TIME SERIES ‘Time series’ data is a bivariate data, where the independent variable is time. We use scatterplot to display the relationship between the dependent variable and time of ‘time series’. The time (years, weeks, months) is always converted to equivalent digits starting from 1. A time series plot differs from a normal scatter plot in that the points will be joined by line segments in time order.

Time series data are often complex and show seemingly wild fluctuations. The fluctuations are generally due to one or more of the following characteristics of the relationship: Trend: upward or downward where the points are increasing or decreasing, respectively as time progresses. increasing decreasing Seasonal variation: the time series graph shows a pattern that repeats at regular intervals which occur within a time period of one year or less.

Cyclic variation:the graph shows longer term movements which last for more than a year. Some cycles repeat regularly, and some do not. Random variation: When the pattern of the series plot is not seasonal or cyclical Example: List the characteristics that are present in each of the time series plots shown below. red: decreasing trend, random variation blue: random and seasonal variation yellow: cyclical and random variation

Constructing a time series plot using Ti- enspire Example:Construct a time series plot from the following data Convert the years to corresponding digits starting from 1. Enter the data on ‘Lists & Spreadsheets’ Construct the scatterplot on ‘data and statistics’. Press menu, 2:Connect Data Points,to join points.

Smoothing a time series plot Smoothing is the process that removes the random fluctuations from time series data. This allows any underlying trend to be more clearly seen, to fit a line and make predictions. Moving mean and median smoothing are techniques that are used to smooth the time series. 3- and 5-moving mean smoothing Example: The following table gives the number of births per month over a calendar year in a small country hospital. Use the 3-moving mean and the 5-moving mean methods, correct to one decimal place, to complete the table. monthnumber of births January10 February12 March6 April5 May22 June18 July13 August7 September9 October10 November8 December15

monthnumber of births January10 February12 March6 April5 May22 June18 July13 August7 September9 October10 November8 December15 3-moving mean ( )/ 3 = 9.3 (12+6+5)/ 3 = 7.7 (6+5+22)/ 3 = 11 ( )/ 3 = 15 ( )/ 3 = 17.7 ( )/3 = 12.7 (13+7+9)/ 3 = 9.7 (7+9+10)/ 3 = 8.7 (9+10+8)/ 3 = 9 ( )/ 3 = 11 Find the mean of the 3 values(number of births) and replace it with the middle

monthnumber of births January10 February12 March6 April5 May22 June18 July13 August7 September9 October10 November8 December15 5-moving mean ( )/ 5 = 11 ( )/ 5 = 12.6 ( )/ 5 = 12.8 ( )/ 5 = 13 ( )/5 = 13.8 ( )/5 = 11.4 ( )/5 = 9.4 ( )/5 = 9.8

If we smooth over an even number of points, we run into a problem. The centre of the set of points is not at a time point belonging to the original series. Usually,we solve this problem by using a process called centring. 4-mean smoothing with centring Use 4- mean smoothing with centring to smooth the number of births per month over a calendar year in a small country hospital.

Example: The table below displays the total rainfall (in mm) in a reservoir over one- year period monthJanFebMarAprMayJuneJulyAugSeptOctNovDec rainf Using 3- mean moving average and 2 - mean moving average with centring, find the smoothed values for the total rainfall in April. Solution: 3- moving mean for April = = Mar & Apr = = moving mean for April = centring Apr = = 77 2 Apr & May = = 87 2

Moving median smoothing monthnumber of births January10 February12 March6 April5 May22 June18 July13 August7 September9 October10 November8 December15 3-moving median Find the 3-moving median smoothing of the following data by finding the median of the 3 values and replacing it with the middle

3-median smoothing using a graphing approach Construct a 3- median smoothed plot of the time series data shown 1 Construct the time series plot 2 Locate on the time series plot the median of the first three points (Jan,Feb,Mar). 3 Continue this process by moving onto the next three points to be smoothed (Feb,Mar,Apr). Mark in their median on the graph, and continue the process until you run out of groups of three. 4 Join up the median points with a line segment

Example: The 5 points shown on the grid have been taken from a time series plot that is to be smoothed using median smoothing. The coordinates of the median of these five points are: A (3,1) B (3,2) C(3,2.4) D(3,2.5) E (3,3) Seasonal indices A seasonal index is a measure of how a particular season compares with the average season. Consider the monthly seasonal indices for unemployment given in the table below: Seasonal indices are calculated so that their average is 1. This means that the sum of the seasonal indices equals the number of seasons. Thus, if the seasons are months, the seasonal indices add to 12. If the seasons are quarters, then the seasonal indices would add to 4, and so on. January has seasonal index of 1.1 which means, January’s unemployment is 10% above average. September’s unemployment is 15% less than average.

Example: Find the seasonal index missing from the following table: Solution: = 0.9 Deseasonalising is the process that is used to remove the seasonal effects from a set of data. This allows any underline trend to be made clearer. We can use seasonal indices to deseasonalise time series. To calculate deseasonalised data, each entry is divided by its seasonal index as follows.

Deseasonalising data Example: Deseasonalise the quarterly sales figures of Summer Year1 using the data and seasonal indices tables below. Solution: Deseasonalised data for ‘Summer 1’ = Summer 1 data = 920 = 893 Summer seasonal index 1.03 Calculating seasonal indices

Example:Mikki runs a shop and she wishes to determine quarterly seasonal indices based on her last year’s sales, which are shown in the table below. Solution: Using the above formula to find the seasonal index seasonal index = value of the quarter yearly average Find the quarter average yearly average = = Find the seasonal index of each season seasonal index Summer = 920 = seasonal index Autumn = 1085 = seasonal index Winter = 1241 = seasonal index Spring = 446 =

Calculating seasonal indices (several years’data) Suppose that Mikki has in fact three years of data, as shown. Use the data to calculate seasonal indices, correct to two decimal places. Solution: The seasonal average of year 1 was found at previews page Find the quarter average for year 2 quarter average year 2 = = Find the seasonal index of each season seasonal index Summer = 1035 = seasonal index Autumn = 1180 = seasonal index Winter = 1356 = seasonal index Spring = 541 =

Find the quarter average for year 3 quarter average year 2 = = Find the seasonal index of each season seasonal index Summer = 1299 = seasonal index Autumn = 1324 = seasonal index Winter = 1450 = seasonal index Spring = 659 = To find the seasonal indices of the 3 years we need to find the average seasonal index of each season.

FORCASTING Example 1 a:Fit a trend line to the data in the following table, which shows the number of government schools in Victoria over the period 1981–92. 1.Construct a time series plot of the data to ensure linearity. 2. Find the equation of least squares regression line. Number of schools = 2169 − 12.5 × year Forcasting is the term that is used in Time series for predicting.

b How many government schools do we predict for Victoria in 2010 if the current decreasing trend continues? Solution: Substitute the appropriate value for year in the equation determined using least squares regression. Since 1981 was designated as year ‘1’, then 2010 is year ‘30’. Number of schools = 2169 − 12.5 × year = 2169 − 12.5 × 30 = 1794 Example2: The following table shows the seasonal indices for the monthly unemployment numbers for workers in a regional town. monthJanFebMarAprMayJunJulAugSepOctNovDec season al index a. The actual number of unemployment in the regional town in September is 330. What is the desaesonalised number of unemployed in September. Solution: deseasonalised = actual = 330 = seasonal index 0.94

b A trend line that can be used to forecast the deseasonalised number of unemployed workers in the regional town for the first 9 months of the year is given by deseasonalised number of unemployed = × month where month 1 is January, month 2 is February, and so on. What is the actual number of unemployed for June? Solution: Find first the deaseasonalised number of unemployed: Replace the month with 6 deseasonalised number of unemployed = × 6 = Use the deseasonalised = actual to find the actual. seasonal index actual = × 0.86 = 303.6

Example 3: The quarterly seasonal indices for mineral water sales are shown in the table below: Quarter 1Quarter 2Quarter 3Quarter 4 Seasonal index When deseasonalised the amount of mineral water sold in quarter 1 is litres. What is the actual amount of mineral water for quarter 1? Solution: Use deseasonalised = actual to find the actual. seasonal index actual = × 1.28 =