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?

When everyone from the team works together and does their part the team is successful. These types of teams know that just one person can carry the team it takes all of them to reach their goal. If the AET faculty would adopt these same teamwork principles they would be more successful.

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.

Null Hypothesis T = 125 Alternate Hypothesis T > 125

Standard Deviation = 2 psi n = 8 Xbar = 127psi alpha = .05%

Critical value = 1.895

Calculated value t = (127 – 125) / (2/√8) t = 2/.707106781187 = 2.83

Since 2.83 > 1.895 we reject the null hypothesis and accept the alternative

a) P value = 95%

b) The alternative is one sided because the alternate hypothesis was only set to be “greater than”

c) 95% Lower CI = 127 - 2.83(2/√8) = 127 - 2.001 = (124.999)

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)?

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

N=10 Standard Deviation = .0303 Xbar = 12.015 α = 0.01

Critical Value = 2.821

Null Hypothesis T > 12oz Alternate Hypothesis T ≤ 12oz

Calculated Value T = (12.015 – 12) / (.0303 / √10) = .015 / .00958170131 = 1.5655

a) calculated 1.5655 < the critical of 2.821 we would accept the null hypothesis that T > 12 oz

b) 95% 2 sided CI = 12.015 +/- (1.5655)(.0303/√10) = 12.015 +/- (.015000153401) = (11.99, 12.03)

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

[pic]

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

WIDTH

Mean = 3.9948, Mode = (my student version doesn’t find mode), Median = 3.994, Variance = 0.0000638, Standard Deviation = 0.00799

GAUGE

Mean = 0.24896, Mode = (N/A), Median = 0.24900, Variance = 0.0000179, Standard Deviation = 0.00423

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

[pic]

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 = -0.160

The correlation is negative, this indicates that there is a negative slope in the data indicating that when the Width increases then the Gauge decreases.

CC 1 = Positive slope of linear relationship

CC -1 = Negative slope of linear relationship

CC 0 = No slope / random relationship

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.

[pic]

Point 5 is out of control

Pbar = 0.01542 UCL = 0.03676 LCL = 0

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.

[pic]

Points 2, 4, 6, and 21 are out of control.

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 nonconfonning is 0.06, determine the central line and control limits.

Central Line P-bar = 0.06 = Σnp/Σn = Σnp/35

Σnp = 35(0.06) = 2.1

UCLnp = np-bar + 3√ (np-bar (1 – p-bar)) = 2.1 + 3√ (2.1 (1 - 0.06)) = 2.1 + 3√ 1.974 = 2.1 + 3(1.405) = 6.315

LCLnp = np-bar – 3√ (np-bar (1 – p-bar)) = 2.1 - 3√ 2.1 (1 – 0.06)) = 2.1 - 3√1.974 = 2.1 – 3(1.405) = -2.115

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?

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