of the mean overpayment is simply a good guess about what the average overpayment for the population is. Investigating all 1‚000 claims and obtaining the overpayment amount for each would either be impractical‚ unfeasible or both. Thus‚ the auditor deems a sample size of 50 claims to be adequate and sufficiently representative of the entire population. The mean overpayment amount of this sample is then calculated in order to obtain a point estimate of the mean overpayment. The estimated mean is then
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• Mean for Population Based on Example 1‚ to calculate mean for population: [pic] • Standard Deviation for Population Based on example 1‚ to calculate standard deviation for population: [pic] |6.1.2 Sampling Distribution | ▪ Sample statistic such as median‚ mode‚ mean and standard deviation
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analyzed the data they collected as though it were at what level of measurement? * a.Nominal * b.Ordinal * c.Interval/ratio * d.Experimental 2. The mean in is 97.12 3. When looking at the information provided the baseline mean was 14.00 and the posttest mean was 13.36. This means that the subjects in the experimental group scored lower on the depression posttest‚ concluding that they were less depressed after the completing the empowerment program. This would
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hyperlipidemia. A study is conducted to estimate the mean cholesterol in children between the ages of 2 - 6 years of age. It also attempted to establish a correlation as to the effect family history has on the onset of the disease. From 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
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Difficulty: Medium 5. In an experiment involving matched pairs‚ a sample of 12 pairs of observations is collected. The degree of freedom for the t statistic is 10. Answer: False Difficulty: Medium 6. When comparing two independent population means‚ if n1 = 13 and n2 = 10‚ degrees of freedom for the t statistic is 22. Answer: False Difficulty: Easy 7. When comparing the variances of two normally distributed populations using independent random samples‚ if [pic]‚ the calculated value of
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“Statistical Comparison among the Smartphones in the Philippines” A Research Paper Presented to: The Faculty of the Decisional Science and Innovation Department Ramon V. Del Rosario College of Business In Partial Fulfillment of the Requirements for the course Decision Science 2 DECSCI2 Submitted by: Velecina‚ Amiel Joshua A. Guiveses‚ Michelle Louis M. Berlanga‚ Henry Daniel B. Kitane‚ Arja Marie S. Submitted to: Ms. Rachel Arcilla Date: December 4‚ 2014 Introduction Background
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steps of your work (when appropriate). Do NOT simply put an answer. Show or explain how you arrived at your answer. 1) Compute the mean‚ median‚ and mode for the following distribution: (5 pts) 1‚2‚2‚3‚3‚3‚3‚3‚3‚4‚4‚5‚6‚7‚7‚8‚8‚8‚8‚8‚8‚9‚9‚10 The mean is what people call the average. (1+2+2+3+3+3+3+3+3+4+4+5+6+7+7+8+8+8+8+8+8+9+9+10 = 127/24 = 5.2916 Mean = 5.29 The median is the middle value. When the middle does not fall neatly in the distribution‚ use the following formula to identify
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Probability and Statistics Research Project Name: Lakeisha M. Henderson ID: @02181956 Spring 2007 Abstract Table of Contents Principle Component Analysis (PCA) Definition .4 Uses of PCA 5 Illustrative Example of PCA 5 Method to Determine PCA ..6 Basic Analysis of Variance (ANOVA) Purpose and Definition of ANOVA 12 Illustrative Example of ANOVA
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and clusters. 4. Shape = Distribution pattern with data B. Summarizing distribution of univariate data. 1. Mean = add up data values and divide by number of data values Median = list data vlues in order‚ locate middle data value 2. Range = Maximum – minimum Interquartile range (IQR) = Q3 –Q1 Standard deviation is the average distance values fall from the mean of graph. 3. Q1(lower quartile) is the 25th percentile of ordered data or median of lower half of ordered data Median
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parameter of interest. 2. A 99% confidence interval for the mean μ of a population is computed from a random sample and found to be 6 ± 3. We may conclude that C. if we took many‚ many additional random samples‚ and from each computed a 99% confidence interval for μ‚ approximately 99% of these intervals would contain μ 3. I collect a random sample of size n from a population and from the data collected compute a 95% confidence interval for the mean of the population. Which of the following would produce
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