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University Of Jordan Faculty Of Engineering and Technology Industrial Engineering Department

University Of Jordan Faculty Of Engineering and Technology Industrial Engineering Department “Quality control project” Done by : Noor Omar Al- khateeb Duha Hani al -baik Ebtehal Al -amr Rand Al -saleh . Zakey industrials . Introduction.

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University Of Jordan Faculty Of Engineering and Technology Industrial Engineering Department

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  1. University Of JordanFaculty Of Engineering and Technology Industrial Engineering Department “Quality control project” Done by : Noor Omar Al-khateeb Duha Hani al -baik Ebtehal Al -amr Rand Al -saleh

  2. Zakey industrials

  3. Introduction Zakey Industries was formed in 1980. This was a unique partnership between two leading Arab entrepreneurial families and Tetra-Pak.Tetra-Pak provided the manufacturing unit, technical support and packaging technology. Zakey Industries provided the investment and the local market knowledge and influence.

  4. Their Vision To manufacture, and to continually improve, the finest quality juice beverages in each segment of the market, made from the finest ingredients sourced from all over the globe, by the use of the finest technology available, at the best possible value and choice for all health conscious consumers.

  5. Their Mission To provide the most professional service to all our distributors, agents and partners from all over the globe, by the means of having well trained, efficient, resourceful and customer orientated employees.

  6. Their Products 1-Zakeyis an innovative fruit juice drink that has become the choice of children and families all over Jordan.

  7. 2-Mounesh is a nectar product that contains not less than 35-50% natural fruit juice.  Full of the integrity, vigor and delight is designed for those who pay special attention to their health and demand a quality, lively and enriching beverage.

  8. 3-Bonanza is a high quality nectar product that contains a minimum of 35-50% Natural Fruit Juice.  Juice drink that is thirst quenching, replenishing and rejuvenating

  9. 4- Catch is an entertaining, appealing fruit flavored drink that is a favorite for children that is a favorite for children.

  10. 5-Fruitolu is the flagship product of Zakey Industries. It is made from 100% Natural Fruit Juice

  11. this video indicates to the process of manufacturing 125 ml juice (catch)

  12. Volume test video

  13. Other tests.. Examine the leakage

  14. Machine 6 Machine 1 128.0 127.0 126.0 126.5 127.0 125.5 127.5 129.0 128.0 127.0 128.5 128.0 127.0 128.5 127.5 128.0 127.0 129.0 129.0 127.0 129.0 127.0 125.5 128.5 125.0 129.0 125.0 126.0 127.5 126.5 127.0 126.5 128.5 127.0 • our data n=1 M=25

  15. Cont. Machine 6 Machine 1 125.0 128.5 128.0 126.5 127.0 126.5 127.5 126.5 126.5 125.0 126.8 127.5 126.5 127.0 126.5 128.0

  16. Histograms: Histogram is compact summery of data used for giving information about shape, variability of the data . Histogram of machine 1

  17. From the previous histogram we can say that the data are approximately normal we say that the data are symmetric if the mean and median are coincide but if the mean , median and mode do not coincide we say that the data are skewed. our data are skewed to the right since ( Mode < Median < Mean ) such that (126.5<127<127.07).

  18. Histogram of machine 6

  19. From the previous histogram we can indicate that the data is approximately normal. • But because( mode =median < mean);(127=127 <127.26) we can’t conclude that our data is symmetry, (our data is skewed to the right ).

  20. Probability plots: From this figure we can see that our data are normally distributed

  21. From the graph above our data can be considered normal.

  22. Stem-and-leaf: Stem and leaf is used to determine some important feature such as shape ,spread ,and middle of the data.

  23. Stem-and-Leaf Display: machine 1 Stem-and-leaf of machine 1 N = 25 Leaf Unit = 0.10   3 125 000 4 125 5 5 126 0 11 126 555558 (5) 127 00000 9 127 55 7 128 0 6 128 55 4 129 0000 From the stem and leaf display we can see that the median of our data=127 And it repeated 5 times

  24. Stem-and-Leaf Display: machine 6 Stem-and-leaf of machine 6 N = 25 Leaf Unit = 0.10 1 125 0 2 125 5 3 126 0 7 126 5555 (6) 127 000000 12 127 555 9 128 00000 4 128 555 1 129 0 • From the stem and leaf display we can see that the median of our data=127 And it repeated 6 times .

  25. Box plots: • Box plot is a graphical display that display many important feature such as variability, outliers ,central tendency, departure from symmetry .

  26. From the box plot we can easily notice that the data is skewed to the right, and the median =127.0 , Q1=126.5, Q3=128.25 ,"No outliers" Q3 Q2 Q1

  27. From the box plot we can easily notice that the data is skewed to the right, and the median =127.0 , Q1=126.5, Q3=128.0 ,"No outliers" Q3 Q2 Q1

  28. also we conclude from this box plot for the two machines(1&6) that whole data distribution is not exactly symmetric around the central value.

  29. Descriptive statistics: Descriptive Statistics: machine 1 Variable Mean SE Mean StDev Variance Minimum Q1 Median Q3 machine 1 127.07 0.253 1.27 1.60 125.00 126.50 127.00 128.25 N for Variable Maximum Range IQR Mode Mode machine 1 129.00 4.00 1.75 126.5; 127 5

  30. Descriptive Statistics: machine 6 Variable Mean SE Mean StDev Variance Minimum Q1 Median Q3 machine 6 127.26 0.196 0.980 0.961 125.00 126.50 127.00 128.00 N for Variable Maximum Range IQR Mode Mode machine 6 129.00 4.00 1.50 127 6

  31. hypothesis tests for the mean: Two-sample T for machine 1 vs machine 6 ( assumed 2-populations) N Mean StDev SE Mean machine 1 25 127.07 1.27 0.25 machine 6 25 127.260 0.980 0.20 Difference = mu (machine 1) - mu (machine 6) Estimate for difference: -0.188 95% CI for difference: (-0.833; 0.457) T-Test of difference = 0 (vs not =): T-Value = -0.59 P-Value = 0.560 DF = 45

  32. According to these results we can conclude : - The p-value = 0.560 which is larger than the p- value of α =0 .05 - Considering the confidence interval we find that the zero lies within the interval these notes shows that we fail to reject the null hypothesis (H0), which mean that the filling volume for machine 1 equals the filling volume for machine 6

  33. One-Sample T: machine 1 We interest in our production of juice that volume of our (catch) juice in the bottle must be 125 ml so we do one sample t test with α=0.05 to ensure that : H0: µ=125 H1: µ>125 Test of mu = 125 vs > 125 95% Lower Variable N Mean StDev SE Mean Bound T P machine 1 25 127.072 1.265 0.253 126.639 8.19 0.000 according to the above test : t= 8.19 , p- value = 0 , degree of freedom = 24 ,and because p-value < 0.05 ,we reject Ho and conclude that the filling volume more than 125ml . from the lower confidence interval (µ>126.925) , H0=125 that doesn't include in the CI.

  34. One-Sample T: machine 6 We interest in our production of juice that volume of our (catch) juice in the bottle must be 125 ml so we do one sample t test with α=0.05 to ensure that : H0: µ=125 H1: µ>125 95% Lower Variable N Mean StDev SE Mean Bound T P machine 6 25 127.260 0.980 0.196 126.925 11.53 0.000 t= 11.53 , p- value = 0 , degree of freedom = 24 , because p-value < 0.05 ,we reject Ho and conclude that the filling volume more than 125ml . from the lower confidence interval (µ>126.925) , H0=125 that doesn't include in the CI.

  35. test hypothesis for the variance : Test for Equal Variances: machine 1, machine 6 95% Bonferroni confidence intervals for standard deviations N Lower StDev Upper machine 1 25 0.954910 1.26508 1.85176 machine 6 25 0.739891 0.98022 1.43479 F-Test (normal distribution) Test statistic = 1.67, p-value = 0.219 From the hypothesis test we use α =0.05, and we get p-value=0.219>.05 so we conclude that filling volume variance is the same for the two machine (fail to reject Ho ).

  36. Control charts: The control chart is a graphical display of a quality characteristic that has been measured or computed from a sample versus the sample number or time. Because sample size n=1, we use moving range control chart (I-MR).

  37. From the previous control chart there is no indication that the process is out of control so we consider that the process is in control, which mean that there is no assignable cause exist in our process.

  38. From the previous control chart there is no indication that the process is out of control so we consider that the process is in control, which mean that there is no assignable cause exist in our process.

  39. Process capability: “Process capability of machine 1”

  40. From the graph we can see that the process mean is approximately the same as the target .and that only 63518.96 ppm (6.35%) are outside the specification limit

  41. “Process capability of machine 6”

  42. From the graph above we can see that the process mean is approximately the same as the target .and that only 14087.58 ppm (1.41%) are outside the specification limit

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