Sunday, May 17, 2020
Confidence Interval for the Difference of Two Population Proportions
Confidence intervals are one part of inferential statistics. The basic idea behind this topic is to estimate the value of an unknown population parameter by using a statistical sample. We can not only estimate the value of a parameter, but we can also adapt our methods to estimate the difference between two related parameters. For example we may want to find the difference in the percentage of the male U.S. voting population who supports a particular piece of legislation compared to the female voting population. We will see how to do this type of calculation by constructing a confidence interval for the difference of two population proportions. In the process we will examine some of the theory behind this calculation. We will see some similarities in how we construct a confidence interval for a single population proportion as well as a confidence interval for the difference of two population means. Generalities Before looking at the specific formula that we will use, lets consider the overall framework that this type of confidence interval fits into. The form of the type of confidence interval that we will look at is given by the following formula: Estimate /- Margin of Error Many confidence intervals are of this type. There are two numbers that we need to calculate. The first of these values is the estimate for the parameter. The second value is the margin of error. This margin of error accounts for the fact that we do have an estimate. The confidence interval provides us with a range of possible values for our unknown parameter. Conditions We should make sure that all of the conditions are satisfied before doing any calculation. To find a confidence interval for the difference of two population proportions, we need to make sure that the following hold: We have two simple random samples from large populations. Here large means that the population is at least 20 times larger than the size of the sample. The sample sizes will be denoted by n1 and n2.Our individuals have been chosen independently of one another.There are at least ten successes and ten failures in each of our samples. If the last item in the list is not satisfied, then there may be a way around this. We can modify the plus-four confidence interval construction and obtain robust results. As we go forward we assume that all of the above conditions have been met. Samples and Population Proportions Now we are ready to construct our confidence interval. We start with the estimate for the difference between our population proportions. Both of these population proportions are estimated by a sample proportion. These sample proportions are statistics that are found by dividing the number of successes in each sample, and then dividing by the respective sample size. The first population proportion is denoted by p1. If the number of successes in our sample from this population is k1, then we have a sample proportion of k1 / n1. We denote this statistic by p̂1. We read this symbol as p1-hat because it looks like the symbol p1 with a hat on top. In a similar way we can calculate a sample proportion from our second population. The parameter from this population is p2. If the number of successes in our sample from this population is k2, and our sample proportion is p̂2 k2 / n2. These two statistics become the first part of our confidence interval. The estimate of p1 is p̂1. The estimate of p2 is p̂2. So the estimate for the difference p1 - p2 is p̂1 - p̂2. Sampling Distribution of the Difference of Sample Proportions Next we need to obtain the formula for the margin of error. To do this we will first consider the sampling distribution of p̂1 . This is a binomial distribution with probability of success p1 and n1 trials. The mean of this distribution is the proportion p1. The standard deviation of this type of random variable has variance of p1 (1 - p1 )/n1. The sampling distribution of p̂2 is similar to that of p̂1 . Simply change all of the indices from 1 to 2 and we have a binomial distribution with mean of p2 and variance of p2 (1 - p2 )/n2. We now need a few results from mathematical statistics in order to determine the sampling distribution of p̂1 - p̂2. The mean of this distribution is p1 - p2. Due to the fact that the variances add together, we see that the variance of the sampling distribution is p1 (1 - p1 )/n1 p2 (1 - p2 )/n2. The standard deviation of the distribution is the square root of this formula. There are a couple of adjustments that we need to make. The first is that the formula for the standard deviation of p̂1 - p̂2 uses the unknown parameters of p1 and p2. Of course if we really knew these values, then it would not be an interesting statistical problem at all. We would not need to estimate the difference between p1 and p2.. Instead we could simply calculate the exact difference. This problem can be fixed by calculating a standard error rather than a standard deviation. All that we need to do is to replace the population proportions by sample proportions. Standard errors are calculated from upon statistics instead of parameters. A standard error is useful because it effectively estimates a standard deviation. What this means for us is that we no longer need to know the value of the parameters p1 and p2. .Since these sample proportions are known, the standard error is given by the square root of the following expression: p̂1 (1 - p̂1 )/n1  p̂2 (1 - p̂2 )/n2. The second item that we need to address is the particular form of our sampling distribution. It turns out that we can use a normal distribution to approximate the sampling distribution of p̂1 - p̂2. The reason for this is somewhat technical, but is outlined in the next paragraph. Both p̂1 and p̂2 have a sampling distribution that is binomial. Each of these binomial distributions may be approximated quite well by a normal distribution. Thus p̂1 - p̂2 is a random variable. It is formed as a linear combination of two random variables. Each of these are approximated by a normal distribution. Therefore the sampling distribution of p̂1 - p̂2 is also normally distributed. Confidence Interval Formula We now have everything we need to assemble our confidence interval. The estimate is (p̂1 - p̂2) and the margin of error is z* [ p̂1 (1 - p̂1 )/n1  p̂2 (1 - p̂2 )/n2.]0.5. The value that we enter for z* is dictated by the level of confidence C.  Commonly used values for z* are 1.645 for 90% confidence and 1.96 for 95% confidence. These values for z* denote the portion of the standard normal distribution where exactly C percent of the distribution is between -z* and z*. The following formula gives us a confidence interval for the difference of two population proportions: (p̂1 - p̂2) /- z* [ p̂1 (1 - p̂1 )/n1  p̂2 (1 - p̂2 )/n2.]0.5
Wednesday, May 6, 2020
Chapter Review 1-3 Principles of Supply Chain Management
Chapter Review 1-3 Principles of Supply Chain Management, 3ed Wisner, Leong, Tan 2012 Chapter Review 1-3 Chapter One: Introduction To Supply Chain Management A Supply Chain is the steps necessary for a manufacturer to procure materials, build a product, and transport the product to consumers. The consumers buy the products based on a combination of cost, quality, availability, maintainability and reputation factors. They hope these products will live up to their needs and expectations. An example of a supply chain that I was involved in while serving in the Air Force would be when I worked in the Supply Squadron. One of our customers, the jet engine maintenance shop, would need a part to repair and F101 engine for use on a†¦show more content†¦This is also likely to lower raw material inventory and total assets. E-Procurement brought many benefits to companies. It has proven to be a time saver and cost effective. It is more accurate as the information is only entered once instead of twice. Before the users had to enter the information and the buyer had to reenter the inf ormation creating another error point in the process. E-procurement is more flexible as it can be used without the restrictions of location or time of day. Status of orders can be looked at without having to check paper trails. There are different reasons why a firm may use a single supplier instead of favoring multiple suppliers. Using a single supplier gives a firm the chance to build a stronger relationship with the supplier. Costs would be lower due to larger purchases keeping the cost per unit down and transportation would be cheaper as the firm can take advantage of full truckloads. Single sourcing would also make sense if the firm’s requirements are too small. It would not be worthwhile to split the order among multiple suppliers. There is a disadvantage with sole sourcing as well. If the purchase was for a proprietary product or process and the supplier holds the patents to them, the firm has no choice but to buy from them. Multiple sourcing is advantageous fo r many reasons. 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Retrofitting and Rehabilitation of Civil Engineering Structures free essay sample
We obviously knew why should it be a non destructive test, because we aren’t allowed to destruct the entire structure in the name of testing. Roadware 10 Minute Concrete Menderâ„ ¢ Roadware needle tip crack injection costs less to install and is more effective than pressure applied epoxy injection methods.  according to top concrete repair contractors. We can get in, permanently repair the cracks, and get out in the same time it use to take just to install the old injection ports. Evaluation and Repair of Fire-Damaged Buildings †¢ Typically, repair materials are similar to the original construction materials. Timber structures may be repaired with new timbers or composites of steel and timber members, and steel structures are normally repaired with steel. †¢ Concrete and masonry structural elements are frequently repaired with fiber reinforced polymers (FRP) or externally bonded steel members using epoxy adhesive. Concrete structures are occasionally repaired with shotcrete as well. †¢ Selecting the appropriate repair material is a critical step in the repair process. We will write a custom essay sample on Retrofitting and Rehabilitation of Civil Engineering Structures or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page The repair material must be compatible with the base material, project needs, technical resources available, financial constraints, and multiple other project specific criteria. Two case studies are presented to illustrate various types of structural damage and repairs. †¢ The first summarizes the assessment and repair for arson damage to a reinforced concrete stadium structure. †¢ The second addresses the evaluation and repair of damage to a steel framed office building. Slab separation and joist spalling Case Study 1: Stadium Fire An arson fire occurred on an elevated level of a reinforced concrete stadium structure. After the fire department approved the area for entry, initial observations were made. The fire severely damaged the concrete in an area approximately 25 feet by 25 feet above the fire; the concrete joists were deeply spalled, the slab separated from the joists, and severe cracking was present The fire was intense to the extent that the seating area directly above the fire suffered heat damage Case Study 2: Occupied Steel High Rise Fire Fire occurred in an occupied space of a steel framed high rise. The steel framing was protected by a sprayed fire resistive material. The exposed metal deck supports a composite concrete floor system.. Heat from the fire caused some buckling of the metal decks near the fire Buckled metal deck. Note separation of concrete from metal deck through exploratory opening The evaluation effort included steel hardness readings of the structural steel framing members. In addition, concrete cores were extracted from the floor slab where the fire occurred and in the deck directly above the fire for compression testing. No testing was considered necessary in the adjacent room with deformed plastic due to the low temperature indication. Test results confirmed that steel hardness was in the expected range to indicate minimal heat damage to the steel frame. Concrete cores indicated that compressive strengths exceeded design requirements The assessment concluded that the steel frame did not require repairs. However, to prevent displacements of the separated concrete slab above the metal deck, structural repairs were required . The repairs included adding new beams to support the metal deck and the addition of grout to fill the voids between the concrete and metal deck FINALLY †¢ All structures should be evaluated in a systematic manner to determine the extent, if any, of required repairs. A variety of testing methods and tools are available to evaluate the effects on both the materials and structural elements. †¢ Evaluations, combined with an engineering analysis, allow effective and economical repair details to be developed and installed as needed.
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