Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Transcript

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? In any successful athletic team they one key ingredient that the team has is trust. If you do not trust your teammate to be where you are supposed to be, or run the right play, then you are not going to be very successful. Of course trust is a two way street and is earned not given. But the most important skill is leadership; leadership is having the ability to draw the line between being a friend and a leader and getting your team to the right spot. I played basketball here at MSU and when we stepped between the lines I had no friends, I did everything I could to lead and make sure my team was ready for the next game. 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. One-Sample Z Test of mu = 125 vs > 125 The assumed standard deviation = 2 95% Lower N Mean SE Mean Bound Z P 8 127.000 0.707 125.837 2.83 0.002 Critcal value=1.895 (found in book n-1) Z value is 2.83 So since 2.83 is greater than 1.895 reject the hypothesis (a) What is the P-value for this test? 0.002 (b) Discuss why a one-sided alternative was chosen in part Because the test was set to be greater than (c) Construct a 95% lower confidence interval on the mean tensile strength. 95% lower= 127-2.83(2SQrt8)=125 (124.999999) 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)? One-Sample T: C1 Test of mu = 12 vs > 12 99.99% Variable N Mean StDev SE Mean Lower Bound T C1 10 12.0150 0.0303 0.0096 11.9575 1.57 0.076 P Since the critical value, 2.82, is less than the calculated value, 1.57, you can accept the hypothesis (b) Construct a 95% two-sided confidence interval on the mean fill volume. CI = 12.015 - (1.5655)(.0303/SQRT10) = 11.99 C2= 12.015 + (1.5655)(.0303/SQRT10) = 12.03 Not sure if this was done correctly 4. Using the data bellow (Table 7.1 at your text) answer the following question (Using Minitab). a. Calculate mean, mode, median, standard deviation and variance for the data in Table 7.1. b. Draw the Histogram for width and analyze the graph. c. 5. Using correlation coefficient formula or Minitab, find the correlation coefficient between width and gauge. What does this mean? 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. P Chart of C1 0.07 1 0.06 Proportion 0.05 0.04 UCL=0.04016 0.03 _ P=0.01747 0.02 0.01 0.00 LCL=0 1 3 5 7 9 11 13 15 17 19 21 23 25 Sample UCL=0.04016 Pbar=0.01747 and LCL=0 There are points out of control so we revisit the data and take out any point that are out of control P Chart of C1 0.04 UCL=0.03676 Proportion 0.03 0.02 _ P=0.01542 0.01 0.00 LCL=0 1 3 5 7 9 11 13 15 17 19 21 23 Sample As you can see this makes a change the values and makes the process in control. 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. Test Results for U Chart of C1 TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 2, 6, 21 * WARNING * If graph is updated with new data, the results above may no * longer be correct. U Chart of C1 1 UCL=5.028 Sample Count Per Unit 5 4 _ U=3.304 3 2 LCL=1.579 1 1 1 0 1 4 7 10 13 16 Sample 19 22 25 28 Points 2, 6, and 21 need to be taken out and the data needs to be revisited. U Chart of C1 5.5 UCL=5.169 Sample Count Per Unit 5.0 4.5 4.0 _ U=3.416 3.5 3.0 2.5 2.0 LCL=1.663 1.5 1 3 5 7 9 11 13 15 Sample 17 19 21 23 25 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. Equations taken from CH. 7 power point npbar = 35(0.06) = 2.1 UCL = 2.1 + 3 SQRT (2.1 (1 - 0.06)) = 2.1 + 3 SQRT (1.974) = 2.1 + 3(1.405) = 6.315 LCL = 2.1 – 3 SQRT (2.1 (1 – 0.06)) = 2.1 – 3 SQRT (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 outof-control points. What is the recommendation for the central line and the revised control limits for a count of nonconformities chart? Since there are no out of control points this could mean that there no real threats within the units themselves and they can still be processed and sold to the general population. If any of the points had been out of control then I would recommend discarding that particular point and revisiting the data.