COMPUTER ASSIGNMENTAUTUMN SESSION 2012SCHOOL OF COMPUTING, ENGINEERING & MATHEMATICS | | | Student Name and ID number | Group Member 1 | Matthew Yee, 17048604. Tuesday 12-1pm | Group Member 2 | Johnathan Morey, 17238696. Wednesday 11am-12pm | Group Member 3 | Linda Yi, 17032764. Tuesday 12-1pm | Group Member 4 | Christopher Shepherd, 16673537. Tuesday 9-10am | Unit Name: Statistics for Business Unit Number: 200032 Due Date: Week 13 tutorial Number of Questions:
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lags=4) #plotting the ACF and PACF plots par(mfrow=c(2,1)) acf(log.ret);pacf(log.ret) #determination of the models library(rugarch) ret.model<-log.ret[1:635] spec1= ugarchspec(variance.model = list(model = 'eGARCH',garchOrder = c(1,1)), distribution = 'ged') egarch.fit <- ugarchfit(data=c(log.ret), spec = spec1) #fitting EGARCH model egarch.fit coef(egarch.fit) spec <- ugarchspec( variance.model = list(model = "fGARCH", submodel = "GARCH", garchOrder = c(1,2))) garch
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| 22 | .581 | | sample 2 | .959 | 21 | .490 | | sample 3 | .959 | 21 | .490 | | sample 4 | .950 | 19 | .395 | | sample 5 | .950 | 19 | .395 | | sample 6 | .950 | 19 | .395 | Answer: The K-S test was significant (p > 0.05) for a normal distribution in all the groups. 1b. Run the analysis. Show table results (directly from SPSS) only for Tests of Within Subjects Effects and Pairwise Comparisons. Tests of Within-Subjects Effects | Measure: Group | Source | Type III Sum of Squares
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Data For Tasks 1-8, consider the following data: 7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5 In Tasks 1-8 you are asked to conduct some computations regarding this data. The computation should be carried out manually. All the steps that go into the computation should be presented and explained. (You may use R in order to verify your computation, but not as a substitute for conducting the manual computations.) A Random Variable In Tasks 9-18 you are
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Goethe University Frankfurt Advanced Econometrics 2, Part 2 Sommersemester 2016 Prof. Michael Binder, Ph.D. I. Vector Autoregressions and Vector Error Correction Models 3. Estimation and Inference with and without Parameter Restrictions Cointegrated VAR – Case of a Single Cointegrating Relationship Special Case of One Cointegrating Relationship: Weak Exogeneity and ARDL Models When the cointegration rank of a cointegrated VAR is one, then under certain conditions it is feasible to work with a notably
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environmental scenario in future. This means the VaR is only useful when it predicts risks accurately (Lucas, 2001). To verify the consistency and reliability of VaR calculations, it is necessary to back test the model with appropriate statistical standards. Back testing entails comparing actual profits and losses to projected VaR estimates as an apt form of reality check (Lucas, 2001). Where the results reveal inaccurate estimates, the models are re-examined to identify the incorrect assumptions, parameters
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small oil press machine adopts the heating temperature control system, according to the environmental temperature changes automatically adjust the temperature of crude oil, in order to achieve the effect of fast and precise filter, so as to ensure the normal production in all kinds of weather. What’s more, the household oil press machine has widely using, it not only can press peanut, sesame, rapeseed, but also can press sesame, walnut and so on more than 20 kinds of oil-bearing crops. Any interests
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GET FILE='\\soliscom.uu.nl \uu\Users\3530647 \1\Capm spss.sav '. DATASET NAME DataSet1 WINDOW=FRONT. NEW FILE. DATASET NAME DataSet2 WINDOW=FRONT. REGRESSION /DESCRIPTIVES MEAN STDDEV CORR SIG N /MISSING LISTWISE /STATISTICS COEFF OUTS CI(95) R ANOVA /CRITERIA =PIN(.05) POUT(.10) /NOORIGIN /DEPENDENT mcdonaldrp /METHOD=ENTER sp500rp /RESIDUALS DURBIN /SAVE RESID. Regression [DataSet2] Descriptive Statistics Mean Std. Deviation N mcdonaldrp -,0017 ,01145 502
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sparksj // BEAT MESSAGE try { window.wixBiSession = { initialTimestamp : Date.now(), viewerSessionId: 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function(c) { var r = Math.random()*16|0, v = c == 'x' ? r : (r&0x3|0x8); return v.toString(16); } ) }; (new Image()).src = 'http://frog.wix.com/bt?src=29&evid=3&pn=1&et=1&v=3
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The population standard deviation is 2. (a) Create a 95% confidence interval for the mean. We know σ, therefore we should use the z − table. This is a one-tailed (lower tail) test, so the 95% confidence interval will be given then by σ x − z.05 √ , ∞ ¯ n 2 19.4 − 1.65 √ , ∞ 40 The 95% confidence interval is µ ∈ [18.878, ∞). (b) What is the p-value? The p-value is the area in the lower tail. First, we calculate the z-value: z= 19.4 − 20 √ = −1.9 2/ 40 Using the normal table with z
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