Finite Populations Correction Factors - Essay Example

Statistics A. The starting salary for new college graduates in computer information is $47,500 on average with an even distribution and a standard deviation of $45,000. To figure out the probability of a new college graduate earning a starting salary of somewhere between $45,000 and $50,000, we must begin by establishing a number sequence representative of standard salary distribution. This could look as follows: $43,000, $47,50 and $52,000. With these possibilities, the mean is obviously $47,50 with the standard deviation of $4,500 represented. Each of these figures allows for a 33% chance of a starting salary. Since the figure bracket we are wondering about is somewhere between $45 and $50k, we must use the product rule and the sum rule to gain the answer. The chances of a graduate earning a $45k salary begins in the first noted price bracket but is only represented by the second half of that bracket. That means we must multiply 50% x 33% = 17%. This is the same for the second and third bracket requiring us to employ the sum rule. 17% + 17% = 33%. B. The probability of a new graduate making a starting salary which exceeds 55k is as follows: The farther away we get from the mean, the more income brackets need to be added in both directions in order to maintain $47,500 as the mean leaving only a 20% chance of a new graduate being hired at a starting salary of 55k or more. C. The percentage of staring salaries which are no more than $42,250 is a 36% chance. This was calculated by representing 5 brackets with 20% chances: 38.5k, 43k, 47.5k, 52k and 56.5k all with a 20% chance each due to even distribution. The bracket between 38.5k and 43.k together represent a 40% chance cumulatively. Those brackets can be divided into smaller brackets leaving the bracket of 42.5 k and everything beneath that amount represented by a 36% chance. D. The cut off for the bottom 5% of the salaries is $38,500. E. The cutoff for the top 3% of salaries is $56,500. In an infinite population, the mean is 150 with a deviation of 40. A. The expected sample mean is going to be 150 when the infinite population has already been determined to have a mean of 150. B. The expected deviation will also be expected to be 40 when the original population has already been determined to have a standard deviation of 40. C. The shape of the sampling distribution is going to resemble a bell curve as the deviation has already been determined to be 40 with a mean of 150. D. The sampling distribution of the sampling shows what we already know to be the case from the original infinite population with mean of 150 and standard deviation of 40. 3. In order to estimate the mean salary of 500 employees, the managing director of a certain company selected at random a sampling of 40 employees. A. Could the finite population correction factor be used in calculating the standard error of the sample mean? If so explain: First, let us allow n to represent the sample size and N to represent the population size. In the event that the value of n is greater than 5% of the finite population N, the population correction factor may be used. In this case, n=40 and N=500. The sample population is equal to 8% of the finite population of 500 which is again, represented by N. B. To establish standard of error utilizing the given information thus far, we can use the formula , which represents the variability of the original data which can divided by the square root n. We may also find the answer by utilizing the following formula: ((N-n)/N-1)) which gives us the number 9.5917 as our standard error. Using the population correction factor to solve the same problem, we find solution of 9.5917 when we employ the population correction factor which is calculated as follows: The Square root all over (N-n)/(N-1). Once the figures for N and n are plugged into this formula, we end up with the standard of error which is again, 9.5917. C. What is the probability that the sample mean salary will be within + or - $200 of the population mean salary? The population of 500 is being represented by a sample group of 40. The variance in pay among the population is $800. The probability of the sample population having the same $ 800 variance as the original population is quite good. The probability of being within + or - what the population’s mean salary is goes down as the figures draw away in either direction for the sample population. As members of the sample population are looked at as their salaries move away in either direction from the $800 deviation, the bell vurve begins to rise or fall on either side of the $800 peak. The probability of seeing a variance of only minus or plus $200 from the $800 is approximately 25%. B. In order to establish a mean regarding the hourly time spent utilizing computers among a certain student population, 10 students were chosen and asked to record their computer time. It was necessary to add all of the recorded times right away and then produce the mean time spent which was 7.5 hours. A margin of error could be then assessed by taking the mean and adding or subtracting it from the product of the Z score which was 1.6 times the standard of error which was 2.37 yielding either a solution of 13.79 or 6.21 as 3.79 was the resulting degree of freedom. C. Assumptions which can be made to support these calculations are as follows: First of all, each student had a varying time allotment for how long they used the computer which lends to a larger degree of freedom among the population as a whole, particularly to the sample population. It is also safe to assume that by making any assumption pertaining to this particular population would be frivolous since there is in fact a small sample group with a large variance in time spent on the computer. This only leads to the notion that inconclusive results were obtained not entirely true of this particular population due to a large degree of freedom and too much variance. A. An admissions officer of a school believes that his school’s test scores are the higher than the state average of 1200. By establishing this, he has set up a null hypothesis to be either accepted or rejected. We can look then at a sample population taken, of about 25 test scores with a mean score of 1375, a boastful 50 points greater than the principles scores of 1300 on average. The standard deviation here is 125. In a null hypothesis, the principle would want to erect a hypothesis meant to be torn down or rejected. In subtracting the test scores average state wide of 1300 from his sample of 25 test scores with a mean of 1375, the null hypothesis could be neither rejected nor accepted with little data and the singular fact that the difference in his school’s scores and the state average was only 75 points making the need for an alternative hypothesis. The officer’s hypothesis was attempting to predict his school as being highest in score within the Nation and thus, had the small sample chosen indicated a higher average than this at half of the deviation number of 125. This officer In an alternative hypothesis, due to the prediction of the first hypothesis and the random selection of tests scores which was in agreement of his predictions, allowed for this to in fact be a null hypothesis not yet to be proven. When using the null hypothesis, we can set up a formula where H = 13oo and H1 = 1375. T0 make this a legitimate alternative hypothesis, we must understand that its outcome will be determined by some sort of non-random cause where a null hypothesis is left to chance and then decidedly accepted or rejected. This is where the alternative hypothesis can come into play as it is not operating as archaically as the null hypothesis. B. Find the rejection reason region of the hypothesis which disproves the null hypotheses. This region could be the numerous students who are bringing down the average. If we can make a chart to lay out some possible scenarios for test results. C. By plugging in the theoretical test scores before and after the mean test score at this particular admissions officer works: 1200, 1250, 1300, 1350, 1400. With the following formula, we will see how to obtain the test state. The square root all over ((1/n1) + ((1/n2)). We can plug this in to accurately identify the test state. The square root all over (1/1300)=((1/1375). The answer to this problem is to first plug in the values for n1 and n2. We plug in the value for n1 as 1300 and value two as 1375. Once the problem is worked out, we have an answer of 2.89. It is now important to use what has been covered to see if identifying a rejection reason Is possible, with a given .025 significance. Level. Given this info we can move on to calculate the p-value and determine whether truly, this fits null hypothesis or not. We also know that p-value represents at least as extreme a number which would be in a given or worked out earlier in the problem. The p-value will serve to allow for human error or luck. In other words, It is always possible to end up lucky within your sample population where every person you choose happens to respond in the way in which the study is experimenting with. This happened in small populations mostly. We can use the t value which we found in an earlier portion of this problem to be 2.89 with a degree of freedom to be the positive integer 125. When we plug these figures into a formula for calculating the p value, we get .0045. This is of significance and shows that the null hypothesis is in fact, not appropriate here as it is a p value below 5%. Works Cited: -------------------------, (2008) “Finite population correction factor”, retrieved from website at: http://www.usm.maine.edu/~smax/course_material/FinitePopulationCorrec tionFactor.pdf -----------------------, (2008) “Confidence intervals”, retrieved from website at: http://www.childrensmercy.org/stats/journal/confidence.asp Handbook of Biological Statistics, (2008) “Standard error of the mean” retrieved from website at: http://udel.edu/~mcdonald/statstderr.html Pieniazek, Adam. (2007) “Finite Populations Correction Factors”, retrieved from website at: http://www.adampieniazek.com/statistics/finite-population- correction-factor/ Price Ph.D, Ian, (2008), Calculating the probability”, retrieved from website at: http://www.une.edu.au/WebStat/unit_materials/c6_common_statistical_tes ts/computing_probability.html Read More