to give the dean some advice after the data analysis. Methodology and Hypothesis The methodology of this report was using descriptive statistics to summarize the data and comments. The descriptive statistics tables were provided as follows. (The sample data were provided in Appendix) Table 1: quiz statistics results for all students All | Internet | Exam | Collaborated | cheater | Y | 23 | 16 | 23 | 48 | N | 67 | 74 | 67 | 42 | Total | 90 | 90 | 90 | 90 | P(Y) | 25.6% | 17.8% | 25.6%
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research (1) Previous studies of X have not dealt with … (2) Researchers have not treated X in much detail. (3) Most studies in the field of X have only focused on … (4) However‚ these studies have used non-validated methods to measure … (5) Small sample sizes have been a serious limitation for many earlier studies. (Can you add some other language of the same function into the list?) Language
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Problem 1. Sampling 4 pieces of precision-cut wire (to be used in computer assembly) every hour for the past 24 hours has produced the following results: |Hour |Average (inches)|R (inches) |Hour |Average[pic] |R (inches) | | | | | |(inches) | | |1 |3.25 |0.71 |13 |3.11 |0.85 | |2 |3.10
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While ‘Backward’ market research’ may sound oxymoronic it has to be one of the most useful things that I learnt throughout my time at University and I must thank my old lecturer Ben Healey for introducing it to me. Why is ‘Backward’ market research so good? – It is good because is it delivers results‚ and if market research cannot deliver results it really is a waste of time and money. ‘Backward’ market research was first postulated by Alan R. Andreasen in the 1980′s. ‘Backward’ market research
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Executive Summery This project gives us a detailed idea of what is stress and also the definition of stress is been defined. For more detailed study the types of the stress is also defined. By looking at the starting of the project you will find: o Introduction to Human resource o Introduction and Definition of stress o Stress in biological terms o What is stress? o Coping with stress at work place. o Stress management o Workplace stress o Reducing of stress. After the theoretical part
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Harrigan’s invitation to enroll 103/178 = 0.5787 = sample mean Binomial distribution Mean = 0.5 = np‚ Standard deviation = {np(1-p)}^0.5 = (0.5*(1-0.5))^0.5 = 0.5 95% confidence intrerval = 0.5787 ± 1.96*0.5/(178)^0.5 =0.5025~0.6521 = 50.25% ~ 65.21% b) less than 330 33/103 = 0.3204 = sample mean 95% confidence interval = 0.3204 ± 1.96*0.5/(103)^0.5 =0.2318~0.4195 = 23.18%~41.95% c) between 330 and 375 d) more than 375 15/103 = 0.1456 = sample mean 95% confidence interval = 0.1456 ± 1.96*0
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Athletic Association (NCAA) reported that the mean number of hours spent per week on coaching and recruiting by college football assistant coaches during the season was 70. A random sample of 50 assistant coaches showed the sample mean to be 68.6 hours‚ with a standard deviation of 8.2 hours. a. Using the sample data‚ construct a 99% confidence interval for the population mean. z = +/- 2.68 68.6 – 2.68 * 8.2 / √50 = 65.49 68.6 + 2.68 * 8.2 / √50 = 71.71 b. Does the 99% confidence
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people associated with businesses are important to communication skills because of the global trades. Hypothesis Testing with communication develops describes the effect it has on cultural differences and results of other researchers studies on sample test. The test results in the business can save money and time by randomly sampling the population. Statistical Hypothesis testing helps the Qce to match up differences from other nations in communication surveys. Businesses are more aware of telecommunication
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information-oriented sampling (extreme‚ maximum variation‚ critical and paradigmatic) (Flyvbjerg‚ 2006). While with random sampling‚ the decisive factor for generalization is the size of cases selected‚ information-oriented technique concern more about the characteristics of specific cases chosen and less about the sample size. Particularly‚ whereas extreme cases offer insights to unusual angle of the problem‚ maximum variation makes use of multiple cases to increase the diversified situations. Furthermore
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Q1. A researcher wishing to estimate the proportion of X-ray machines that malfunction and produce excess radiation. A random sample of 40 machines is taken and 12 of the machines malfunction. The problem is to compute the 95% confidence interval on π‚ the proportion that malfunction in the population. Solution: The value of p is 12/40 = 0.30. The estimated value of σp is = 0.072. A z table can be used to determine that the z for a 95% confidence interval is 1.96. The limits of the confidence
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