following for the data in Column K‚ “The degree of agreement among patrons that Remington’s has large portions‚” on the Remington Data worksheet of the Remington’s Data Set workbook: Mean -3.26 Standard deviation-0.911 Range -3 4 Mean 3.261306533 Standard Error 0.064596309 Median 4 Mode 4 Standard Deviation 0.911243075 Sample Variance 0.830363941 Kurtosis -1.16899198 Skewness -0.663704706 Range 3 Minimum 1 Maximum 4 Sum 649 Count 199 Largest(1) 4 Smallest(1) 1 Confidence Level(95.0%) 0.12738505
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order is long and uncertain. This time gap is called “lead time.” From past experience‚ the materials manager notes that the company’s demand for glue during the uncertain lead time is normally distributed with a mean of 187.6 gallons and a standard deviation of 12.4 gallons. The company follows a policy of placing an order when the glue stock falls to a predetermined value called the “reorder point.” Note that if the reorder point is x gallons and the demand during lead time exceeds x gallons
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Remington’s Steakhouse Project Brian Jones Research Methods & Applications Dr. Jones August 25‚ 2011 Table of Contents Table of Contents 2 List of Tables 3 Introduction 4 The Research Objectives 4 The Research Questions 5 Literature Review 6 Answers to Research Questions 8 Recommendations to Remington’s 15 References 18 Annotated Bibliography 19 Appendix(ces) 22 List of Tables Table 1 Demographic Description of the Average Remington’s Patron9 Table 2 Reported Income by Remington’s
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Mean and Standard Deviation The mean‚ indicated by μ (a lower case Greek mu)‚ is the statistician ’s jargon for the average value of a signal. It is found just as you would expect: add all of the samples together‚ and divide by N. It looks like this in mathematical form: In words‚ sum the values in the signal‚ xi‚ by letting the index‚ i‚ run from 0 to N-1. Then finish the calculation by dividing the sum by N. This is identical to the equation: μ =(x0 + x1 + x2 + ... + xN-1)/N. If you are not
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the actuarial field and finds the average salary to be $41‚000. The population standard deviation is $3000. Can her claim be supported at 0.05? x¯=14.7‚ μx¯=13.77‚ ox¯=5.34‚ n=29‚ α=.01 3. Monthly Home Rent. The average monthly rent for a one bedroom in San Francisco is $ 1229. A random sample of 15 one bedroom homes about 15 miles outside of San Francisco had a mean rent of $1350. The population standard deviation is $250. At a=0.05 can we conclude that the monthly rent outside San Francisco
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Standard deviation is the square root of the variance (Gravetter & Wallnau‚ 2013). It uses the mean of the distribution as a reference point and measures variability by considering the distance of each score from the mean. It is important to know the standard deviation for a given sample because it gives a measure of the standard‚ or average‚ range from the mean‚ and specifies if the scores are grouped closely around the mean or are widely scattered (Gravetter & Wallnau‚ 2013). The standard deviation
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1. Convert prices to total return (% change in the price) = (Pt – Pt-1) / Pt-1 2. Remove outliers – sort data and remove anything +/- 20% 3. Calculate historical average and historical risk X-BAR = Σx/n Calculate the sum of the total return and divide by the number of observations • Variance = σ2 = Σ(x – x bar) 2 / (n-1) Fix X-BAR‚ double click to apply to all dates‚ get the sum‚ divide by (n-1) Risk = σ = √σ = SQRT(Variance) = standard deviation 4. Average Matrix Excel Options
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x can be included in y f. Represents the simplest measure of spread (or variability) g. Indicates the most frequent observation in a frequency distribution h. Represents a theoretical family of distributions that may have any mean or any standard deviation i. Indicates how widely the out around the measures of central tendency j. Measure an event over time B. Match the following data display tools with their descriptions. (Descriptions may be used more than once or not at all.) A horizontal
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Investment Returns: These data are the annual returns on shareholders’ funds of 97 of Australian’s top 100 companies for the years 1990 and 1998. (i) Produce a histogram of the 1990 returns. (ii) Produce a histogram of the 1998 returns. (iii) Find the mean‚ median‚ range and standard deviation for the 1990 returns. Annual Returns % (1990) Mean 12.91865979 Median 11.38 Standard Deviation 9.297513067 Range 75.01 (iv) Repeat part (iii) for the 1998 returns. Annual Returns % (1998)
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13. Variance and Standard Deviation (expected). Using the data from problem 13‚ calculate the variance and standard deviation of the three investments‚ stock‚ corporate bond‚ and government bond. If the estimates for both the probabilities of the economy and the returns in each state of the economy are correct‚ which investment would you choose considering both risk and return? Why? ANSWER Variance of Stock = 0.10 x (0.25 – 0.033)2 + 0.15 x (0.12 – 0.033)2 + 0.50 x (0.04 – 0.033)2 + 0
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