mind that a statistic is only worthwhile when it satisfies the assumptions on the test. Knowing whether the assumptions are met is dependent on the competence of the person running the test. Just because two things seem to have a relationship‚ could it have been by pure chance? It cannot be determined by causation and effect. The two variables have no effect on each other at all. Chapter 9 – How to Statisticulate Statisticulate is the process of misleading people using statistics. It is also
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QMT2033 BUSINESS STATISTICS Assignment for mid -term break 1. Most of the information technology developers claim that wireless connection would give speed of at least 11 Mbps compared to wired connection. A sample of 105 computers using wireless connection shows it gives a mean of 11.7 Mbps and a standard deviation of 2.3 Mbps. a) Do you think that the wireless usage should be implemented? Test at 7% significance level). b) Repeat the test by using p-value approach. 2. A jack is usually used
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Case Problem 1: National Health Care Association(Descriptive Statistics) The National Health Care Association is concerned about the shortage of nurses the health care profession is projecting for the future. To learn the current degree of job satisfaction among nurses‚ the association has sponsored a study of hospital nurses throughout the country. As part of this study‚ a sample of 50 nurses was asked to indicate their degree of satisfaction in their work‚ their pay and their opportunities for
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(-3.5)/1.6396 = -2.1347 Critical Value: Zα/2= Z0.05/2= @qnorm(1-0.05/2)= 1.96 When comparing the test statistic to the critical value: Z=2.1347>1.96‚ we reject the null hypothesis. We can calculate the P-value using the EViews command: Show @tdist (t‚ d.f) In this EViews command‚ t stands for the appropriate test statistic and d.f are the degrees of freedom. The appropriate test statistic was calculated above‚ namely Z=2.1347. For the degrees of freedom‚ we can insert NA+NB-2. Show @tdist (2
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It is known that there are two data types that are utilized to evaluate and draw meaningful conclusions through statistics‚ population and sample data. These two data types are utilized to formulate end conclusions of data that is to be collected and data that is to be reviewed. The description of population data can best be explained‚ as the complete collection of all data that is to be queried/collected and reviewed. Sample data‚ a subset of population data‚ is the partial collection and review
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study. In order to get a true representation of the car population‚ we have used the data from a sample observing 20‚001 cars of various classes‚ engine sizes‚ types‚ makes and models between the years 1978 to 2008. To fully incorporate the impact of weight on fuel efficiency (lurking variable) we also noted down the number of doors‚ passengers and luggage present in the car. The sample also includes the type of fuel used to see its effect on fuel efficiency. Our goal will be to analyze the relationship
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BE01106 - BUSINESS STATISTICS ASSIGNMENT – PART I SEMESTER 2‚ 2014 _________________________________________________________________________ The complete BEO1106 assignment (all three parts) accounts for 20% of the overall assessment in the unit. This first part of the BEO1106 assignment is to be completed in your own time and must be submitted for correction in the tutorial of week 6. No time extensions will be granted although special consideration applications may be accepted from students
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Omkar & Yaying Wednesday 5-6pm WEEK 3 BES PASS Descriptive Statistics Population - a set of all possible observations. Sample - a portion of a population. We often use information concerning a sample to make an inference (conclusion) about the population. Parameter - describes a characteristic of the population‚ eg: the population variance Statistic- describes a characteristic of a sample‚ eg: the sample variance Frequency Distribution and Histograms Class - a collection of
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(109925 – (807^2/6)/6-1 = (109925 – 108541)/5 = 1384/5 = 276.8. Standard Deviation = √276.8 = 16.6373. ***the 109925 is all values of x individually squared and then summed together. ***the 6-1 is because it is a sample‚ if this were a population it would just be 6. ***the 807 is the sum off all x. Coefficient of Variation = (16.63/134.5)*100 = 12.3643. Calculate estimates of tolerance intervals containing 68.26‚ 95.44‚ and 99.73 percents. Mean ± 1 SD (68.26%) = 134.5 ± 16.63 = [117.87‚ 151.13]
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rewarded in most cases as long as manager is willing to do whatever is necessary to achieve success. Manager needs some essential skills to carry out duties and issues. Robert L. Katz (1993) has identified three basic skills for manager to possess and how these skills bring rewards to manager. These skills are a big challenge for manager because managers need to make an effort to develop the skills and make them become natural. A top level manager is supposed to make good use of the three basic skills
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