# 3 1. The first data set contains the weekly closing price of AT&T common shares for year 1979. a. Plot the time series plot of the observations. Make comments of the time series plot. 60 58 52 54 56 C lo s in g P ric e 62 64 Plot of Closing Price of AT&T 0 10 20 30 40 50 Time As you can see, the series on a steady, down. The mean of series appears to be changing and moving down, hence the series is likely not stationary. b. Plot SACF and SPACF, and comment on the plots. 0 .4 - 0 .2 0 .0 0 .2 ACF 0 .6 0 .8 1 .0 Plot of Sample Autocorrelation Function (ACF) 0 5 10 15 Lag The first elevens lags are quite strong. And the fact that most of them pierce the 1.96 standard error line is clearly proof that the series is not white noise and also non stationary. Since the lags in declined slowly, that means that terms in the series are correlated serval periods in the past. 0 .4 0 .2 -0 .2 0 .0 P a rtia l A C F 0 .6 0 .8 Plot of Sample Partial Autocorrelation Function (SPACF) 5 10 15 Lag The lag 1 is pierce those dashed lines, then the lag is significantly different from zero and the series is not white noise. c. Take the differencing of the data. Now, plot the SACF and SPACF of the differenced data and comment on the plots. -2 -1 0 x 1 2 3 Run-Order Plot of Differenced Data 0 10 20 30 40 50 Sequence The plot probably indicates stationary. 0 .4 -0 .2 0 .0 0 .2 ACF 0 .6 0 .8 1 .0 Autocorrelation of Differenced Data With 95 % Confidence Bands 0 5 10 Lag The AFC plot can be considered as a white noise. 15 20 0 .0 -0 .2 -0 .1 P a rtia l A C F 0 .1 0 .2 Partial Autocorrelation of Differenced Data 5 10 15 20 Lag 95 % Confidence Bands As we can see, this series is white noise because there is no lag pierce those dashed lines. Therefore, this time series is stationary. d. Based on the results obtained from part b and c, what would be your choice of initial modal to begin with as you move to modeling stage? I would choose the model (c) because when we have determined that we have stationary, naturally we want to model it. Model (b) is non-stationary. It assumes that the data becomes stationary after differencing. 2. The second data set contains quarterly change in business inventories, stated at annual rates in billions of dollars, from the first quarter of 1955 through the fourth quarter of 1969. a. Plot the time series plot of the observations. Make comments of the time series plot. 15 10 5 -5 0 C h a n g e i n B u s i n e s s In v e n to ri e s Plot of Change in Business Inventories 0 10 20 30 40 50 60 Time The series is likely stationary. The time series plot shows that the series' mean and variance both seen to hold fairly constant for duration of the data set. b. Plot SACF and SPACF, and comment on the plots. 0 .4 -0 .2 0 .0 0 .2 ACF 0 .6 0 .8 1 .0 Plot of Sample Autocorrelation Function (SACF) 0 5 10 15 Lag The sample ACF exhibit significant autocorrelation at lag 1, 2, 3, 4, 5, 6, 7, 11, 12, 13, and 14. 0 .4 0 .2 -0 .2 0 .0 P a rti a l A C F 0 .6 Plot of Sample Partial Autocorrelation Function (SPACF) 5 10 15 Lag The \"upward trend\" in the time series plot suggests nonstationary of some kind. c. Take the differencing of the data. Now, plot the SACF and SPACF of the differenced data and comment on the plots. -1 0 -5 x 0 5 Run-Order Plot of Differenced Data 0 10 20 30 40 50 60 Sequence Now, the series appears to be stationary. 0 .4 -0 .2 0 .0 0 .2 ACF 0 .6 0 .8 1 .0 Autocorrelation of Differenced Data With 95 % Confidence Bands 0 5 10 Lag The AFC plot can be considered as a white noise. 15 20 0 .0 -0 .2 -0 .1 P a rtia l A C F 0 .1 0 .2 Partial Autocorrelation of Differenced Data 5 10 15 20 Lag 95 % Confidence Bands The SPACF are not statistically significantly different from zero. The series looks like white noise. d. Based on the results obtained from part b and c, what would be your choice of initial modal to begin with as you move to modeling stage? I would choose the model (c) because when we have determined that we have stationary, naturally we want to model it. Model (b) is non-stationary. It assumes that the data becomes stationary after differencing. 3. The third data set contains observations adapted from a series provided by a large U.S. corporation. There are 90 weekly OBSERVAINS SHOINHG HE PERENTAGE OF THE TIME THAT PARS FOR AN INDUSRAIL PRODUCT ARE AVALABE HEN NEEDED. 80 82 84 86 88 Plot of Parts for an Industrial Product are Available 76 78 P a r ts fo r a n In d u s tr i a l P r o d u c t a r e A v a i l a b l e a. Plot the time series plot of the observations. Make comments of the time series plot. 0 20 40 60 80 Time He series appears to be stationary. In this plots he o observations \"hang together\" too much for the white noise, the plot is too smooth. Furthermore, the value of 76 at order number 19, 79 at order number 85, and 90 at order number 85 look rather unusual relative the other values. b. Plot SACF and SPACF, and comment on the plots. 0 .4 -0 .2 0 .0 0 .2 ACF 0 .6 0 .8 1 .0 Plot of Sample Autocorrelation Function (SACF) 0 5 10 15 Lag 0 .1 -0 .2 -0 .1 0 .0 P a rtia l A C F 0 .2 0 .3 Plot of Sample Partial Autocorrelation Function (SPACF) 5 10 15 Lag In the SACF and SPACF graphs, the \"upward trend\" in the time series plot suggest nonstationary of some kind. c. Take the differencing of the data. Now, plot the SACF and SPACF of the differenced data and comment on the plots. 0 .4 -0 .2 0 .0 0 .2 ACF 0 .6 0 .8 1 .0 Autocorrelation of Differenced Data With 95 % Confidence Bands 0 5 10 15 20 Lag The AFC plot can be considered as a white noise. 0 .1 0 .0 -0 .2 -0 .1 P a rti a l A C F 0 .2 Partial Autocorrelation of Differenced Data 5 10 15 20 Lag 95 % Confidence Bands As we can see, this series is white noise because there is no lag pierce those dashed lines. Therefore, this time series is stationary. d. Based on the results obtained from part b and c, what would be your choice of initial modal to begin with as you move to modeling stage? I would choose the model (c) because when we have determined that we have stationary, naturally we want to model it. Model (b) is non-stationary. It assumes that the data becomes stationary after differencing. R CODE: x = read.table("C:/Users/Rawiyah/Desktop/r/H3.txt") # import .txt file x=ts(x) #this makes sure R knows that x is a time series plot(x, type="o", xlab="Time",ylab="Parts for an Industrial Product are Available")#time series plot of x with points marked as \"o\" title(" Plot of Parts for an Industrial Product are Available") install.packages("astsa") library(astsa) # start using the package. acf(x, main = "Plot of Sample Autocorrelation Function (SACF)")# Plots the ACF pacf(x, main = "Plot of Sample Partial Autocorrelation Function (SPACF)")# Plots the PACF ## Take first differences. dx = diff(x) plot(dx, ylab="x", xlab="Sequence", main="Run-Order Plot of Differenced Data") ## Generate autocorrelation plot of first differenced data. acf(dx, type = c("correlation"), lag.max=20, main="Autocorrelation of Differenced Data With 95 % Confidence Bands", ylim=c(-.2,1)) ## Generate partial autocorrelation plot of first differenced data. pacf(dx, lag.max=20, main="Partial Autocorrelation of Differenced Data", sub="95 % Confidence Bands") Stock 61 61.625 61 64 63.75 63.375 63.875 61.875 61.5 61.625 62.125 61.625 61 61.875 61.625 59.625 58.75 58.75 58.25 58.5 57.75 57.125 57.75 58.875 58 57.875 58 57.125 57.25 57.375 57.125 57.5 58.375 58.125 56.625 56.25 56.25 55.125 55 55.125 53 52.375 52.875 53.5 53.375 53.375 53.5 53.75 54 53.125 51.875 52.25 change 4.4 5.8 6.7 7.1 5.7 4.1 4.6 4.3 2 2.2 3.6 -2.2 -5.1 -4.9 0.1 4.1 3.8 9.9 0 6.5 10.8 4.1 2.7 -2.9 -2.9 1.5 5.7 5 7.9 6.8 7.1 4.1 5.5 5.1 8 5.6 4.5 6.1 6.7 6.1 10.6 8.6 11.6 7.6 10.9 14.6 14.5 17.4 11.7 5.8 11.5 11.7 5 10 8.9 7.1 8.3 10.2 13.3 6.2 Percentage 80.4 83.9 82.6 77.9 83.5 80.8 78.5 79.3 81.2 81.1 78.2 80.9 77.9 81.5 81.4 78.9 76.2 79.4 81.4 80 79.9 80.5 79.7 81.4 82.4 83.1 80.4 82.9 82.7 84.9 84.4 83.7 84.5 84.6 85.2 85.2 80.1 86.5 81.8 84.4 84.2 84.1 83.2 83.9 86 82.2 81.2 83.7 82.7 84.8 81.2 83.8 86.4 81.6 83.6 85.9 79.8 80.8 78.7 80.6 79.4 77.9 80.4 79.4 83.2 81 81.7 81.2 79.1 80 81.5 83.8 82.2 82.4 79.9 82.3 83.2 81.3 82.4 82.2 82 83.7 84.6 85.7 85.1 84.5 85.6 84.7 79.9 88.9