SCSA Mathematics Applications Time series analysis

12 sample questions with marking guides and sample answers · Avg. score: 54.5%

Q1
2021
QCAA
Paper 1
1 mark
Q1
1 mark

The second smoothed value for the 3-point moving average is

Day1234567
Value5101832527090
A

32

B

25

C

20

D

18

Reveal Answer
A

32

This is the raw data value for Day 4, not the calculated moving average.

B

25

This value is incorrect; the average of the second window of data points (10, 18, and 32) is 20, not 25.

C

20

Correct Answer

The second smoothed value is calculated by averaging the second set of three data points (Days 2, 3, and 4): 10+18+323=603=20\frac{10 + 18 + 32}{3} = \frac{60}{3} = 20.

D

18

This is the raw data value for Day 3, not the calculated moving average.

Q25
2024
QCAA
Paper 1
5 marks
Q25

The table shows Darwin’s actual rainfall (mm) each season for two years.

 20222023
Autumn410390
Winter3020
Spring205150
Summer11351100
Q25a
3 marks

Calculate the seasonal index for each season in Darwin.

Reveal Answer

2022 mean rainfall =(410+30+205+1135)/4=445= (410 + 30 + 205 + 1135)/4 = 445
2023 mean rainfall =(390+20+150+1100)/4=415= (390 + 20 + 150 + 1100)/4 = 415

 20222023
Autumn410/445=0.9213410/445 = 0.9213390/415=0.9398390/415 = 0.9398
Winter30/445=0.067430/445 = 0.067420/415=0.048220/415 = 0.0482
Spring205/445=0.4607205/445 = 0.4607150/415=0.3614150/415 = 0.3614
Summer1135/445=2.55061135/445 = 2.55061100/415=2.65061100/415 = 2.6506
 Seasonal index
Autumn(0.9213+0.9398)/2=0.9306(0.9213 + 0.9398)/2 = 0.9306
Winter(0.0674+0.0482)/2=0.0578(0.0674 + 0.0482)/2 = 0.0578
Spring(0.4607+0.3614)/2=0.4111(0.4607 + 0.3614)/2 = 0.4111
Summer(2.5506+2.6506)/2=2.6006(2.5506 + 2.6506)/2 = 2.6006
Marking Criteria
DescriptorMarks

correctly calculates the 2022 mean rainfall and 2023 mean rainfall

1

calculates seasonal ratios for 2022 and 2023

1

calculates seasonal index for each season

1
Q25b
2 marks

This table shows Hobart’s actual rainfall (mm) each season for 2023 and the long-term seasonal indices.

 AutumnWinterSpringSummer
2023 rainfall (mm)130145155132
Seasonal index0.921.021.120.94

Deseasonalise the Hobart rainfall data to identify the 2023 season with the highest seasonally adjusted rainfall.

Reveal Answer

 AutumnWinterSpringSummer
Deseasonalised rainfall130/0.92=141.30130/0.92 = 141.30145/1.02=142.16145/1.02 = 142.16155/1.12=138.39155/1.12 = 138.39132/0.94=140.43132/0.94 = 140.43

Winter has the highest seasonally adjusted rainfall.

Marking Criteria
DescriptorMarks

correctly calculates the deseasonalised rainfall for each season

1

identifies season with highest seasonally adjusted rainfall

1
Q16
2025
VCAA
Paper 1
1 mark
Q16
1 mark

The seasonal index for the number of meat pie sales in winter is 1.75

To correct for seasonality, the actual number of meat pie sales for winter should be reduced, to the nearest whole percentage, by

A

25%

B

43%

C

57%

D

75%

Reveal Answer
A

25%

Incorrect. A 25% reduction would mean multiplying the actual sales by 0.75, which does not match the required deseasonalizing factor of 1/1.750.571/1.75 \approx 0.57.

B

43%

Correct Answer

Correct. To correct for seasonality, you divide the actual sales by the seasonal index (1/1.750.57141/1.75 \approx 0.5714). This means the deseasonalized value is about 57% of the actual value, requiring a reduction of 100%57%=43%100\% - 57\% = 43\%.

C

57%

Incorrect. This is the percentage that the deseasonalized sales represent of the actual sales (1/1.7557%1/1.75 \approx 57\%), rather than the percentage by which the actual sales must be reduced.

D

75%

Incorrect. While a seasonal index of 1.75 means sales are 75% above the seasonal average, reversing this increase requires dividing by 1.75, not subtracting 75%.

Q18
2021
QCAA
Paper 1
4 marks
Q18

The table shows the profit made each year (in thousands of dollars) by a small business.

YearProfit ($'000s)
201542.1
201636.9
201748.4
201852.3
201956.1
202059.8
Q18a
2 marks

Use a mathematical model to determine the equation of the least-squares line to fit this data.

Reveal Answer

Let x=x = the number of years since 2014
Let y=y = the business's annual profit (in $'000s)

y=4.286x+34.267y = 4.286x + 34.267

Marking Criteria
DescriptorMarks

correctly defines the variables

1

correctly determines the equation of the least-squares line

1
Q18b
2 marks

Use the least-squares line to forecast the profit in 2021, to the nearest hundred dollars.

Reveal Answer

For 2021, x=7x = 7
y=4.286×7+34.267\therefore y = 4.286 \times 7 + 34.267
=64.269= 64.269

The business will make $64 300.

Marking Criteria
DescriptorMarks

correctly determines the xx value

1

determines profit

1
Q16
2022
QCAA
Paper 1
3 marks
Q16

The table shows the number of sales for a small business in their first six months of trading.

Time in months, ttNumber of sales, nn
186
2180
3160
4226
5240
6335
Q16a
1 mark

Use your calculator to determine the equation of the least-squares line.

Reveal Answer

n=42.6t+55.4n = 42.6t + 55.4

Marking Criteria
DescriptorMarks

Correctly determines the equation of the least-squares line

1
Q16b
2 marks

Use the equation from Question 16a) to predict the number of sales in the 21st month.

Reveal Answer

Let t=21t = 21
n=42.6(21)+55.4n = 42.6(21) + 55.4
=950= 950

The predicted number of sales is 950.

Marking Criteria
DescriptorMarks

Substitutes into equation from Question 16a)

1

Predicts number of sales

1
Q6
2024
QCAA
Paper 1
1 mark
Q6
1 mark

The table shows the maximum daily temperature (°C) for a week.

MonTueWedThuFriSatSun
24.425.224.625.225.625.725.9

If the simple 5-point moving average for Wednesday is 25.0 °C, what is the simple 5-point moving average (°C) for Friday?

A

25.4

B

25.5

C

25.6

D

26.0

Reveal Answer
A

25.4

Correct Answer

The 5-point moving average for Wednesday was calculated using the 5 days centered on it (Mon–Fri). Similarly, the average for Friday uses the 5 days centered on Friday (Wed–Sun): 24.6+25.2+25.6+25.7+25.95=127.05=25.4\frac{24.6 + 25.2 + 25.6 + 25.7 + 25.9}{5} = \frac{127.0}{5} = 25.4.

B

25.5

This is incorrect. The sum of the temperatures from Wednesday to Sunday is 127.0127.0, which results in an average of 25.425.4, not 25.525.5.

C

25.6

This is the actual temperature recorded on Friday (25.6C25.6^\circ\text{C}), rather than the calculated 5-point moving average.

D

26.0

This value is incorrect. It is significantly higher than the actual average of the temperatures from Wednesday to Sunday.

Q10
2023
QCAA
Paper 1
1 mark
Q10
1 mark

Annual sales data and related quarterly indices are shown. The quarterly indices were calculated by applying the average percentage method using the mean.

QuarterQ1Q2Q3Q4
Sales160xx128200
Index1.00.95yy1.25

Determine the values for xx and yy.

A

122 | 0.8

B

122 | 3.2

C

152 | 0.8

D

152 | 3.2

Reveal Answer
A

122 | 0.8

While the value for yy (0.8) is correct, the value for xx is incorrect. The average quarterly sales is 160 (calculated as 160/1.0160/1.0), so xx should be 160×0.95=152160 \times 0.95 = 152.

B

122 | 3.2

Both values are incorrect. The sum of quarterly indices must equal 4, making y=0.8y=0.8, and the sales value xx is calculated by multiplying the average sales (160) by the index (0.95).

C

152 | 0.8

Correct Answer

The sum of quarterly indices is 4, so y=4(1.0+0.95+1.25)=0.8y = 4 - (1.0 + 0.95 + 1.25) = 0.8. The deseasonalized average sales is 160 (derived from 160/1.0160/1.0), so x=160×0.95=152x = 160 \times 0.95 = 152.

D

152 | 3.2

The value for xx (152) is correct, but yy is incorrect. The sum of the indices must equal 4, not the partial sum of the other indices (3.2).

Q13
2024
QCAA
Paper 1
1 mark
Q13
1 mark

The table shows time series data for a company's quarterly sales.

Quarter1234
Sales ($)2700360045007200
Seasonal index0.60.81.0

Determine the seasonally adjusted sales ($) for the fourth quarter.

A

4500

B

6000

C

8640

D

11 520

Reveal Answer
A

4500

Correct Answer

The sum of seasonal indices for four quarters must equal 4. The Q4 index is 4(0.6+0.8+1.0)=1.64 - (0.6 + 0.8 + 1.0) = 1.6. Seasonally adjusted sales are calculated by dividing actual sales by the index: 72001.6=4500\frac{7200}{1.6} = 4500.

B

6000

This assumes the seasonal index follows a linear pattern (0.6, 0.8, 1.0, 1.2) and divides sales by 1.2. However, seasonal indices must sum to the number of periods (4), making the correct Q4 index 1.6.

C

8640

This result likely comes from estimating the index incorrectly as 1.2 (based on a pattern) and multiplying it by the sales, rather than dividing by the correct index.

D

11 520

This is the result of multiplying the actual sales by the seasonal index (7200×1.67200 \times 1.6). To deseasonalize data (find seasonally adjusted sales), you must divide the actual sales by the seasonal index.

Q5
2022
QCAA
Paper 1
1 mark
Q5
1 mark

The table lists the number of books sold per month by an online bookstore. If the simple 3-point moving average in October is 54, what is the simple 3-point moving average in May?

JanFebMarAprMayJunJulAugSepOctNovDec
455268658965533340457792
A

69

B

73

C

74

D

89

Reveal Answer
A

69

This value represents the centered moving average for June (calculated as 89+65+533=69\frac{89 + 65 + 53}{3} = 69), rather than May.

B

73

Correct Answer

The context provided for October (40+45+773=54\frac{40 + 45 + 77}{3} = 54) indicates a centered moving average. For May, the calculation uses the values from April, May, and June: 65+89+653=2193=73\frac{65 + 89 + 65}{3} = \frac{219}{3} = 73.

C

74

This result comes from using a trailing moving average (March, April, May: 68+65+893=74\frac{68 + 65 + 89}{3} = 74), but the problem context establishes that a centered moving average is used.

D

89

This is the actual number of books sold in May, not the 3-point moving average.

Q2
2023
QCAA
Paper 2
4 marks
Q2
4 marks

Buffalo fly bites cause skin wounds on cattle. The table shows the average number of skin wounds per animal in a herd for two years.

 AutumnWinterSpringSummer
202128528195460
202227622170392

Deseasonalise the data.

Reveal Answer

YearSeasonNumber of skin woundsYearly averageNumber / Yearly averageSeasonal indicesDeseasonalised number
2021Autumn2852421.1776...1.2307...232
 Winter28 0.1157...0.1090...257
 Spring195 0.8057...0.7982...244
 Summer460 1.9008...1.8620...247
2022Autumn2762151.2837...1.2307...224
 Winter22 0.1023...0.1090...202
 Spring170 0.7906...0.7982...213
 Summer392 1.8232...1.8620...211
Marking Criteria
DescriptorMarks

correctly determines the yearly averages

1

determines number/yearly average values

1

determines seasonal indices

1

determines deseasonalised numbers

1
Q16
2023
VCAA
Paper 1
1 mark
Q16
1 mark

The number of visitors each month to a zoo is seasonal.

To correct the number of visitors in January for seasonality, the actual number of visitors, to the nearest percent, is increased by 35%.

The seasonal index for that month is closest to

A

0.61

B

0.65

C

0.69

D

0.74

E

0.77

Reveal Answer
A

0.61

Incorrect. This is the result of calculating 1/1.651 / 1.65, which would correspond to a 65% increase rather than a 35% increase.

B

0.65

Incorrect. This incorrectly assumes the seasonal index is found by simply subtracting the percentage increase from 1 (10.35=0.651 - 0.35 = 0.65).

C

0.69

Incorrect. This is a miscalculation. The correct seasonal index is found by solving 1.35=1/Seasonal Index1.35 = 1 / \text{Seasonal Index}.

D

0.74

Correct Answer

Correct. Increasing the actual value by 35% means multiplying by 1.35. Since the deseasonalised value equals the actual value divided by the seasonal index, the seasonal index is 1/1.350.741 / 1.35 \approx 0.74.

E

0.77

Incorrect. This value would be the seasonal index if the actual number of visitors was increased by about 30% (1/1.300.771 / 1.30 \approx 0.77), not 35%.

Q1
2022
QCAA
Paper 2
4 marks
Q1
4 marks

The table shows a swimwear company’s seasonally adjusted swimsuit sales (in thousands).

 SpringSummerAutumnWinter
Seasonally adjusted swimsuit sales (in thousands)33.334.836.435.8

The long-term seasonal indices for spring, summer and winter are 1.11, 1.42 and 0.62 respectively.

Determine the actual swimsuit sales for autumn.

Reveal Answer

Let x=x = autumn's seasonal index

Total of seasonal indices:
1.11+1.42+0.62+x=41.11 + 1.42 + 0.62 + x = 4

x=0.85\therefore x = 0.85

Actual value for autumn

actual value=deseasonalised value×seasonal index\text{actual value} = \text{deseasonalised value} \times \text{seasonal index}
=36.4×0.85= 36.4 \times 0.85
=30.94= 30.94

In autumn they had actual sales of 30 940 swimsuits.

Marking Criteria
DescriptorMarks

correctly identifies the sum of all the seasonal indices

1

correctly determines autumn's seasonal index

1

uses an appropriate method for determining actual value

1

determines actual sales for autumn

1

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