper. Assume any out -of-control points have assignable causes.

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Trial

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Revised

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7. An np chart is to be established on a painting process that is in statistical control. If 35 pieces are to be inspected every 4 hours, and the fraction nonconforming is 0.06, determine the central line and control limits.

Total number of parts = 35*4 = 140

UCL = 140 + (140*.06) = 148.4

Center line = 140

LCL = 140- (140*.06) = 131.6

8. A quality technician has collected data on the count of rivet nonconformities in four meters travel trailers. After 30 trailers, the total count of non-conformities is 316. Trial control limits have been determined and a comparison with the data shows no out-of-control points. What is the recommendation for the central line and the revised control limits for a count of nonconformities chart?

The recommendation should be to take the highest nonconformities and revise the limits to a tighter dispersion the rerun the test to see if the revision gives a better idea what is in control.

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