same set‚ because the audit team could request a comprehensive documents package from the client. The sample size table in Appendix A indicates that a sample size of 58 is acceptable for the first control‚ but the higher estimated population deviation rate for the second control suggests that a larger sample size of 77 is necessary in order to gather sufficient evidence and obtain reasonable assurance over the control. The larger of the possible sample sizes should be selected for testing‚ because
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manufacturing processes. In one particular application‚ a client gave Quality Associates a sample of 800 observations taken during a time in which that client’s process was operating satisfactorily. The sample standard deviation for these data was .21; hence‚ with so much data‚ the population standard deviation was assumed to be .21. Quality Associates then suggested that random samples of size 30 be taken periodically to monitor the process on an ongoing basis. By analyzing the new samples‚ the client could
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with a probability of over 99%; these plans are: CTC‚ HIC‚ and RCNC1. CTC costs are estimated to be the lowest when considering the average of $13.5m and standard deviation of only ~$2m which reflects a lower spread of the costs. Additionally‚ CTC cost savings average ~$14.5m when compared with the other plan offerings‚ with a low standard deviation which reflects more predictable savings. Similar results are obtained when we consider a safer accident rate (i.e.‚ 1 in 6‚6m flights). 2. Cost analysis
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Arturo Delgado Jr. April 23‚ 2013 ESSM 301-‐507 RAINFALL‚ INFILTRATION‚ AND INFLUENCING FACTORS Introduction In this study‚ a rainfall simulation was conducted in order to measure infiltration rates at the Low Stony Hill site in Gatesville‚ Texas. Rainfall simulations are important contributors to understanding infiltration and factors‚ which influence infiltration rate‚ such as bulk density‚ plant biomass and percent cover‚ and topography. The objective
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Project in Math Statistical Data Concerning the Height of BSDAS Academy Department Students S.Y. 2013-2014 Made by: Kent Irvin Pastrana Miguel Pump Adawan Christian Yango Presented to Wayne B. Valera on the day of March 24‚ 2014 I. Abstract Many times‚ people would ask what another person’s height is‚ and joke or say wow depending on the answer. Height is important in our lives; it allows us to reach higher (yes‚ literally)‚ see farther (yes‚ literally again) and
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STATISTICS - Lab #6 Statistical Concepts: Data Simulation Discrete Probability Distribution Confidence Intervals Calculations for a set of variables Open the class survey results that were entered into the MINITAB worksheet. We want to calculate the mean for the 10 rolls of the die for each student in the class. Label the column next to die10 in the Worksheet with the word mean. Pull up Calc > Row Statistics and select the radio-button corresponding to Mean. For Input variables:
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because most of the salaries are less than $1‚000‚000. Standard Salaries Data Mean- $2‚006‚742.122 Median- $1‚016‚630 Standard Deviation- $2‚741‚746.033 Salaries with $100‚000 Bonus Data Mean- $2‚106‚742.122 Median- $1‚116‚630 Standard Deviation- $2‚741‚746.033 With the $100‚000 bonus‚ the mean and the median increased by $100‚000‚ because we increased all the salaries by $100‚000. However‚ the standard deviation was not affected‚ because the distribution of data was not
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data collected as shown in Table 1 of the spreadsheet attached‚ a sample size of 9 (n=9) participants enrolled in the study. Total cholesterol levels measured in children between ages 2 – 6 years was summarized at 1‚765. The sample mean (X) and standard deviation (S) computed as (1765/90) =196.1 and square root summation (X-X) square / n-1 =29.0 respectively. Now to generate a 95% confidence interval for the true mean total cholesterol levels in children from data collected‚ we used the z value for
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a. x-bar = (157+132+109+145+125+139)/6=$134.5 s^2 = ((157-134.5)^2+(132-134.5)^2+(109-134.5)^2+(145-134.5)^2+(125-134.5)^2+(139-134.5)^2)/5 = $276.7 s=sqrt(s^2) = sqrt(276.7) =$ 16.634 b. 69.26% of the expense would fall within 1 standard deviation from the mean -> 134.5+/- 16.634 =
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comparison between results‚ as initial values may have been different. Standard Deviation This measures the spread of data‚ and it can only be calculated when there is more than one repeat. The larger the standard deviation‚ the more widely spread the data is and therefore you have a less reliable mean. You can compare the standard deviation of two different sets of data‚ if the two standard deviation values
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