------o0o------ CHAPTER 14: CHI-SQUARE TESTING STATISTICS FOR BUSINESS TAs: Vo Vuong Van Anh‚ Le Phuoc Thien Thanh‚ and Le Nhat Ho December 21‚ 2013 TABLE OF CONTENTS • PART I: CHI-SQUARE TESTING FOR GOODNESS-OF-FIT. • PART II: CHI-SQUARE TESTING FOR NORMAL DISTRIBUTION. • PART III: CHI-SQUARE TESTING FOR INDEPENDENCE. December 2013 Powered by Vo Vuong Van Anh 2 1 12/21/2013 Hypothesis Testing Procedure for Chi-Square Testing 5 Steps to Perform an Chi-Square Testing STEP 01 State the null and
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order to make the most appropriate recommendation to Management‚ a Managerial report has been prepared that addresses following issues: 1. Normal Probability Distribution sketch used to approximate the demand distribution with the its mean and Standard Deviation 0.4750 0.4750 0.95 10‚000 20‚000 30‚000 10‚000 20‚000 30‚000 Figure 1: Normal Probability Distribution Curve for Expected demand Expected Demand = 20‚000 Hence‚ Mean () = 20‚000 The probability of demand units
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Chapter 9 Monte Carlo methods 183 184 CHAPTER 9. MONTE CARLO METHODS Monte Carlo means using random numbers in scientific computing. More precisely‚ it means using random numbers as a tool to compute something that is not random. For example1 ‚ let X be a random variable and write its expected value as A = E[X]. If we can generate X1 ‚ . . . ‚ Xn ‚ n independent random variables with the same distribution‚ then we can make the approximation A ≈ An = 1 n n Xk . k=1
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Original Marks 44 40 61 52 32 44 70 41 67 72 53 72 Marks after the course 53 38 69 57 46 39 73 48 73 74 60 78 Was the course useful? Consider these 12 participants as a sample from a population. Write short notes on Bernoulli Trials Standard Normal distribution Central Limit theorem www.ignousolvedassignments.com Course Code Course Title Assignment Code Assignment Coverage MS - 8 Quantitative Analysis for Managerial Applications MS-8/SEM - II /2013 All Blocks 1. “Statistical
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Math 221 Quiz Review for Weeks 5 and 6 1. Find the area under the standard normal curve between z = 1.6 and z = 2.6. First we look for the area of both by doing “2nd ‚Vars‚ NORMALCDF” and inputting “-1000‚ “Z‚” 0‚ 1 then find the difference between both. 2. A business wants to estimate the true mean annual income of its customers. It randomly samples 220 of its customers. The mean annual income was $61‚400 with a standard deviation of $2‚200. Find a 95% confidence interval for the true mean
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visual acuity is sharpest at the fovea and decreases with increasing angle of eccentricity (Cowey & Rolls‚ 1974). This has been attributed to factors such as a decline in cone density and an increased receptive field size (Millodot et al.‚ 1975). Normal measures of visual acuity as a function of retinal eccentricity can be best described in reference to results obtained by Millodot et al. (1975) who plotted the minimum angle of resolution (MAR) as a function of target distance from fixation (eccentricity)
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have knowledge of unauthorized aid on this or [assignment‚ quiz‚ paper‚ test]. 2 Pg 212 #14) A normal population has a mean of 12.2 and a standard deviation of 2.5. A) Compute the z value associated with 14.3. - z = (14.3 – 12.2)/2.5 z value is .84 B) What proportion of the population is between 12.2 and 14.3? - Looking for .84 on the curve
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deviation--”average” difference between the mean of a sample and each data value in the sample 14+ 2.51 = Mean of 14 and SD of 2.51 Distribution Shape • Normal • Skewed • Multimodial NORMAL CURVE • Mean is the focal point from which all assumptions made • Area under curve = 100% • Total area divided into segments (these %’s are always the same in the normal curve) – Between Mean & One SD = 68.26% – Between Mean & Two SD = 95.45% – Between Mean & Three SD = 99.7% Distribution Shape Skewed • Most scores
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statistical test(s) to answer this question. * These histogram are clearly asymmetrical and the superimposed normal curve is a little bit do not fit to them well‚ infact the histogram are skewed to the right so the sample is non-normal distribution * The sample mean is far above the median. Other important measures of the normality are the skewness and kurtosis. (In normal distribution the values of skew and kurtosis are close to 0). However in this sample‚ both of them are large positive
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Inferential Statistics • Descriptive statistics (mainly for samples) • Our objective is to make a statement with reference to a parameter describing a population • Inferential statistics does this using a two-part process: • (1) Estimation (of a population parameter) • (2) Hypothesis testing Inferential Statistics • Estimation (of a population parameter) - The estimation part of the process calculates an estimate of the parameter from our sample (called a statistic)‚ as a kind of “guess” as to
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