Exercise 1 Suppose That 73 61 38 84 69 71 53 82 49 A

Calculate the sample mean, sample variance, and an approximate 95 percent confidence interval for the population mean (µ) based on the given data: 7.3, 6.1, 3.8, 8.4, 6.9, 7.1, 5.3, 8.2, 4.9, and 5.8. Then, perform a hypothesis test for H0: µ = 6 at the 0.05 significance level using the sample data.

Sample Paper For Above instruction

In statistical analysis, understanding the data's central tendency and variability is fundamental to making inferences about the population from which the data are drawn. The data provided—7.3, 6.1, 3.8, 8.4, 6.9, 7.1, 5.3, 8.2, 4.9, and 5.8—are a sample of size 10 from a distribution that is not highly skewed. To analyze these observations, the first step involves calculating the sample mean (\(\bar{x}\)) and the sample variance (\(s^2\)), which serve as estimators for the population parameters.\n\nThe sample mean is computed as follows:\n\n\[\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i\]\n\nwhere \(n=10\) and \(x_i\) represents each data point. Substituting the values:\n\n\[\bar{x} = \frac{7.3 + 6.1 + 3.8 + 8.4 + 6.9 + 7.1 + 5.3 + 8.2 + 4.9 + 5.8}{10} = \frac{63.8}{10} = 6.38\]\n\nNext, the sample variance measures the variability of the data points around the mean and is calculated as:\n\n\[\mathrm{s}^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2\]\n\nCalculating the squared deviations:\n\n\[(7.3 - 6.38)^2 = 0.8521\n\]\n\[(6.1 - 6.38)^2 = 0.0784\n\]\n\[(3.8 - 6.38)^2 = 6.6084\n\]\n\[(8.4 - 6.38)^2 = 4.084\n\]\n\[(6.9 - 6.38)^2 = 0.2704\n\]\n\[(7.1 - 6.38)^2 = 0.5184\n\]\n\[(5.3 - 6.38)^2 = 1.177\n\]\n\[(8.2 - 6.38)^2 = 3.3124\n\]\n\[(4.9 - 6.38)^2 = 2.1664\n\]\n\[(5.8 - 6.38)^2 = 0.3316\n\]\n\nSum of squared deviations:\n\n\[\Sigma (x_i - \bar{x})^2 = 0.8521 + 0.0784 + 6.6084 + 4.084 + 0.2704 + 0.5184 + 1.177 + 3.3124 + 2.1664 + 0.3316 = 19.9985\]\n\nSample variance:\n\n\[\mathrm{s}^2 = \frac{19.9985}{9} \approx 2.222\]\n\nand the sample standard deviation is:\n\n\[\mathrm{s} = \sqrt{2.222} \approx 1.491\]\n\nTo construct a 95% confidence interval (CI) for the population mean \(\mu\), we use the t-distribution because the sample size is small, and the population variance is unknown. The CI is given by:\n\n\[\bar{x} \pm t_{(1-\alpha/2, df)} \times \frac{s}{\sqrt{n}}\]\n\nwhere \(df = n - 1 = 9\), and \(t_{0.975,9}\) is the critical t-value from the t-table. For 9 degrees of freedom, \(t_{0.975,9} \approx 2.262\).\n\nCalculating the margin of error:\n\n\[\mathrm{ME} = 2.262 \times \frac{1.491}{\sqrt{10}} \approx 2.262 \times 0.472 = 1.068\]\n\nThe confidence interval is:\n\n\[\left( 6.38 - 1.068, 6.38 + 1.068 \right) = (5.312, 7.448)\]\n\nThis interval suggests with 95% confidence that the true population mean \(\mu\) lies between approximately 5.31 and 7.45.\n\nNext, to test the null hypothesis \(H_0: \mu = 6\) against the alternative \(H_A: \mu \neq 6\), we perform a t-test:\n\n\[\text{Test statistic} = t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{6.38 - 6}{1.491 / \sqrt{10}} \approx \frac{0.38}{0.472} \approx 0.805\]\n\nUsing the t-distribution with 9 degrees of freedom, the critical value for a two-tailed test at \(\alpha=0.05\) is approximately 2.262. Since \(|0.805|

Exercise 2 Parts arrive at a single workstation system according to an exponential interarrival distribution with mean 21.5 seconds; the first arrival is at time 0. Upon arrival, the parts are initially processed. The processing-time distribution is TRIA(16, 19, 22) seconds. There are several easily identifiable visual characteristics that determine whether a part has a potential quality problem. These parts, about 10% (determined after the initial processing), are sent to a station where they undergo a thorough inspection. The remaining parts are considered good and are sent out of the system. The inspection-time distribution is 95 plus a WEIB(48.5, 4.04) random variable, in seconds. About 14% of these parts fail the inspection and are sent to scrap. The parts that pass the inspection are classified as good and are sent out of the system (so these parts didn't need the thorough inspection, but you know what they say about hindsight). Run the simulation for 10,000 seconds to observe the number of good parts that exit the system, the number of scrapped parts, and the number of parts that received the thorough inspection. Animate your model. Put a text box in your model with the output performance measures requested, and make just one replication.

To simulate the described parts processing system accurately, a discrete event simulation model should be built using software such as Arena, Simul8, or similar. The key components include arrival processes, processing, inspection, quality assessment, and output tracking. The simulation begins with parts arriving according to an exponential distribution with a mean of 21.5 seconds, modeled through the interarrival times. Each part undergoes an initial processing modeled by a triangular distribution TRIA(16, 19, 22). After initial processing, the parts are classified based on visual characteristics, with approximately 10% identified as potential quality issues requiring thorough inspection. These parts are sent to an inspection station, where the inspection times follow a distribution of 95 seconds plus a Weibull variant (WEIB(48.5, 4.04)). Parts passing inspection are deemed good and exit the system; those failing are scrapped, comprising about 14% of inspected parts.

The simulation duration of 10,000 seconds encompasses the entire process, ensuring sufficient data for reliable performance measures. During the simulation, the following outputs are monitored and recorded: total number of good parts exiting the system, total scrapped parts, and total parts receiving thorough inspection. To add clarity and better understanding, an animation of the process provides a visual flow of parts through the system, illustrating the flow and processing steps. Including a text box displaying key performance measures summarizes the results, which include the counts of good parts, scrapped parts, and inspected parts, as well as potential metrics like throughput and scrap rate.

Implementing this model involves setting up resources (processing stations, inspection station), defining arrival processes, processing times, inspection times, classification logic, and counters for output measures. Once the model is complete, running the simulation for the specified duration and analyzing the outputs provides insights into system performance and potential bottlenecks, guiding process improvements and quality control strategies.

Exercise 3 An acute-care facility treats non-emergency patients (cuts, colds, etc.). Patients arrive according to an exponential interarrival-time distribution with a mean of 11 (all times are in minutes). Upon arrival they check in at a registration desk staffed by a single nurse. Registration times follow a triangular distribution with parameters 6, 10, and 19. After completing registration, they wait for an available examination room; there are three identical rooms. Data show that patients can be divided into two groups with regard to different examination times. The first group (55% of patients) has service times that follow a triangular distribution with parameters 14, 22, and 39. The second group (45%) has triangular service times with parameters 24, 36, and 59. Upon completion, patients are sent home. The facility is open 16 hours each day. Make 200 independent replications of 1 day each and observe the average total time patients spend in the system. Put a text box in your Arena file with the numerical results requested.

Modeling the patient flow in an acute-care facility involves simulating arrival, registration, examination, and departure processes over multiple days to capture variability and average system performance. Patients arrive following an exponential distribution with a mean interarrival time of 11 minutes, which can be modeled using the exponential interarrival time distribution in simulation software. After arrivals, each patient undergoes registration with a triangular distribution of times (6, 10, 19), representing the variability in registration durations.

Following registration, patients wait for an available examination room among three identical ones. The examination times differ based on patient groups: approximately 55% of patients have service times modeled by a triangular distribution (14, 22, 39), indicating quicker examinations, while the remaining 45% experience longer service times (24, 36, 59). These service times are assigned probabilistically, reflecting real-world variability. The simulation runs for one full day (16 hours), and this process is replicated independently 200 times to obtain reliable average metrics.

Key outputs include the average total time patients spend in the system (from arrival to departure), the number of patients processed, and the distribution of waiting and service times. To facilitate analysis, display the results in a text box within the simulation model, summarizing the key performance indicators such as average wait time, total time in system, and throughput. Multiple replications account for stochastic variability, providing statistically robust estimates of system performance.

The modeling process requires careful setup of resources, patient flow logic, probabilistic service time assignment based on patient groups, and data collection points for performance measures. After executing the simulations, analyzing the results helps identify bottlenecks, improve resource utilization, and enhance patient flow efficiency in the healthcare setting.