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Data Analysis, GEN2 2024 VCAA 4

The time series plot below shows the gold medal-winning height for the women's high jump, \(\textit{Wgold}\), in metres, for each Olympic year, \(year\), from 1952 to 1988.
 

A five-median smoothing process will be used to smooth the time series plot above.

The first two points have been placed on the graph with crosses (X) and joined by a dashed line (---).

  1. Complete the five-median smoothing by marking smoothed values with crosses (X) joined by a dashed line (---) on the time series plot above.   (1 mark)

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  2. Identify two qualitative features that best describe the time series plot above.  (1 mark)

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Show Answers Only

a.     

b.    \(\text{Random fluctuations, increasing trend.}\)

Show Worked Solution

a.    \(\text{Medians are:}\)

\(\textbf{1960}: 1.67, 1.76, \colorbox{lightblue}{1.82}, 1.85, 1.90\)

\(\textbf{1964}: 1.76, 1.82, \colorbox{lightblue}{1.85}, 1.90, 1.92\)

\(\textbf{1968}: 1.82, 1.85, \colorbox{lightblue}{1.90}, 1.92, 1.93\)

\(\textbf{1972}: 1.82, 1.90, \colorbox{lightblue}{1.92}, 1.93, 1.97\)

\(\textbf{1976}: 1.82, 1.92, \colorbox{lightblue}{1.93}, 1.97, 2.02\)

\(\textbf{1980}: 1.92, 1.93, \colorbox{lightblue}{1.97}, 2.01, 2.02\)
  

b.   \(\text{Qualitative features:}\)

\(\text{- Random fluctuations}\)

\(\text{- Increasing trend.}\)

♦♦♦ Mean mark (b) 24%.

Filed Under: Time Series Tagged With: Band 4, Band 5, smc-266-40-Time Series Trends, smc-266-70-MEDIAN Smoothing

Data Analysis, GEN2 2023 VCAA 4

The time series plot below shows the average monthly ice cream consumption recorded over three years, from January 2010 to December 2012.

The data for the graph was recorded in the Northern Hemisphere.

In this graph, month number 1 is January 2010, month number 2 is February 2010 and so on.
 

  1. Identify a feature of this plot that is consistent with this time series having a seasonal component.  (1 mark)

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  2. The long-term seasonal index for April is 1.05
  3. Determine the deseasonalised value for average monthly ice cream consumption in April 2010 (month 4).
  4. Round your answer to two decimal places.  (1 mark)

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  5. The table below shows the average monthly ice cream consumption for 2011 . 
Consumption (litres/person)
Year Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec
2011 0.156 0.150 0.158 0.180 0.200 0.210 0.183 0.172 0.162 0.145 0.134 0.154
  1. Show that, when rounded to two decimal places, the seasonal index for July 2011 estimated from this data is 1.10.  (2 marks)

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Show Answers Only

a.    \(\text{Maximum values are 12 months apart.}\)

b.    \(\text{Deseasonalised (Apr)}\ = \dfrac{0.180}{1.05} = 0.1714… = 0.17\ \text{(2 d.p.)}\)

c.    \(\text{2011 monthly mean}\ =\dfrac{2.004}{12}=0.167\)
 

\(\text{Seasonal index (July)}\) \(=\dfrac{0.183}{0.167}\)  
  \(=1.095…\)  
  \(=1.10\ \text{(2 d.p.)}\)  
Show Worked Solution
a.    \(\text{Maximum values are 12 months apart.}\)
♦♦♦ Mean mark (a) 26%.

 
b.
    \(\text{Deseasonalised (Apr)}\ = \dfrac{0.180}{1.05} = 0.1714… = 0.17\ \text{(2 d.p.)}\)

 
c.
    \(\text{2011 monthly mean}\)

\(=(0.156+0.15+0.158+0.18+0.2+0.21+0.183+0.172+\)

\(0.162+0.145+0.134+0.154)\ ÷\ 12 \)

\(=\dfrac{2.004}{12}\)

\(=0.167\)
 

\(\text{Seasonal index (July)}\) \(=\dfrac{0.183}{0.167}\)  
  \(=1.095…\)  
  \(=1.10\ \text{(2 d.p.)}\)  
♦♦ Mean mark (c) 36%.

Filed Under: Time Series Tagged With: Band 4, Band 5, smc-266-10-Seasonal Index from a Table, smc-266-20-(De)Seasonalising Data, smc-266-40-Time Series Trends

CORE, FUR1 2021 VCAA 12 MC

The time series plot below shows the quarterly sales, in thousands of dollars, of a small business for the years 2010 to 2020.
 

The time series plot is best described as having

  1. seasonality only.
  2. irregular fluctuations only.
  3. seasonality with irregular fluctuations. 
  4. a decreasing trend with irregular fluctuations.
  5. a decreasing trend with seasonality and irregular fluctuations.
Show Answers Only

`D`

Show Worked Solution

`text{The peaks and troughs both show a decreasing trend.}`

♦♦ Mean mark 34%.

`text{Fluctuations within any year are irregular and don’t}`

`text{display any seasonality.}`

`=> D`

Filed Under: Time Series Tagged With: Band 5, smc-266-40-Time Series Trends

CORE, FUR1 2017 VCAA 13-15 MC

The wind speed at a city location is measured throughout the day.

The time series plot below shows the daily maximum wind speed, in kilometres per hour, over a three-week period.
  

 

Part 1

The time series is best described as having

  1. seasonality only.
  2. irregular fluctuations only.
  3. seasonality with irregular fluctuations.
  4. a decreasing trend with irregular fluctuations.
  5. an increasing trend with irregular fluctuations.

 

Part 2

The seven-median smoothed maximum wind speed, in kilometres per hour, for day 4 is closest to

  1. `22`
  2. `26`
  3. `27`
  4. `30`
  5. `32`

 

Part 3

The table below shows the daily maximum wind speed, in kilometres per hour, for the days in week 2.

A four-point moving mean with centring is used to smooth the time series data above.

The smoothed maximum wind speed, in kilometres per hour, for day 11 is closest to

  1. `22`
  2. `24`
  3. `26`
  4. `28`
  5. `30`
Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ D`

`text(Part 3:)\ D`

Show Worked Solution

`text(Part 1)`

`text(The time series plot shows no obvious trend and)`

`text(is over too short a period to show seasonality.)`

`=> B`

 

`text(Part 2)`

`text(Consider the 7 values where day 4 is the middle)`

`text(data point.)`

`text(By inspection of the graph, the 4th highest point = 30.)`

`=> D`

 

`text(Part 3)`

`text(Mean for Day 9 – 12)`

`= (22 + 19 + 22 + 43)/4 = 26.5`

 

`text(Mean for Day 10 – 13)`

`= (19 + 22 + 43 + 37)/4 = 30.25`

 

`:. 4text(-point moving mean with centring)`

`= (26.5 + 30.25)/2`

`= 28.375`

`=> D`

`

Filed Under: Time Series Tagged With: Band 4, Band 5, smc-266-40-Time Series Trends, smc-266-60-MEAN Smoothing, smc-266-70-MEDIAN Smoothing

CORE, FUR1 2016 VCAA 13 MC

Consider the time series plot below.
 


 

The pattern in the time series plot shown above is best described as having

  1. irregular fluctuations only.
  2. an increasing trend with irregular fluctuations.
  3. seasonality with irregular fluctuations.
  4. seasonality with an increasing trend and irregular fluctuations.
  5. seasonality with a decreasing trend and irregular fluctuations.
Show Answers Only

`C`

Show Worked Solution

`text(The graph shows definite seasonality)`

`text(and no noticeable trend.)`

`=> C`

Filed Under: Time Series Tagged With: Band 3, smc-266-40-Time Series Trends

CORE, FUR2 2009 VCAA 2

The time series plot below shows the rainfall (in mm) for each month during 2008.
 

CORE, FUR2 2009 VCAA 2
 

  1. Which month had the highest rainfall?   (1 mark)

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  2. Use three-median smoothing to smooth the time series. Plot the smoothed time series on the plot above.
  3. Mark each smoothed data point with a cross (×).   (2 marks)

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  4. Describe the general pattern in rainfall that is revealed by the smoothed time series plot.   (1 mark)

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Show Answers Only
  1. `text(November)`
  2.  
    CORE, FUR2 2009 VCAA 2 Answer
     
  3. `text(Until April, the rainfall increases each month)`
    `text(and then it remains relatively constant until)`
    `text(November).`
  4.  
Show Worked Solution

a.   `text(November)`

 

b.    CORE, FUR2 2009 VCAA 2 Answer
MARKER’S COMMENT: Locate medians graphically by inspection. Explaining a general pattern with more than one trend proved challenging.

c.   `text(Until April, there is an increase in rainfall)`

`text(and then it remains relatively constant until)`

`text(November.)`

 

Filed Under: Time Series Tagged With: Band 3, Band 4, smc-266-40-Time Series Trends

CORE, FUR2 2010 VCAA 3

Table 2 shows the Australian gross domestic product (GDP) per person, in dollars, at five yearly intervals for the period 1980 to 2005.
 

CORE, FUR2 2010 VCAA 31 

  1. Complete the time series plot above by plotting the GDP for the years 2000 and 2005.   (1 mark)

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  2. Briefly describe the general trend in the data.   (1 mark)

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In Table 3, the variable year has been rescaled using 1980 = 0, 1985 = 5 and so on. The new variable is time.

CORE, FUR2 2010 VCAA 32

  1. Use the variables time and GDP to write down the equation of the least squares regression line that can be used to predict GDP from time. Take time as the independent variable.   (2 marks)

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  2. In the year 2007, the GDP was $34 900. Find the error in the prediction if the least squares regression line calculated in part c. is used to predict GDP in 2007.   (2 marks)

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Show Answers Only
  1.  
    CORE, FUR2 2010 VCAA 31 Answer
  2. `text(An increasing trend.)`
  3. `text(GDP = 20 000 + 524 × time)`
  4. `752\ text(below the actual GDP)`
Show Worked Solution
a.    CORE, FUR2 2010 VCAA 31 Answer

 

b.   `text(An increasing trend.)`

 

c.   `text(By calculator,)`

MARKER’S COMMENT: Students are expected to use the variables in the diagram rather than just `x` and `y`.

`text(GDP) = 20\ 000 + 524 × text(time)`
 

d.   `text(In 2007, time = 27,)`

`text{GDP (Predicted)}\ = 20\ 000 + 524 xx 27= 34\ 148`

`:.\ text(Error)\ = 34\ 900-34\ 148= 752\ \ text{(less than real GDP)}`

Filed Under: Time Series Tagged With: Band 3, Band 4, smc-266-40-Time Series Trends

CORE, FUR2 2011 VCAA 3

The following time series plot shows the average age of women at first marriage in a particular country during the period 1915 to 1970.
 

CORE, FUR2 2011 VCAA 31

  1. Use this plot to describe, in general terms, the way in which the average age of women at first marriage in this country has changed during the period 1915 to 1970.   (1 mark)

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During the period 1986 to 2006, the average age of men at first marriage in a particular country indicated an increasing linear trend, as shown in the time plot below.

CORE, FUR2 2011 VCAA 32

A three-median line could be used to model this trend.

  1. On the graph above
  2.  i. clearly mark with a cross (×) the three points that would be used to fit a three-median line to this time series plot.   (2 marks)

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  3. ii. draw in the three-median line.   (1 mark)

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Show Answers Only
  1. `text(Constant between 1915 and 1935 and then)`
    `text(decreasing between 1935 and 1970.)`
  2.  
    CORE, FUR2 2011 VCAA 3 Answer
Show Worked Solution

a.   `text(The average age of women at first marriage was fairly)`

`text(constant between 1915 and 1935, and then decreased)`

`text(between 1935 and 1970.)`

 

b.i. & ii.

CORE, FUR2 2011 VCAA 3 Answer

Filed Under: Time Series Tagged With: Band 4, Band 5, smc-266-40-Time Series Trends, smc-266-70-MEDIAN Smoothing

CORE, FUR1 2007 VCAA 11-13 MC

The following information relates to Parts 1, 2 and 3.

The time series plot below shows the revenue from sales (in dollars) each month made by a Queensland souvenir shop over a three-year period.

Part 1

This time series plot indicates that, over the three-year period, revenue from sales each month showed

A.   no overall trend.

B.   no correlation.

C.   positive skew.

D.   an increasing trend only.

E.   an increasing trend with seasonal variation.

 

Part 2

A three median trend line is fitted to this data.

Its slope (in dollars per month) is closest to

A.   `125`

B.   `146`

C.   `167`

D.   `188`

E.   `255`

 

Part 3

The revenue from sales (in dollars) each month for the first year of the three-year period is shown below.

If this information is used to determine the seasonal index for each month, the seasonal index for September will be closest to

A.   `0.80`

B.   `0.82`

C.   `1.16`

D.   `1.22`

E.   `1.26`

Show Answers Only

`text (Part 1:)\ E`

`text (Part 2:)\ C`

`text (Part 3:)\ E`

Show Worked Solution

`text (Part 1)`

`text(The time series plot clearly shows an increasing)`

`text(trend and a seasonal spike and drop over the)`

`text(Summer months.)`

`rArr E`

 

`text (Part 2)`

♦ Mean mark 37%.
MARKERS’ COMMENT: A common error was to incorrectly use the 6 month and 30 month data points as the median.

`text(From the graph, the median of the bottom third)`

`text(of data points is)\ \ (6.5, 3000).`

`text(From the graph, the median of the top third of)`

`text(data points is)\ \ (30.5, 7000).`

`:.\ text(Gradient of the three median line)`

`=(7000 – 3000)/(30.5 – 6.5)`

`=166.66…`

`rArr C`

 

`text (Part 3)`

`text(Average monthly sales)\ ` `= (43\ 872)/12`
  `=3656`

`:.\ text(Seasonal index for September)`

`=4597/3656`

`=1.257…` 

`rArr E`

Filed Under: Time Series Tagged With: Band 2, Band 4, Band 5, smc-266-10-Seasonal Index from a Table, smc-266-40-Time Series Trends

CORE, FUR1 2008 VCAA 11-13 MC

The time series plot below shows the number of users each month of an online help service over a twelve-month period.
 

2008 11-13

Part 1

The time series plot has

A.   no trend. 

B.   no variability.  

C.   seasonality only.

D.   an increasing trend with seasonality.

E.   an increasing trend only.

 

Part 2

The data values used to construct the time series plot are given below.

2008 12

A four-point moving mean with centring is used to smooth timeline series.
The smoothed value of the number of users in month number 5 is closest to

 

A.   `357`

B.   `359`

C.   `360`

D.   `365`

E.   `373`

 

Part 3

A least squares regression line is fitted to the time series plot.
The equation of this least squares regression line is

number of users = 346 + 2.77 × month number

Let month number 1 = January 2007, month number 2 = February 2007, and so on.

Using the above information, the regression line predicts that the number of users in December 2009 will be closest to

A.   `379`

B.   `412`

C.   `443`

D.   `446`

E.   `448`

Show Answers Only

`text(Part 1:)\ E`

`text(Part 2:)\ C`

`text(Part 3:)\ D`

Show Worked Solution

`text(Part 1)`

♦ Mean mark 39%.
MARKERS’ COMMENT: 50% of students incorrectly read the three large random fluctuations in monthly sales as seasonality, which can’t be determined over only 12 months.

`text(The time series is clearly trending upwards with)`

`text(higher lows and higher highs occurring.)`

`text(The large fluctuations are random and should)`

`text(not be confused with seasonality.)`

`=>E`

 

`text(Part 2)`

`text(Mean for months 3-6)`

`=(354+356+373+353)/4`

`=359`

`text(Mean for months 4-7)`

`=(356+373+353+364)/4`

`=361.5`

 

`:.\ text(Four point moving mean with centring)`

`=(359+361.5)/2`

`=360.25`

`=>  C`

 

`text(Part 3)`

`text(December 2009 will be month number 36.)`

`:.\ text(Number of users)` `= 346+2.77xx36`
  `= 445.72`

`=>  D`

Filed Under: Time Series Tagged With: Band 4, Band 5, smc-266-40-Time Series Trends, smc-266-60-MEAN Smoothing

CORE, FUR1 2013 VCAA 12-13 MC

The time series plot below displays the number of guests staying at a holiday resort during summer, autumn, winter and spring for the years 2007 to 2012 inclusive.
 

CORE, FUR1 2013 VCAA 12-13 MC_1
 

 Part 1

Which one of the following best describes the pattern in the time series?

A.  random variation only

B.  decreasing trend with seasonality

C.  seasonality only

D.  increasing trend only

E.  increasing trend with seasonality 

 

Part 2

The table below shows the data from the times series plot for the years 2007 and 2008. 
 

CORE, FUR1 2013 VCAA 12-13 MC_2
 

Using four-mean smoothing with centring, the smoothed number of guests for winter 2007 is closest to

A.  `85`

B.  `107`

C.  `183`

D.  `192`

E.  `200` 

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`text(The pattern in the time series is seasonal only,)`

`text(with peaks appearing in Summer. There is no)`

`text(apparent year-to-year trend.)`

`=> C`

 

`text(Part 2)`

`text{Mean of guests (Season 1-4)}`

`=(390+126+85+130)/4`
`=182.75`

 

`text{Mean of guests (Season 2-5)}`

`=(126+85+85+130+460)/4`
`=200.25`

 

`:.\ text(Four-mean smoothing with centring)`

`=(182.75+200.25)/2`
`=191.5`

`=> D`

Filed Under: Time Series Tagged With: Band 3, Band 4, smc-266-40-Time Series Trends, smc-266-60-MEAN Smoothing

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