Business Analytics Quiz: Moving Average And Smoothing

10 questions · exam conditions

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Moving Average And SmoothingQuestion 1 of 10

A service center uses simple exponential smoothing. Its forecast for one week was $80$ calls, actual volume was $100$ calls, and the updated forecast for the next week became $85$ calls. The next week's actual volume was then $105$ calls.

Which combination correctly identifies the smoothing constant and the forecast made after observing $105$ calls?

The smoothing constant is

0.20

, and the new forecast is

89

calls.

The smoothing constant is

0.25

, and the new forecast is

90

calls.

The smoothing constant is

0.75

, and the new forecast is

100

calls.

The smoothing constant is

0.80

, and the new forecast is

101

calls.

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Business Analytics Quiz

Business Analytics Quiz: Moving Average And Smoothing

Practice Moving Average And Smoothing in Business Analytics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Moving Average And Smoothing, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Analytics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which combination correctly identifies the smoothing constant and the forecast made after observing $105$ calls?

The smoothing constant is $0.20$ , and the new forecast is $89$ calls.
The smoothing constant is $0.25$ , and the new forecast is $90$ calls. (correct answer)
The smoothing constant is $0.75$ , and the new forecast is $100$ calls.
The smoothing constant is $0.80$ , and the new forecast is $101$ calls.

Explanation: Whenever you see a simple exponential smoothing question, your first move should be to recover the smoothing constant

\alpha

from the information given, then apply it forward. The exponential smoothing formula is:

F_{t+1} = F_t + \alpha(A_t - F_t)

, where

F_t

is the prior forecast,

A_t

is actual demand, and

\alpha

is the smoothing constant. Here, the forecast was

80

, actual was

100

, and the updated forecast became

85

. Plugging in:

85 = 80 + \alpha(100 - 80)

, so

5 = 20\alpha

, giving

\alpha = 0.25

. Now apply this to the next period:

F = 85 + 0.25(105 - 85) = 85 + 0.25(20) = 85 + 5 = 90

. That confirms answer B — the smoothing constant is

0.25

and the new forecast is

90

calls. Answer A uses

\alpha = 0.20

, which doesn't satisfy the first equation:

80 + 0.20(20) = 84 \neq 85

. It's a plausible-looking round number, but it fails the back-check. Answer C claims

\alpha = 0.75

, which would produce

80 + 0.75(20) = 95

, far from

85

— this reflects a misreading of how heavily the actual demand is weighted. Answer D uses

\alpha = 0.80

, giving

80 + 0.80(20) = 96

, also inconsistent with the given updated forecast of

85

. Your key study tip: always solve for $\alpha$ first using the given forecast update before doing any forward calculation. Questions like this often provide enough data to pin down

\alpha

exactly — use it as your anchor, then the second calculation becomes straightforward.

Question 2

Four product teams have forecasts of $100$ units immediately before observing a one-time actual demand of $140$ units. Before the increase, all observations in each team's relevant moving-average window were also $100$ units. Team 1 uses a three-period moving average, Team 2 uses a six-period moving average, Team 3 uses exponential smoothing with $\alpha=0.20$ , and Team 4 uses exponential smoothing with $\alpha=0.50$ .

Which team's next forecast will show the largest immediate upward revision?

Team 1, because its three-period moving average increases by about $13.33$ units.
Team 2, because its six-period moving average increases by about $6.67$ units.
Team 3, because exponential smoothing increases its forecast by exactly $8$ units.
Team 4, because exponential smoothing increases its forecast by exactly $20$ units. (correct answer)

Explanation: When comparing forecasting methods, the key question is: how much weight does each method place on the most recent observation? That single weight determines how aggressively a forecast reacts to new data. For a moving average, each period in the window receives equal weight of

\frac{1}{n}

. Team 1's three-period average gives the new observation weight

\frac{1}{3}

, so its revision is

\frac{1}{3} \times (140 - 100) = 13.33

units. Team 2's six-period average gives weight

\frac{1}{6}

, producing a revision of only

6.67

units. For exponential smoothing, the revision equals

\alpha \times (\text{actual} - \text{forecast})

. Team 3 revises by

0.20 \times 40 = 8

units, and Team 4 revises by

0.50 \times 40 = \mathbf{20}

units. Team 4 shows the largest jump, confirming D is correct. Choice A is tempting because

13.33

is larger than both exponential smoothing values and Team 2's revision — but it's not the largest overall. Choice B is the smallest revision of all four, not the largest. Choice C correctly calculates Team 3's revision but misidentifies it as the winner;

8 < 20

. The trap in A and C is failing to compare all four methods before selecting an answer. Study tip: Always map each forecasting method to its "effective weight on the newest data point" —

\frac{1}{n}

for moving averages and

\alpha

for exponential smoothing. Whichever method has the highest effective weight will react most strongly to any surprise in demand.

Question 3

Monthly subscriptions have increased steadily, with the four most recent actual values equal to $100$ , $110$ , $120$ , and $130$ . Both a simple moving average and simple exponential smoothing have repeatedly produced forecasts below actual subscriptions. A manager proposes increasing the moving-average window and reducing the exponential-smoothing constant.

Which assessment of the proposal is most appropriate?

Both changes will likely increase lag; shorter windows and larger smoothing constants react faster, although neither method explicitly models trend. (correct answer)
Both changes will likely reduce lag; longer windows and smaller smoothing constants place greater emphasis on the latest observations.
Only the larger window will reduce lag; the smaller smoothing constant will make exponential smoothing respond more slowly.
Only the smaller smoothing constant will reduce lag; the larger window will make the moving average respond more slowly.

Explanation: Whenever you see forecasting questions involving lag, focus on how quickly each method responds to recent changes. A forecast that consistently falls below rising actual values is exhibiting lag — the method is too slow to follow an upward trend. The fix is to make the model more responsive, not less. For a simple moving average, a shorter window increases responsiveness because fewer past observations dilute the influence of recent data. Extending the window does the opposite — it averages in more older, lower values, pulling the forecast further behind reality. For simple exponential smoothing, a larger smoothing constant (

\alpha

closer to 1) places heavier weight on the most recent observation, reducing lag. A smaller

\alpha

weights history more heavily, slowing the model's reaction. The manager proposes the exact opposite of both corrective actions — a longer window and a smaller

\alpha

— which will each worsen the lag problem. Answer A correctly identifies this: both changes increase lag, and crucially, neither method inherently models trend regardless of parameter choice. Answer B is wrong because it reverses the logic entirely — longer windows and smaller smoothing constants do not emphasize recent observations more. Answer C is wrong because it correctly diagnoses the smoothing constant but incorrectly claims a larger window reduces lag, when it actually increases it. Answer D is wrong for the mirror reason — it correctly identifies that a larger window slows the moving average but incorrectly claims a smaller smoothing constant reduces lag. A reliable rule of thumb: more history = more lag. Longer windows and smaller

\alpha

values both incorporate more historical data, always moving in the direction of slower response.

Question 4

A store normally sells $100$ units per week. Immediately before a one-week promotion, both a three-week moving-average model and an exponential-smoothing model with $\alpha=0.25$ forecast $100$ units. Promotion-week sales were $180$ units, and sales in the following week returned to $100$ units. Before the promotion, all observations in the moving-average window were $100$ units.

After the post-promotion actual value of $100$ is incorporated, what will the two models forecast for the next week?

The moving-average forecast will be $100$ , and the exponential-smoothing forecast will be $115$ .
The moving-average forecast will be about $126.67$ , and the exponential-smoothing forecast will be $120$ .
The moving-average forecast will be about $126.67$ , and the exponential-smoothing forecast will be $115$ . (correct answer)
The moving-average forecast will be $120$ , and the exponential-smoothing forecast will be $126.67$ .

Explanation: When comparing forecasting models, always track how each one handles a temporary spike — they "forget" the anomaly at different speeds, which is exactly what this question tests. For the three-week moving average, after the post-promotion week you incorporate the actual value of 100. The three most recent observations are now: the promotion week (180), the post-promotion week (100), and one pre-promotion week (100). The forecast is

\frac{180 + 100 + 100}{3} = \frac{380}{3} \approx 126.67

. The spike is still sitting in the window, so the model remains distorted. For exponential smoothing (

\alpha = 0.25

), you apply the update formula twice. After the promotion week:

F_{\text{promo+1}} = 0.25(180) + 0.75(100) = 45 + 75 = 120

. Then, after incorporating the post-promotion actual of 100:

F_{\text{next}} = 0.25(100) + 0.75(120) = 25 + 90 = 115

. The model has partially recovered but still carries residual memory of the spike. This confirms C as correct. Choice A is wrong because it gives the moving average as 100, as if the spike never happened — the 180 is still inside the three-week window. Choice B gets the moving average right but stops exponential smoothing one period too early, reporting 120 instead of applying the second update to get 115. Choice D swaps the two model results entirely, which would only make sense if you confused which formula belongs to which model. The key study tip: moving averages retain a distortion for exactly as many periods as the window length, while exponential smoothing dilutes it gradually. Know both update formulas cold, and always count how many times you need to apply each one.

Question 5

To smooth monthly revenue, an analyst computes a centered three-month average for June using May, June, and July revenue. The analyst then labels this value as the forecast for June that would have been available at the end of May.

Which evaluation of the analyst's procedure is most accurate?

It is invalid because realized June and July revenue would not have been available at the end of May. (correct answer)
It is valid because equal weighting makes a centered average equivalent to a real-time forecast.
It is invalid only because the most recent month should receive a larger weight than earlier months.
It is valid whenever the revenue series has stable seasonality and relatively little random variation.

Explanation: When evaluating any forecasting or smoothing procedure, the critical question is: what information was actually available at the time the forecast was supposedly made? This is the concept of data availability, and violating it produces what analysts call "look-ahead bias." Here, the analyst uses May, June, and July revenue to compute a centered three-month average, then claims this value represents a forecast available at the end of May. The fatal flaw is that June and July haven't happened yet at the end of May — those values exist only in hindsight. The procedure smuggles future data into a "past" forecast, making any backtest or historical evaluation artificially optimistic. Answer A correctly identifies this: the centered average is mathematically fine for smoothing historical data, but it cannot be relabeled as a real-time forecast without a fundamental logical error. Answer B is wrong because equal weighting has nothing to do with the problem. Whether weights are equal or unequal doesn't make unavailable data suddenly available. Answer C misidentifies the issue — the weighting scheme is irrelevant here. The flaw isn't how months are weighted but which months are included. A forecast using only heavier weights on May still wouldn't fix the July data problem. Answer D is wrong because stable seasonality and low variance don't retroactively make future data observable; those are conditions about the series' behavior, not about when data becomes known. Study tip: On any forecasting question, always trace the timeline — ask "what did the analyst know, and when?" If a model uses data from period t+1 to forecast period t, it's look-ahead bias regardless of how elegant the math looks.

Question 6

After incorporating June's actual sales, a simple exponential-smoothing model produces a July forecast of $210$ units. The company needs forecasts for July, August, and September immediately, before any actual sales for those months become available. The model contains no trend or seasonal component.

What forecasts should the model produce for the three months?

It should forecast July at $210$ but leave August and September undefined until actual values arrive.
It should forecast $210$ for July and then reduce later forecasts by the smoothing constant.
It should forecast $210$ for July and then increase later forecasts using June's forecast error.
It should forecast $210$ for July, $210$ for August, and $210$ for September. (correct answer)

Explanation: When you see a question about simple exponential smoothing (SES), the key concept to anchor on is that SES produces a flat forecast: without a trend or seasonal component, every future period gets the same predicted value — the most recently smoothed level. Here's the core logic: the SES update formula produces a single number, the smoothed level, which becomes your best estimate of the future. Once you've incorporated June's actual sales, the model yields a smoothed level of

210

. Because the model has no mechanism to project growth, decline, or seasonal patterns, that

210

is simply carried forward indefinitely. July forecast =

210

, August forecast =

210

, September forecast =

210

. The model doesn't "wait" for new data — it gives you its best guess right now, and that guess is the same flat value for all future horizons. D is correct. Answer A is wrong because SES doesn't leave future forecasts undefined — it always produces a value, even without new actuals. Leaving forecasts blank would make the model operationally useless for multi-period planning. Answer B reflects a misunderstanding: the smoothing constant

\alpha

determines how quickly past observations are discounted in the update step, not how future forecasts decay period by period. Answer C confuses SES with trend-adjustment methods like Holt's exponential smoothing, where a trend component is explicitly updated using forecast errors — SES has no such mechanism. A useful study tip: whenever a question specifies "no trend or seasonal component," that's your signal that all multi-step-ahead forecasts collapse to a single flat value. Recognizing model assumptions is half the battle on forecasting questions.

Question 7

A distributor applies simple exponential smoothing according to $F_{t+1}=\alpha A_t+(1-\alpha)F_t$ , where $A_t$ is actual demand in period $t$ . The smoothing constant is $\alpha=0.30$ . The forecast for April was $120$ units, actual April demand was $150$ units, and actual May demand was $135$ units.

After incorporating both actual observations, what is the forecast for June?

The June forecast is $129.0$ units.
The June forecast is $130.8$ units. (correct answer)
The June forecast is $135.0$ units.
The June forecast is $139.5$ units.

Explanation: Simple exponential smoothing questions test your ability to apply a recursive formula — meaning each new forecast feeds into the next. The key insight is that you must update the forecast twice here: once using April's actual demand to get the May forecast, then again using May's actual demand to get the June forecast. Starting with what you're given: the April forecast is

F_{April} = 120

and actual April demand is

A_{April} = 150

. Plug into the formula to find the May forecast:

F_{May} = 0.30(150) + 0.70(120) = 45 + 84 = 129.0

Now use the May forecast alongside actual May demand (

A_{May} = 135

) to forecast June:

F_{June} = 0.30(135) + 0.70(129.0) = 40.5 + 90.3 = 130.8

That confirms B is correct. Choice A (129.0) is the May forecast, not June — a classic "stopped one step too early" trap. Choice C (135.0) simply uses actual May demand as the June forecast, ignoring the smoothing mechanism entirely. Choice D (139.5) likely results from mistakenly applying the weights in reverse — using

(1-\alpha)

on the actual demand and

\alpha

on the prior forecast, which swaps the formula's logic. A useful habit: whenever a problem says "after incorporating both observations," mentally count how many times you need to apply the formula. Write out each step sequentially so you don't confuse an intermediate result for the final answer.

Question 8

A wholesaler forecasts next month's orders using a weighted three-month moving average. From most recent to oldest, actual monthly orders were $200$ , $160$ , and $120$ units. The corresponding weights are $0.50$ , $0.30$ , and $0.20$ .

How does the weighted forecast compare with an unweighted three-month moving-average forecast?

The weighted forecast is $148$ units, which is $12$ units below the unweighted forecast.
The weighted forecast is $160$ units, which is equal to the unweighted forecast.
The weighted forecast is $172$ units, which is $12$ units above the unweighted forecast. (correct answer)
The weighted forecast is $176$ units, which is $16$ units above the unweighted forecast.

Explanation: Forecasting questions like this one require you to compute two different averages and compare them — don't stop after finding just one. The weighted moving average multiplies each period's value by its assigned weight, then sums the results. Using the given data (most recent to oldest: 200, 160, 120) with weights 0.50, 0.30, and 0.20:

\text{Weighted Forecast} = (200 \times 0.50) + (160 \times 0.30) + (120 \times 0.20) = 100 + 48 + 24 = 172

The unweighted (simple) moving average treats all three months equally, dividing their sum by 3:

\text{Unweighted Forecast} = \frac{200 + 160 + 120}{3} = \frac{480}{3} = 160

The difference is

172 - 160 = 12

units, confirming that C is correct — the weighted forecast exceeds the unweighted forecast by 12 units. A gets the direction backwards: 148 would result from weighting the oldest month most heavily (0.50 on 120), which inverts the weight assignment. B claims both forecasts are equal at 160, which is only true for the simple average — assigning heavier weight to the most recent (and highest) value must pull the forecast upward. D uses plausible-looking numbers but doesn't correspond to any standard calculation with the given weights; it likely results from a weight-application or arithmetic error. A useful study tip: when weights are ordered from most-recent to oldest, heavier recent weights always pull the forecast toward the most recent value. If the trend is rising, the weighted forecast will exceed the simple average — use that as a quick sanity check.

Question 9

A retailer uses a trailing three-week moving average to forecast weekly demand. Demand during the four most recent weeks, from oldest to newest, was $82$ , $94$ , $88$ , and $100$ units. Actual demand in the following week was $106$ units. Forecast error is defined as actual demand minus forecast demand.

What were the forecast and forecast error for the week in which actual demand was $106$ units?

The forecast was $94$ units, and the error was $12$ units. (correct answer)
The forecast was $91$ units, and the error was $15$ units.
The forecast was $94$ units, and the error was $-12$ units.
The forecast was $88$ units, and the error was $18$ units.

Explanation: When you see a moving average question, your first job is to identify which periods feed into the forecast. A trailing three-week moving average uses the three most recent weeks of actual demand to predict the next week. Here, the four weeks of demand (oldest to newest) are 82, 94, 88, and 100. To forecast the week when actual demand was 106, you use the three weeks immediately before that week — which are 94, 88, and 100 (not 82, since that data is now too old). The forecast is:

\frac{94 + 88 + 100}{3} = \frac{282}{3} = 94 \text{ units}

Forecast error is defined as actual minus forecast:

106 - 94 = +12 \text{ units}

That confirms A is correct. B is wrong because it averages all four weeks — 82, 94, 88, and 100 — giving

\frac{364}{4} = 91

. That's a four-week average, not three, and it misapplies the model. C uses the right forecast of 94 but flips the sign of the error, reporting

-12

instead of

+12

. This would mean the forecast exceeded actual demand, which isn't the case here — the retailer under-forecast. D averages the wrong three weeks — 82, 94, and 88 — ignoring the most recent period (100). Always count from the most recent data backward. A useful habit: underline or list the data points in order and cross off any that fall outside the moving-average window before you calculate.

Question 10

Two analysts use the same simple exponential-smoothing model with $\alpha=0.40$ and subsequently observe exactly the same actual values. Their initial forecasts differ by $20$ units. Each analyst updates the forecast once after each new actual value.

After three forecast updates, by how much will their forecasts differ?

Their forecasts will differ by $3.20$ units.
Their forecasts will differ by $4.32$ units. (correct answer)
Their forecasts will differ by $7.20$ units.
Their forecasts will differ by $12.00$ units.

Explanation: When you see a question about simple exponential smoothing (SES), remember the core update formula:

F_{t+1} = \alpha \cdot A_t + (1-\alpha) \cdot F_t

. The key insight is that any difference in initial forecasts shrinks geometrically with each update, because each new forecast retains only

(1-\alpha)

of the previous forecast's influence. Here,

\alpha = 0.40

, so

(1-\alpha) = 0.60

. Both analysts observe identical actual values, meaning the only source of divergence is their differing starting forecasts. After each update, the gap between their forecasts is multiplied by

0.60

. Starting with a difference of

20

units, after three updates the gap becomes:

20 \times (0.60)^3 = 20 \times 0.216 = 4.32 \text{ units}

That confirms B is correct. Now for the distractors: A (3.20 units) results from mistakenly multiplying by

(1-\alpha)^3

using

\alpha

itself — specifically,

20 \times (0.40)^3 = 20 \times 0.032 \times 5

... actually it reflects using only

\alpha

once, like

20 \times 0.40 \times (0.40) = 3.20

, confusing which weight drives persistence. C (7.20 units) comes from applying the decay factor only twice instead of three times:

20 \times (0.60)^2 = 7.20

. This is an off-by-one error — a very common trap when counting update periods. D (12.00 units) applies the factor only once:

20 \times 0.60 = 12.00

, as if only a single update occurred. Your study tip: always track how many updates have occurred — each update multiplies the initial difference by

(1-\alpha)

, so after

n

updates, the gap is

\Delta_0 \times (1-\alpha)^n

. Off-by-one errors on

n

are the most common mistake here.