A time study has been conducted. The steps on producing one bean bag were documented. Performance rating and observed time are summarized below. What is the standard time for producing one bean bag, given that allowances total to 10%? Steps to Produce One Bean Bag: Steps Performance Rating (%) Mean Observed Time (mins) A (Cutting) 95 1.02 B (Sewing) 90 5.30 C (Stuffing) 105 1.15 D (Sealing) 100 2.53 E (Attach accesories) 105 1.95 F (Cleaning) 95 2.65 G (Packing) 90 1.73
Recall that time study is a work measurement technique for recording the time of performing a certain specific job (or its elements) carried out under specified conditions and analyzing the data so as to obtain the time necessary for an operator to carry it out at a defined rate of performance The process of time study can be summarized by the following steps: 1. Selecting the operator. 2. Securing and recording information about the operation and operator being studied. 3. Analyzing the job and breaking it down into its elements. 4. Recording the elapsed elemental values. 5. Determining the number of cycles to be timed. 6. Rating the operator’s performance. 7. Assigning appropriate allowances. 8. Determining the standard time for the operation. From the problem above, steps 1 to 7 has already been done. We now proceed with the computation and determination of the standard time for the operation, which in this case is the production of one bean bag. The standard time is defined as the time required for a fully qualified, trained operator, working at a standard pace and exerting average effort, to perform the operation is termed the standard time. It is given by the formula: ST = NT ( 1 + allowance) where ST = standard time, NT = normal time, and allowance = allowance as a fraction of the normal time. NT can be computed from the formula: NT = OT (R/100) where OT = mean observed time and R = performance rating of the operator expressed as a percentage, with 100 percent being a standard performance by a qualified operator. The basic principle of performance rating is to adjust the mean observed time for each element performed during the study to normal time that would be required by the qualified operator to perform the same work. Thus, ST = [ OT (R/100) ] [ 1 + allowance ] A. Normal Time for Each Step Steps Normal Time (mins) A (Cutting) 1.02 (95/100) = 0.969 B (Sewing) 5.30 (90/100) = 4.77 C (Stuffing) 1.15 (105/100) = 1.2075 D (Sealing) 2.53 (100/100) = 2.53 E (Attach accesories) 1.95 (105/100) = 2.0475 F (Cleaning) 2.65 (95/100) = 2.5175 G (Packing) 1.73 (90/100) = 1.557 B. Standard Time for Each Step Steps Standard Time (mins) A (Cutting) 0.969 (1 + 0.10) = 1.0659 B (Sewing) 4.77 (1 + 0.10) = 5.247 C (Stuffing) 1.2075 (1 + 0.10) = 1.32825 D (Sealing) 2.53 (1 + 0.10) = 2.783 E (Attach accesories) 2.0475 (1 + 0.10) = 2.25225 F (Cleaning) 2.5175 (1 + 0.10) = 2.76925 G (Packing) 1.557 (1 + 0.10) = 1.7127 C. Standard Time for Production of One Bean Bag Steps Standard Time (mins) A (Cutting) 1.0659 B (Sewing) 5.247 C (Stuffing) 1.32825 D (Sealing) 2.783 E (Attach accessories) 2.25225 F (Cleaning) 2.76925 G (Packing) 1.7127 Total Standard Time: 17.15835 Conclusion: The standard time for the production of one bean bag is 17.15835 mins.
Ten measurements of impact energy (J) on specimens of A238 steel cut at 60 degrees C are listed as follows: - 64.1 - 64.7 - 64.5 - 64.6 - 64.5 - 64.3 - 64.6 - 64.8 - 64.2 - 64.3 Assume that impact energy is normally distributed with a variance of 1 J^2. Test the hypothesis that the mean impact energy is 65 J.
For the introduction to hypothesis testing, we can divide the general procedure into the following 8 steps. The steps are the following: 1. Identify the Parameter of Interest. 2. State the Null Hypothesis. 3. State the Alternative Hypothesis. 4. State the Level of Significance. 5. Identify Test Statistic. 6. Determine the Rejection Region. 7. Compute the Test Statistic. 8. State the Conclusion. Following the steps mentioned above, we solve the problem. 1. Identify the Parameter of Interest. - That is, 𝜇 or the mean impact energy. 2. State the Null Hypothesis (Ho). - The null hypothesis (Ho): 𝜇 = 65J. 3. State the Alternative Hypothesis. - The alternative hypothesis (Ho): 𝜇 ≠ 65J. 4. State the Level of Significance (𝛼). - Since the level of significance is not given, we can assume it to be 0.05. This assures us of 95% confidence level. That is, out of 100 samples, 95 of the samples will contain the true value of the population. 5. Identify Test Statistic. - Since 𝜎^2 (variance) is known, we use the z-statistic: z0 = (X-bar - u0) / (𝜎/sqrt(n)) where X-bar is the sample mean, 𝜎 is the standard deviation, and n is the number of measurements in a sample. Note: In z0 and u0, 0 is a subscript. 6. Determine the Rejection Region. - Based on the problem, it can be identified the test should be two-sided. Thus: 𝑅𝑒𝑗𝑒𝑐𝑡 𝑖𝑓 |𝑧0| > 𝑧𝛼/2 𝑜𝑟 𝑧0.025 Looking at the z-table. 𝑅𝑒𝑗𝑒𝑐𝑡 𝑖𝑓 |𝑧0| > 1.96 Note: In z0, 0 is a subscript. In z𝛼/2, 𝛼/2 is a subscript. 7. Compute the Test Statistic. - Substitute on values on the test statistic: z0 = (X-bar - u0) / (𝜎/sqrt(n)). X-bar can be computed by getting average or mean of the given values: - 64.1 - 64.7 - 64.5 - 64.6 - 64.5 - 64.3 - 64.6 - 64.8 - 64.2 - 64.3 a. How do we get the mean (X-bar)? Summation of all the values divided by the number of values. i. What is the sum of the given values? 644.6. ii. How many given values? 10 Thus, the mean (X-bar) is 64.46. b. How do we get the standard deviation (𝜎)? Get the sqrt of the variance (𝜎^2). Given that the variance (𝜎^2)? 1 J^2. The standard deviation (𝜎) is 1 J. c. How many measurements in a sample (n)? n = 10. d. Substituting values on the test statistic. i. z0 = (X-bar - u0) / (𝜎/sqrt(n)). ii. z0 = (64.46 - 65) / (1/sqrt(10)) iii. z0 = - 1.70762993649 iv. | z0 | = 1.70762993649 8. State the Conclusion. - We state the conclusion by comparing the test statistic to the rejection criteria. Since 𝑧0 < 1.96, we do not reject H0 at 𝛼 = 0.05. There is no sufficient evidence to say that the mean impact energy is not equal to 65 J.
There are two bottles that contain 4% and 16% acid solutions. Two liters of a 7% acid solution is needed. How much of each bottle should be mixed together to obtain the needed solution?
Let x be the liters of solution from the first bottle. Since the mixed (or needed) solution should be 2 liters, the liters of solution from the second bottle is given by 2 - x. Given: % Concentration of Solution 1: 4% or 0.04 Liters of Solution 1: (x) liters % Concentration of Solution 2: 16% or 0.16 Liters of Solution 2: (2 - x) liters % Concentration of Mixed Solution: 7% or 0.07 Liter of Mixed Solution: 2 liters Based on the given, we can express the solution to the problem by the equation of the concentration of acid: (Eq. 1) 0.04x + 0.16(2 - x) = 0.07(2) Simplifying the second term of the left-hand side (LHS): (Eq. 2) 0.04x - 0.16x + 0.32 = 0.07(2) (Eq. 3) - 0.12x + 0.32 = 0.07(2) Transposing the second term from Eq. 3 to the right-hand side (RHS) and simplifying the RHS, we get: (Eq. 4) - 0.12x = 0.14 - 0.32 (Eq. 5) - 0.12x = - 0.18 Solving for x, that is by dividing both sides by the coefficient of x in the LHS which is -0.12, we have: (Eq. 6) (-0.12x)/(-0.12) = (-0.18)/(-0.12) (Eq. 7) x = 1.5 liters Thus, the amount of solution 1 (4%) should be 1.5 liters. On the other hand, the amount of solution 2 (16%) should be (2 - x) or (2 - 1.5), which is 0.5 liter. Conclusion: One and a half liters of the 4 percent solution and a half liter of the 16 percent solution are needed.