A textbook of Computer Based Numerical and Statiscal Techniques part 51 ppsx

486 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Negative LCL being taken as zero. Also for drawing the control chart we mark the sample No.’s along the horizontal axis and control limits and central line marked along the vertical axis. Finally the number of defects. (c i ) per inspection units are marked in the c- chart. From the control chart, we observe that the point corresponding to 6th inspection unit goes beyond UCL showing a out-of-control situation. So for computing revised control limits we omit this unit and use the remaining 15 inspection units for the purpose. The average number of defects in the remaining 15 units is 15 1 134 2.27 15 15 i i cc = === ∑ so the revised limits for c-chart are: UCL 3 2.27 3 2.27 2.27 4.52 6.79 CL 2.27 LCL 3 2.27 3 2.27 2.27 4.52 2.25 0 cc c cc =+=+ =+= == =−=− =−=−≈ Negative LCL being as zero. Example 12. A food company puts mango juice into cans advertised as containing 10 ounces of the juice. The weights of the juice drained from cans immediately after filling for 20 samples are taken by a random method (at an interval of every 30 minutes). Each of the samples includes 4 cans. The samples are tabulated in the following table. The weights in the table are given in units of 0.01 ounces in excess of 10 ounces. For example, the weight of juice drained from the first can of the sample is 10.15 ounces whch is in excess of 10 ounces being 0.15 ounces (10.15 – 10 = 0.15) since the unit in the table is 0.01 ounce, the excess is recorded as 15 units in the table. Construct an x — -chart to control the weights of mango juice for the filling. Weight of each can (4 cans in each sample, x, n = 4) Sample Number x 1 x 2 x 3 x 4 11512 1320 210 8 814 3 8 15 17 10 41217 1112 51813 15 4 62016 1420 71519 2317 81323 1416 998185 10 6 10 24 20 11 5 12 20 15 12 3 15 18 18 13 6 18 12 10 14 12 9 15 18 STATISTICAL QUALITY CONTROL 487 15 15 15 6 16 16 18 17 8 15 17 13 16 5 4 18 10 20 8 10 19 5 15 10 12 20 6 14 12 14 Sol. Total Sample Sample Weight of each can weight of Mean Range (4 cans in each sample, x, n = 4) 4 cans x 1 x 2 x 3 x 4 ∑x x x= 4 ∑ R = x max – x min 1 15 12 13 20 60 15.0 8 2 10 8 8 14 40 10.0 6 3 8 15 17 10 50 12.5 9 4 12 17 11 12 52 13.0 6 5 18 13 15 4 50 12.5 14 6 20 16 14 20 70 17.5 6 7 15 19 23 17 74 18.5 8 8 13 23 14 16 66 16.5 10 9 9 8 18 5 40 10.0 13 10 6 10 24 20 60 15.0 18 11 5 12 20 15 52 13.0 15 12 3 15 18 18 54 13.5 15 13 6 18 12 10 46 11.5 12 14 12 9 15 18 54 13.5 9 15 15 15 6 16 52 13.0 10 16 18 17 8 15 58 14.5 10 17 13 16 5 4 38 9.5 12 18 10 20 8 10 48 12.0 12 19 5 15 10 12 42 10.5 10 20 6 14 12 14 46 11.5 8 Total ∑ x — = 263.0 ∑R = 211 2 UCL xAR =+ = 13.15 + 0.729 × 10.55 (A 2 = 0.729 for n = 4) 2 UCL 20.84095 CL 13.15 LCL x xAR = == =− = 13.15 – 0.729 × 10.55 = 5.46 Sample Number 488 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES The values in above computation are expressed in units of 0.01 ounces in excess of 10 ounces. The actual value of UCL = 10.2084, and LCL = 10.0546 ounces. Since all points are falling with in control limits the process is in a statistical control. Now since standards are not given calculating 1. The mean of the sample mean x is given by 263 13.15 20 20 x x === ∑ 2. The mean of the Range values R is given by 211 10.55 20 20 R R === ∑ 3. Trial control limits for x -chart 0 5 10 15 20 25 20 15 10 5 0 FIG. 11.27 PROBLEM SET 11.1 1.1. 1.1. 1. A machine is set to deliver packets of a given weight 10 samples of size 5 each were recorded. Data being given below: Sample no. 12345678910 Mean x — 15 17 15 18 17 14 18 15 17 16 Range R 774987124115 Calculate the values for the central line and control limits of mean chart and the range and then comment on the state of control. Given for 5,n = 2 0.58, A = 3 0, D = 4 2.115. D = Ans. UCL 15.614 UCL 20.492 LCL 0 LCL 11.908 CL 7.4 CL 16.2 R x Rx Rx  ==   ==    ==  STATISTICAL QUALITY CONTROL 489 2.2. 2.2. 2. The data below give the number of defective bearing in samples of size 150. Construct p-chart for these data and state your comment. . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 . 12754159015674136810527 Sample no No of defective Compute control limits for p-chart. [Ans. UCL = 0.08650, CL = 0.03905, LCL = 0] 3.3. 3.3. 3. A process produces rubber belts in lots of size 2300. Inspection of the last 20 lots reveals the following data: . 1 2 3 4 5 6 7 8 9 1011121314151617181920 . 308 342 311 285 327 230 346 221 435 230 407 221 269 131 414 198 331 285 394 456 Lot no No of defective belts Compute control limits for p-chart. [Ans. UCL p = 0.1548, CL p = 0.1335, LCL p = 0.1122] 4.4. 4.4. 4. The following figure give the number of defectives in 20 samples, each sample containing 2,000 items. 425 430 216 341 225 322 280 306 337 305 356 402 216 264 126 409 193 326 280 389 Calculate the control limits for fraction defective chart (p-chart). Draw the p-chart and state the comment. [Ans. UCL p = 0.178, CL p = 0.154, LCL p = 0.130] 5.5. 5.5. 5. An inspection of 10 samples of size 400 each from 10 lots revealed the following no. of defective units; 17, 15, 14, 26, 9, 4, 19, 12, 9, 15. Calculate control limits for the no. of defective units. Plot the control limits and the observations and state whether the process is under control or not. [Ans. UCL np = 25.02679, CL np = 14, LCL np = 2.97231] 6.6. 6.6. 6. The following data refer to visual defects found during inspection of the first 10 samples of size 100 each. Use them to obtain upper and lower control limits for percentage defective in sample of 100. . 12345678 910 . 4811311771612 6 Sample no No of defective [Ans. UCL np = 16.87, CL np = 8.5, LCL np = 0.13] 490 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 7.7. 7.7. 7. The pieces of cloth out of the different rolls of equal length contained the following number of defects: 3028421371 prepare a c-chart and state whether the process is in a statistical control? [Ans. UCL c = 8.38, CL c = 3.1, LCL c = 0] 8.8. 8.8. 8. The following table gives the no. of defects in carpets manufactured by a company. .12345678910 . 345633536 2 Carpet serial no No of defective Determine the control line and the control limits for c-chart. 9.9. 9.9. 9. The following data relate to the number or break downs in the rubber covered wires in 24 successive lengths of 10,000 feet each. 8113712611 10 5 0 19 16 20 1 6 12 4 2375 Draw c-chart and state your comment. [Ans. UCL c = 13.0715, CL c = 5.875, LCL c = 0 (Process out of control)] 10.10. 10.10. 10. A drilling machine bores holes with a mean diameter of 0.5230 cm. and a standard deviation of 0.0032 cm. Calculate the 2-sigma and 3-sigma upper and lower control limits for mean of sample of 4. Ans. 2–sigma UCL 0.5262 cm LCL 0.5198 cm CL 0.5230 cm =   =   =  3-sigma UCL 0.5278 cm LCL 0.5182 cm CL 0.5230 cm =   =   =  GGG STATISTICAL QUALITY CONTROL 491 FACTORS USEFUL IN THE CONSTRUCTION OF CONTROL CHARTS Mean Chart Standard Deviation Chart Range Chart Sample Factors for Factors for Factors for control limits Factors for Factors for control limit size control limits central line central line n A A 1 A 2 C 2 B 1 B 2 B 3 B 4 d 2 D 1 D 2 D 3 D 4 2 2.121 3.760 1.880 0.5642 0 1.843 0 3.267 1.128 0 3.686 0 3.267 3 1.732 2.394 1.023 0.7236 0 1.858 0 2.568 1.693 0 4.358 0 2.575 4 1.500 1.880 0.729 0.7979 0 1.808 0 2.266 2.059 0 4.698 0 2.282 5 1.342 1.596 0.577 0.8407 0 1.756 0 2.089 2 326 0 4.918 0 2.115 6 1.225 1.410 0.483 0.8686 0.026 1.711 0.030 1.970 2.534 0 5.078 0 2.004 7 1.134 1.277 0.419 0.8882 0.105 1.672 0.118 1.882 2.704 0.205 5.203 0.076 1.924 8 1.061 1.175 0.373 0.9027 0.167 1.638 0.185 1.815 2.847 0.387 5.307 0.136 1.864 9 1.000 1.094 0.337 0.9139 0.219 1.609 0.239 1.761 2.970 0.546 5.394 0.184 1.816 10 0.949 1.028 0308 0.9227 0.262 1.584 0.284 1.716 3.078 0.687 5.469 0.223 1.777 11 0.905 0.973 0.285 0.9300 0.299 1.561 0.321 1.679 3.173 0.812 5.534 0.256 1.744 12 0.866 0.925 0.266 0.9359 0.331 1.541 0.354 1.646 3.258 0.924 5.592 0.284 1.716 13 0.832 0.884 0.249 0.9410 0.359 1.523 0.382 1.618 3.336 1.026 5.646 0.308 1.692 14 0.802 0.848 0.235 0.9443 0.384 1.507 0.406 1.594 3.407 1.121 5.693 0.329 1.671 15 0.775 0.816 0.223 0.9490 0.406 1.492 0.428 1.572 3.472 1.207 5.737 0.348 1.652 16 0.750 0.788 0.212 0.9523 0.427 1.478 0.448 1.552 3.532 1.285 5.779 0.364 1.636 17 0.728 0.762 0.203 0.9551 0.445 1.465 0.466 1.534 3.588 1.359 5.817 0.379 1.621 18 0.707 0.738 0.194 0.9576 0.461 1.454 0.482 1.518 3.640 1.426 5.854 0.392 1.608 19 0.688 0.717 0187 0.9599 0.477 1.443 0.497 1.503 3.689 1.490 5.888 0.404 1.596 20 0.671 0.697 0.180 0.9619 0.491 1.433 0.510 1.490 3.735 1.548 5.922 0.414 1.586 21 0.655 0.679 0.173 0.9638 0.504 1.424 0.523 1.477 3.778 1.606 5.950 0.425 1.575 22 0.640 0.662 0.167 0.9655 0.516 1.415 0.534 1.466 3.819 1.659 5.979 0.434 1.566 23 0.626 0.647 0.162 0.9670 0.527 1.407 0.545 1.455 3.858 1.710 6.006 0.443 1.557 24 0.612 0.632 0.157 0.9684 0.538 1.399 0.555 1.445 3.395 1.759 6.031 0.452 1.548 25 0.600 0.619 0.153 0.9696 0.548 1.392 0.565 1.435 3.931 1.804 6.058 0.459 1.541 CHAPTER 12 Testing of Hypothesis 12.1 INTRODUCTION Suppose some business concern has an average sale of Rs. 10000/- daily estimated over a long period. A salesman claims that he will increase the average sales by Rs. 700/- a day. The concern is interested in an increased sale no doubt, but how to know whether the claim of the man is justified or not? For this some such a mathematical model for the population of increased sales is assumed which agrees to the maximum with the practical observations. In the example given, let us assume that the claim of the girl about her sales is justified and that the increase in sales is normally distributed with mean µ = 700 and variance σ 2 . This assumption is called statistical hypothesis. Thereafter the suitability of the assumed model is examined on the basis of the sale observations made. This procedure is called testing of hypothesis. A statistical hypothesis is some statement or assertion about a population or equivalently about the probability distribution characterising a population which we want to verify on the basis of information available from a sample. If the statistical hypothesis specifies the population completely then it is termed as a simple statistical hypothesis, otherwise it is called a composite statistical hypothesis. Example: If X 1 , X 2 , , X n is a random sample of size n from a normal population with mean µ and variance σ 2 , then the hypothesis. H 0 : µ= µ 0 , σ 2 = σ 0 2 is a simple hypothesis, whereas each of the following hypothesis is a composite hypothesis: (1) µ = µ 0 (2) σ 2 = σ 0 2 (3) µ = µ 0 , σ 2 < σ 0 2 (4) µ < µ 0 , σ 2 > σ 0 2 (5) µ < µ 0 , σ 2 = σ 0 2 (6) µ = µ 0 , σ 2 > σ 0 2 (7) µ > µ 0 , σ 2 = σ 0 2 A hypothesis which does not specify completely ‘r’ parameters of a population is termed as a composite hypothesis with r degrees of freedom. 492 TESTING OF HYPOTHESIS 493 12.2 SOME IMPORTANT DEFINITIONS Test of a Statistical Hypothesis: A test of a statistical hypothesis is a two action decision problem after the experimental sample values have been obtained, the two actions being the acceptance or rejection of the hypothesis under consideration. Null Hypothesis: The statistical hypothesis tested under the assumption that it is true is called null hypothesis. It is tested on the basis of the sample observations and is liable to be rejected as well, depending upon the outcome of the statistical test applied. There are many occasions where null hypothesis is formulated for the sole purpose of rejecting it. In other words, null hypothesis is statement of zero or no change. If the original claim includes equality (< =, =, or > = ), it is the null hypothesis. If the original claim does not include equality (<, not equal, >) the null hypothesis is the complement of original claim. The null hypothesis always includes the equal sign. The decision is based on the null hypothesis. The null hypothesis is denoted by H 0 . Alternative Hypothesis: Statement which is true if the null hypothesis is false is known as alternative hypothesis. In other words a possible or the acceptable alternative to the null hypothesis called alternative hypothesis, and is denoted by H 1 . It testing if H 0 is rejected, then H 1 is accepted. The type of test (left, right, or two tail) is based on the alternative hypothesis. Type I Error and Type II Error: When a null hypothesis H 0 is tested against an alternative H 1 , then there can be either of the following two types of errors: (a) Rejecting the null hypothesis H 0 when actually it is true (b) Failing to reject the null hypothesis when it is false These are called errors of Type I and Type II and denoted by α and β respectively. The other two possible outcomes of testing are: (c) Rejection of H 0 when it was wrong and (d) Acceptance of H 0 when it was true. H 0 is true H 1 is true (H 0 is false) Accept H 0 Correct decision Type II error (β) Accept H 1 (reject H 0 ) Type I error (α) Correct decision Alpha: The probability of rejecting H 0 , when it was true = The probability of committing type I error = The size of type I error = α. Beta: The probability of accepting H 0 , when it was wrong = The probability of committing type II error = The size of type II error = β. Level of Significance: Alpha, the probability of type I error is known as the level of significance of the test. It is also called the size of the critical region. In other words, the maximum value of type I error which we would be willing to risk is called level of significance of the test. In general, 0.05 and 0.01 are the commonly accepted values of the levels of the significance. When the level of significance is 0.05, it simply means that on the average in 5 chances out of 100 we are likely to reject a correct H 0 . Probability (P-Value) Value: The probability of getting the results obtained if the null hypothesis is true. If this probability is too small (smaller than the level of significance), then we 494 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES reject the null hypothesis. If the level of significance is the area beyond the critical values, then the probability value is the area beyond the test statistic. Test Statistic: “Sample statistic used to decide whether to reject or fail to reject the null hypothesis”. Critical Region: Set of all values which would cause us to reject H 0 . Suppose the sample values x 1 , x 2 , x n determine a point E on the n-dimensional sample space S which would be the set of the various sample points corresponding to the all possible outcomes of the experiment. The testing of statistics hypothesis is made on the basis of the division of this sample space into two mutually exclusive regions: (1) Acceptance region (2) Rejection (critical region) region of H 0 The null hypothesis H 0 is rejected as soon as the sample points falls in the critical region of the sample space S. The region of rejection is denoted either by R or by C. Critical region R ` S acceptance region A _ S A + R = S FIG. 12.1 The null hypothesis is accepted as soon as the sample point falls in the acceptance region, which is denoted by A. The values which separates the critical region from the non-critical region is known as critical values. The critical values are determined independently of the sample statistics. Decision: Decision is a statement based upon the null hypothesis. It is either “reject the null hypothesis” or ”fail to reject the null hypothesis” we will never accept the null hypothesis. Conclusion: Conclusion is a statement which indicates the level of evidence (sufficient or insufficient), at what level of significance, and whether the original claim is rejected (null) or supported (alternative). Unbiased Critical Region: A critical region is said to be unbiased if the size of type II error β comes out to be less than the size of type I error. 12.3 UNDERSTANDING THE TYPE OF TEST The type of test is determined by the Alternative Hypothesis (H 1 ). The following way explain how to determine if the test is a left tail, right tail, or two tail test. (a) Left Tailed Test H 1 : Parameter < value Notice that the inequality points to the left. A R S TESTING OF HYPOTHESIS 495 Decision Rule: Reject H 0 if t.s.< c.v. Critical region Non Critical region Critical value α FIG. 12.2 (b) Right Tailed Test H 1 : Parameter > value Notice that the inequality points to the right. Decision Rule: Reject H 0 if t.s. > c.v. Critical region Non Critical region Critical value α FIG. 12.3 (c) Two-Tailed Test H 1 : Parameter not equal value another way to write not equal is < or > Notice that the inequality points to both sides. Decision Rule: Reject H 0 if t.s. < c.v. (left) or t.s. > c.v. (right) Critical region Non Critical region Critical value Critical value α/2 α/2 FIG. 12.4 Critical region . 486 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Negative LCL being taken as zero. Also for drawing the control chart we mark the sample No.’s along the horizontal axis and control. Number 488 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES The values in above computation are expressed in units of 0.01 ounces in excess of 10 ounces. The actual value of UCL = 10.2084, and LCL =. into cans advertised as containing 10 ounces of the juice. The weights of the juice drained from cans immediately after filling for 20 samples are taken by a random method (at an interval of every