Fit a time trend and forecast

A fitted trend gives a deseasonalised forecast for the target period. What is the next step when an actual seasonal forecast is required?

Reuse the Year 10 time-series experiment. Change the day 3 observation while its neighbours remain fixed. Decide whether the evidence shows one isolated departure or a lasting new level; the graph does not establish a cause

At a day 3 reading of 60 kL, what should happen before changing the record?

Fitting a trend and forecasting

A time trend uses consecutive integer indices for the observations. With seasonal data, divide by the relevant index before fitting and multiply by the target index after forecasting

1. 1Index time consistently Set the first observed period to t=1 and increase the index by one for every subsequent period \(\displaystyle t=1,2,3,\ldots,N\)
2. 2Remove seasonality before fitting Divide each actual observation by the seasonal index for its period \(\displaystyle D=\frac{A}{I}=\frac{660}{1.1}=600\)
3. 3Fit and interpret the line Use the model y-hat equals a plus bt, where b is the average response change per time period \(\displaystyle \hat{y}=380+12t,\quad b=12\text{ units per period}\)
4. 4Forecast and restore the target season Substitute the target index, multiply by its seasonal index when needed and consider whether the target is an extrapolation \(\displaystyle \hat{A}_6=(360+20\times6)\times1.1=528\)

Choose and verify a smoothing method

A constructed nine-day water record in kL is 18,20,19,21,60,22,23,24,26 for days 1 through 9. Construct the complete time plot. Describe the overall movement and isolated feature without claiming its cause. Choose and justify a method for showing the local typical level when one unusually high observation may dominate a mean

1. 1Build complete comparisons Calculate and plot every three-term moving mean and every three-term moving median at their correct central days. Also construct the five-term moving-median series. Explain the changed endpoint loss and why a wider window is not automatically better
2. 2Check a counterexample As a sensitivity experiment only, replace day 5 by 22 kL and compare the three-term mean and median at day 5. Keep the original record separate. State what evidence you would need before actually correcting an observation
3. 3Save your independent work Save your tables, labelled plots, tool choice, calculations and conclusions

Model response: Choose and verify a smoothing method

Compare your saved work with the calculations and reasoning. Other suitable tools and plotting windows are acceptable when they preserve time, units and all observations

1. 1Plot and describe Use day on the horizontal axis, water in kL vertically and proportional daily spacing. A suitable window is days 0 to 10 and 0 to 65 kL. Join consecutive observations in time order. Apart from the isolated 60 kL day 5 reading, the level generally rises from 18 towards 26 kL. Check the original record and measurement circumstances before correcting the day 5 reading
2. 2Complete moving series At central days 2..8, the three-term means are 19,20,100/3,103/3,35,23,73/3 kL and the three-term medians are 19,20,21,22,23,23,24 kL. At central days 3..7, the five-term medians are 20,21,22,23,24 kL. Keep the central time even though each median’s response values are sorted. Seven three-term points and five five-term points are supported; unsupported ends remain blank
3. 3Choice and sensitivity For the stated local-typical-level purpose, a three-term median limits the influence of the isolated high observation while retaining more end positions than the five-term median. A mean is still useful when every measurement’s magnitude should contribute. In the hypothetical replacement, the day 5 mean changes from 103/3 to 65/3, a decrease of 38/3 kL; the day 5 median remains 22 kL. This calculation does not authorise changing the original observation

Which description distinguishes an isolated reading from a structural change?

Given information

\(\displaystyle 21,60,22\)

Calculate the three-term mean centred at day 5, to two decimal places

Given information

\(\displaystyle 21,60,22\)

Find the three-term median centred at day 5.

Given information

\(\displaystyle N=9,\quad k=5\)

How many five-term moving medians are supported by nine observations?

After sorting the response values in a window, what happens to the time coordinate?

Construct seasonal and smoothed records

Constructed quarterly water use in kL is Year 1:(88,144,156,112), Year 2:(120,192,204,144), Year 3:(152,240,252,176), in Q1,Q2,Q3,Q4 order. Assign t=1 to Year 1 Q1. Construct the full raw time plot, describe the trend and calendar-linked repetition, and justify a four-quarter smoothing window

1. 1Centre and check Calculate all nine preliminary four-term means at half-integer times 2.5 through 10.5, then all eight centred values at t=3 through 10. Plot the centred series and explain the lost endpoints
2. 2Calculate seasonal indices Use this stated convention: divide each quarter by its own year’s mean and average the three ratios for each quarter. Show all 12 ratios, the four indices and their sum/mean check. Explain why pooling the raw years first is a different calculation when annual levels vary
3. 3Save your independent work Save your tables, labelled plots, tool choice and rationale, formulas or commands, checks and written conclusions before opening the model response. Complete construction, explanations and actual external-tool operation are not automatically marked; review them yourself or with a teacher

Model response: Construct seasonal and smoothed records

Compare your saved work with the calculations and reasoning. Other suitable tools and plotting windows are acceptable when they preserve time, units and all observations

1. 1Plot and centred means Time indices 1..12 correspond to consecutive quarters. A four-quarter window spans one annual seasonal cycle. Annual levels rise and Q2/Q3 are relatively high in each of the three years; this supports a seasonal description but not a causal explanation. The nine preliminary means are 125,133,145,157,165,173,185,197,205 kL at times 2.5..10.5. Adjacent-pair averaging gives 129,139,151,161,169,179,191,201 kL at t=3..10. No centred values are supported at 1,2,11,12.
2. 2Cycle-normalised ratios The own-year means are 125,165,205 kL. The ratio rows are (88,144,156,112)/125, (120,192,204,144)/165 and (152,240,252,176)/205. Average down the same-quarter columns. Exact indices Q1..Q4 are 122488/169125,196544/169125,209356/169125,148112/169125, approximately 0.724245,1.162123,1.237877,0.875755
3. 3Interpret and distinguish The exact indices sum to 4 and average 1. Under this convention Q1 is about 27.6% below the average-season level, and Q3 is about 23.8% above it. The pooled Q1 calculation 120/165=8/11 is not the own-year-normalised average 122488/169125. With only three years, changing seasonal behaviour or a structural change would weaken use of the same indices in later years

Which description is supported by the full quarterly plot?

Given information

\(\displaystyle A_{2.5}=125,\quad A_{3.5}=133\)

Find the centred four-quarter mean at t=3.

Why are there no centred points at t=1,2,11,12?

Given information

\(\displaystyle I_1=\frac13\left(\frac{88}{125}+\frac{120}{165}+\frac{152}{205}\right)\)

Report the Q1 seasonal index to three decimal places

Given information

\(\displaystyle \text{Four quarters; each complete-cycle ratio row has mean 1}\)

What is the exact sum of four seasonal indices calculated by this convention?

Complete a seasonal forecast and report

Using quarterly water use (kL) of Year 1: (88,144,156,112), Year 2: (120,192,204,144), and Year 3: (152,240,252,176), in Q1–Q4 order, and exact Q1–Q4 indices of 122488/169125, 196544/169125, 209356/169125, and 148112/169125, construct every deseasonalised value D=A/I. Choose a suitable regression tool and record why it suits a multi-stage reproducible table. Fit D=a+bt using t=1..12, without rounding intermediate values. Produce the adjusted plot, fitted line and residual checks

1. 1Forecast and verify Map Year 4 Q1 and Year 6 Q1 to consecutive time indices. Forecast the adjusted and actual water use for Year 4 Q1, using the target Q1 index rather than the last observed Q4 index. State the units and meaning of the slope, and the precision of each reported result. Compare the caution needed for the two horizons
2. 2Separate validation and conclusion After fixing the model, an additional constructed Year 4 Q1 observation is 163 kL. Evaluate its original-scale residual without silently refitting. Report remaining patterns, the small number of cycles, assumptions behind continuing a linear trend and seasonal pattern, and evidence that would make you revise the model
3. 3Save your independent work Save your tables, labelled plots, tool choice and rationale, formulas or commands, checks and written conclusions before opening the model response. Complete construction, explanations and actual external-tool operation are not automatically marked; review them yourself or with a teacher

Model response: Complete a seasonal forecast and report

Compare your saved work with the calculations and reasoning. Other suitable tools and plotting windows are acceptable when they preserve time, units and all observations

1. 1Reproducible table and fit A spreadsheet can store columns t,actual,year mean,season ratio,assigned seasonal index,D=actual/index,fitted D and residual D−fitted D. Record the actual formulas or regression commands used; a calculator is also suitable if all rows and settings are retained. The 12 adjusted values to six decimals are 121.505780,123.911185,126.022182,127.889705,165.689700,165.214914,164.798238,164.429621,209.873620,206.518642,203.574295,200.969537. Fit using full values derived from exact indices, not this rounded display
2. 2Coefficients and checks The fitted line is D-hat approximately 106.845227728+8.951983163t. Independently verify b=Σ(t−t̄)(D−D̄)/Σ(t−t̄)² and a=D̄−bt̄. The full-precision line passes through the mean point; residuals and time-weighted residuals sum to zero. The slope is about 8.952 deseasonalised kL per quarter, not a guaranteed change in raw water use or a total change over 12 quarters. The intercept refers to t=0, outside the observed interval
3. 3Target season and independent row Year 4 Q1 is t=13, while Year 6 Q1 is t=21. At t=13, the full fitted adjusted forecast is approximately 223.221009 kL. Multiply by the exact Q1 index 122488/169125 to obtain 161.666785 kL, reported as 161.7 kL. The additional observation has original-scale residual 163−161.666785 approximately 1.3 kL. Do not subtract the adjusted forecast from a raw observation or refit before this check
4. 4Qualified report The model gives a one-quarter-ahead estimate, not a guaranteed measured value. The adjusted plot still groups by year and is not exactly linear; this warns against claiming that seasonal adjustment has removed every pattern. Year 6 Q1 is nine quarters beyond the final observed quarter and needs substantially more caution. Three constructed cycles and one separate check do not establish future reliability or causation. Check later observations, changes in use or measurement, and whether the seasonal pattern and trend remain plausible before wider use

What does the adjusted plot tell you about the fitted line?

Given information

\(\displaystyle A_1=88,\quad I_1=122488/169125\)

Deseasonalise the first observation to two decimal places

Given information

\(\displaystyle \text{Four quarters per year; Year 1 Q1 is }t=1\)

Find the time index of Year 4 Q1 when Year 1 Q1 is t=1.

Given information

\(\displaystyle \hat D_{13}\approx223.221008851675,\quad I_1=122488/169125\)

Report the actual Year 4 Q1 forecast to one decimal place, using the full stored regression from the 12-row task

Given information

\(\displaystyle A_{13}=163,\quad\hat A_{13}\approx161.666784521649\)

Report the original-scale residual for the separate 163 kL observation to one decimal place

The adjusted forecast for period 8 is 540 and its seasonal index is 1.2. Which line gives the actual forecast?