CA-time series exercises 1 PDF

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Time Series revision exercises Dec 2010 The annual report of a Greenspot Departmental Store showed the following sales figure for the past three years Year 2002 2003 2004

Quarter 1 3.5 3.3 2.9

Sales (RM million) Quarter 2 Quarter 3 4.7 5.3 4.6 5.0 4.1 4.8

Quarter 4 8.4 7.5 7.1

(a)

Plot the data on the graph. marks)

(6

(b)

Use an additive model and calculate the trend values and the appropriate seasonal factors for the time series. (12 marks)

(c)

Forecast the sales for the first two quarters of the year 2005 and comment on the accuracy of your results. (3 marks)

(d)

Find the seasonally adjusted value for year 2004. marks)

(4

(Total: 25 marks)

Dec 2011 (a)

Explain why it is necessary to adjust seasonally time series data.

(b)

The following data shows the output of a factory over a 3-week period. The factory works a 5-day week.

Week 1 Week 2 Week 3

Monday 77 90 93

Tuesday 92 99 102

Wednesday 104 110 106

Thursday 98 102 110

(5 marks)

Friday 78 80 83

(i) By means of a moving average, find the trend and the average daily variations. (15 marks) (ii) Seasonally adjust the data for Week 3. What does this show?

(5 marks) (Total: 25 marks)

Dec 2010 (a)

Sales (RM million)

Time series graph 9 8 7 6 5 4 3 2 1 0 Q1

Q2

Q3

Q4

Q1

Q2

Q3

Q4

Q1

Q2

Q3

Q4

Quarters

the title (1 mark), the x/y axis (1 mark) and the graph (4 marks) (b) Year 2002

1

Sales 3.5

2

4.7

Sum of 4

Sum of 8

Trend

Deviation

43.6

5.45

-0.15

43.3

5.4125

2.9875

42.9

5.3625

-2.0625

41.7

5.2125

-0.6125

40.4

5.05

-0.05

39.5

4.9375

2.5625

38.8

4.85

-1.95

38.2

4.775

-0.675

(2 marks) (2 marks) (2 marks)

(2 marks)

21.9 3

5.3 21.7

4

8.4 21.6

2003

1

3.3

2

4.6

21.3 20.4 3

5.0 20.0

4

7.5

1

2.9

19.5 2004

19.3 2

4.1 18.9

3

4.8

4

7.1

Seasonal Variation table Year Qtr 1 2002 2003 -2.0625 2004 -1.95 Total -4.0125 Average -2.00625 adjustment S Factor

-0.00625 -2.0125 1 mark

Qtr2 -0.6125 -0.675 -1.2875 -0.64375

Qtr3 -0.15 -0.05 -0.20 -0.1

Qtr4 2.9875 2.5625 5.55 2.775

-0.00625 -0.6500 1 mark

-0.00625 -0.10625 1 mark

-0.00625 2.76875 1 mark

Total Ave=0.025 Total sf =0

Adjustment = 0.025/-4 = -0.00625 (c)

(d)

Average projection = (4.775 -5.45) / 7 = -0.0964 Forecast (2005, Q1) = 4.775 + 3(-0.0964) – 2.0125 = 2.4733(RM million)

(1 mark) (½ mark)

Forecast (2005, Q2) = 4.775 + 4 (-0.0964) – 0.65 = 3.7394 (RM million)

(1 mark) (½ mark)

Q1 : 2.9 – (-2.0125) = 4.9 Q2: 4.1 – (-0.65) = 4.8 Q3: 4.8 – (-0.10625) = 4.9 Q4: 7.1 – (2.76875) = 4.3

(4 marks) (Total: 25 marks)

Dec 2011

(a)

Explain why it is necessary to adjust seasonally time series data. One problem with interpreting data over time is that many data series exhibit movements (1M) that recur every year in the same month or quarter. For example, housing permits increase every spring when the weather improves, while toy sales usually peak in December (1M) . This dynamic makes it hard for economists to interpret (1M) the underlying trend in some data series. For instance, were sales better this December or was it just the usual holiday run up ? Economists want to know if sales were better than the normal seasonal increase. To understand what the data are really saying about economic growth, statisticians and economists remove such predictable fluctuations (1M) —or seasonality (1M) —from the data.

(b)

.. (iii)

By means of a moving average, find the trend and the average daily variations.

week no.

data(sales) add in fives 77 92

1

2

3

Trend, ÷ 5

Data - trend

-

104

449

89.8

14.2

98

462

92.4

5.6

78

469

93.8

-15.8

90

475

95.0

-5.0

99

479

95.8

3.2

110

481

96.2

13.8

102 80

484

96.8

5.2

487

97.4

-17.4

93

483

96.6

-3.6

102

491

98.2

3.8

106

494

98.8

7.2

110 83

-

(3M)

(3M)

(3M)

Week 1 2 3

Mon -5.0 -3.6

Tue

Wed

-

14.2 13.8 7.2

3.2 3.8

Adjusted value Daily adjustment

5.2

-15.8 -17.4 -33.2000

-8.6000 -4.3000 0.0533 4

7.0000 3.5000 0.0533 4

0.05334

-4.2467

3.5533

11.7867

5.4533

-16.5467

-4

4

12

5

-17

(1M)

(1M)

(1M)

35.2000 11.7333

-16.6000 0.05334

(1M)

Adjustment = - [ ∑ average / 5 ] = -[-0.2667 /5] = 0.05334 (iv)

Fri

10.800 0 5.4000 0.0533 4

Total Average Adjustment

Thu 5.6

(1M)

(1M)

Seasonally adjust the data for Week 3. What does this show? Monday: 93 – (-4) = 97units Tuesday: 102 –(4) = 98 units Wednesday: 106 – (12)= 94 units Thursday: 110 - (5) = 105 units Friday: 83-(-17)= 100 units

(1M) (1M) (1M) (1M) (1M)...