Statistics Exam

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Mathematics

ENGR 2080

The Pennsylvania State University

ENGR

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Note: Your submission must be one PDF file. Please scan clearly. Unclear presentation may result in point deduction. NAME: SIGN: Total: /40 1-1 1-2 /2 2-2 1-3 /2 2-3 /2 4-2 3-1 4-3 5-4 /2 /2 /2 5-3 /2 7-1 /2 /2 4-1 5-2 6-2 /2 /2 /2 /2 2-1 3-2 5-1 6-1 /2 /2 /2 /2 1-4 /2 7-2 /2 /2 1 1. An article shows data to compare several methods for predicting the shear strength for steel plate girders. Data for two of these methods, the Karlsruhe and Lehigh procedures, when applied eight times to a girder, are shown in the following table. We wish to determine whether there is any difference between the two methods. Assume the same standard deviations. (Note that we will perform unpaired comparison. We do not perform a paired test.) Karlsruhe Method 1.224487 1.170828 1.229321 1.200906 1.28087 1.201107 1.287458 1.186672 1.227283 1.211963 Lehigh Method 1.140994 1.207858 1.190368 1.148473 1.167855 1.179984 1.196009 1.154533 1.166998 1.212248 1.1. (2 points) Draw the normal probability plots for two samples using Microsoft Excel Data Analysis Toolpak. What are their R-squared? 2 1.2. (2 points) For a sample standard deviation, use Excel function STDEV.S(), not STDEV.P(). Fill in the blanks below: Karlsruhe Method Lehigh Method 𝑥𝑥̅1 = 𝑥𝑥̅2 = 𝑛𝑛1 = 𝑛𝑛2 = 𝑠𝑠1 = 𝑠𝑠2 = 1.3. (2 points) Do the data suggest that the two methods provide the same mean shear strength? Use α = 0.05. Find the P-value rounded to 6 decimal places (using Microsoft Excel.) 1.4. (2 points) Find a 95% confidence interval on the mean difference between the two methods and use it to answer the question in 1.3. 3 2. An article reports a comparison of several methods for predicting the shear strength for steel plate girders. Data for two of these methods, the Karlsruhe and Lehigh procedures, when applied to six specific girders, are shown in the following table. We wish to determine whether there is any difference (on the average) between the two methods. Use the paired t-test. Girder S1/1 S2/1 S3/1 S4/1 S5/1 S2/1 S2/2 Karlsruhe 1.186 1.151 1.322 1.339 1.2 1.402 1.365 Lehigh 1.061 0.992 1.063 1.062 1.065 1.178 1.037 2.1. (2 points) Draw the normal probability plot for the difference using Microsoft Excel Data Analysis Toolpak. What is the R-squared? 4 2.2. (2 points) Do the data suggest that the two methods provide the same mean shear strength? Use α = 0.01. Find the P-value. 2.3. (2 points) Construct a 99% confidence interval and comment on why the methods are the same or different in mean shear strength. 5 3. An article shows data to compare several methods for predicting the shear strength for steel plate girders. Data for two of these methods, the Karlsruhe and Lehigh procedures, when applied eight times to a girder, are shown in the following table. We wish to determine whether there is any difference between the two methods. Assume the population standard deviations are known to be 𝜎𝜎 1 = 𝜎𝜎 2 = 0.027. Karlsruhe Method 1.224487 1.170828 1.229321 1.200906 1.28087 1.201107 1.287458 1.186672 1.227283 1.211963 Lehigh Method 1.140994 1.207858 1.190368 1.148473 1.167855 1.179984 1.196009 1.154533 1.166998 1.212248 3.1. (2 points) Do the data suggest that the two methods provide the same mean shear strength? Use α = 0.01. Find the P-value rounded to 6 decimal places (using Microsoft Excel.) 3.2. (2 points) Find a 99% confidence interval on the mean difference between the two methods and use it to answer the question in 3.1. 6 4. Wonder Shed Inc. is a manufacturer of storage sheds. The manufacturing process involves the procurement of sheets of steel that will be used to form both the roof and the base of each shed. The first step involves separating the material needed for the roof from that needed for the base. Then the roof and the base can be fabricated in parallel, or simultaneously. Roof fabrication involves first punching and then forming the roof to shape. Base fabrication entails the punching-and-forming process plus a subassembly operation. Fabricated roofs and bases are then assembled into finished sheds that are subsequently inspected for quality assurance. A list of activities needed to fabricate a roof, fabricate a base, and assemble a shed is given in Table 4.1. A flowchart of the process is shown in Figure 4.1. 7 4.1 (2 points) We use a beta distribution to represent an activity duration. (Note that it is a BETA DISTRIBUTION.) We assume that the activity durations are independent. We denote the optimistic, the most likely and the pessimistic times by a, c and b. Calculate the mean time and the variance of each activity duration and fill in the blanks of the table below. Activity a c b 1 15 20 30 2 25 35 40 3 20 25 30 4 5 10 20 5 10 20 25 6 25 30 40 7 10 15 30 8 35 40 60 Mean Variance 8 4.2. (2 point) The process contains two paths: Path 1 : Start → 1 → 3 → 5 → 7 → 8 → End Path 2 : Start → 1 → 2 → 4 → 6 → 7 → 8 → End What are the mean and the variance of the duration of each path? Fill in the following table. Path 1 Path 2 Mean Variance 4.3. (2 points) Based on Central Limit Theorem, we assume that the duration of each path is normally distributed with mean and variance given in 4.2. We also assume that the durations of the two paths are independent random variables. What is the probability that both paths are completed within 170 minutes? 9 5. Overhead Door (OD) Corporation’s founder, C. G. Johnson, invented the upward-lifting garage door in 1921 and the electric garage door opener in 1926. Since then OD has been a leading supplier of commercial, industrial, and residential garage doors sold through a nationwide network of more than 450 authorized distributors. They have built a solid reputation as a premier door supplier, commanding 15 % share of the market. Suppose that customers assess door quality first in terms of the ease of operation, followed by its durability. The quality improvement team (QIT) might then assign an engineering team to determine the factors that contribute to these two main problems. Smooth operation of a garage door is a critical quality characteristic that affects both problems: If a door is too heavy, it’s difficult and unsafe to balance and operate; if it’s too light, it tends to buckle and break down frequently or may not close properly. Suppose the design engineers determine that a standard garage door should weigh a minimum of 75.5 kg. and a maximum of 85.5 kg., which thus specifies its design quality specification. QIT is inspecting if there is evidence that the percentage of defective doors does not exceed 10%. Suppose the QIT decides to collect data on the actual weights of 60 standard garage doors sampled randomly from their monthly production of almost 2,000 doors. See the tables on the next page. 5.1 (2 point) What is the sample defective rate ? 5.2 (2 points) Formulate and test an appropriate set of hypotheses to determine if the machine can be qualified. Use α = 0.05. Find the P-value. 10 5.3 (2 points) Is there evidence to support the claim that a standard garage door weigh 80 kg.? What is the P-value? Use α = 0.01. 5.4 (2 points) What is the 99% confidence interval? Table 2: T\D 1 2 3 4 5 6 7 8 9 10 9 81 82 80 74 75 81 83 86 88 82 1 73 77 83 81 76 76 82 83 79 84 5 85 78 76 81 82 83 76 82 86 79 T\D 11 12 13 14 15 16 17 18 19 20 9 80 78 84 75 84 78 77 79 84 84 1 80 84 82 83 75 81 78 85 85 80 5 76 76 78 72 84 76 74 85 82 79 11 6. Benzene is a toxic chemical used in the manufacturing of medicinal chemicals, dyes, artificial leather, and linoleum. A manufacturer claims that its exit water meets the federal regulation with a mean of less than 7980 ppm of benzene. To assess the benzene content of the exit water, 10 independent water samples were collected and found to have an average of 7906 ppm of benzene. Assume a known standard deviation of 80 ppm and use a significance level of 0.05. 6.1. (2 points) What is the 𝛽𝛽 -value if the true mean is 7950? 6.2. (2 points) What sample size would be necessary to detect a true mean of 7950 with a probability of at least 0.99? 12 7. Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart will last and operate almost indefinitely once it is implanted in the patient’s body, but the battery pack needs to be recharged about every 8 hours. A random sample of 60 battery packs is selected and subjected to a life test. The average life of these batteries is 8.04 hours. Assume that battery life is normally distributed with standard deviation σ = 0.15 hour. The claim is that mean battery life exceeds 8 hours. Use α = 0.01. 7.1 (2 points) What is the P-value? 7.2 (2 points) Is there evidence to support the claim? Explain by constructing a one-sided confidence bound on the mean life. 13
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