Moving averages in ML Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Understanding Simple Moving Average (SMA)

Which of the following best describes the Simple Moving Average (SMA) in time series data?

A It calculates the average of all past data points up to the current point.

B It calculates the average of a fixed number of the most recent data points.

C It weights recent data points more heavily than older ones.

D It predicts future values by fitting a linear trend to the data.

💡 Hint

Think about how SMA uses a window of data points.

The Simple Moving Average calculates the average of a fixed number of recent data points, smoothing short-term fluctuations.

❓ Predict Output

intermediate

Output of Exponential Moving Average Calculation

What is the output of the following Python code that calculates an exponential moving average (EMA) with alpha=0.5 on the data [10, 20, 30]?

ML Python

data = [10, 20, 30]
alpha = 0.5
ema = [data[0]]
for x in data[1:]:
    ema.append(alpha * x + (1 - alpha) * ema[-1])
print([round(v, 2) for v in ema])

A [10, 15.0, 22.5]

B [10, 20.0, 30.0]

C [10, 15.0, 25.0]

D [10, 17.5, 23.75]

💡 Hint

EMA updates by weighting the new value and previous EMA.

EMA starts at 10, then 0.5*20 + 0.5*10 = 15, then 0.5*30 + 0.5*15 = 22.5.

❓ Model Choice

advanced

Choosing Moving Average Type for Noisy Data

You have a noisy sensor data stream and want to smooth it while giving more importance to recent readings. Which moving average method is best?

A Exponential Moving Average (EMA)

B Cumulative Moving Average (CMA)

C Simple Moving Average (SMA)

D Weighted Moving Average with equal weights

💡 Hint

Consider which method weights recent data more.

EMA weights recent data more heavily, making it better for noisy data where recent changes matter.

❓ Hyperparameter

advanced

Effect of Alpha in Exponential Moving Average

In an Exponential Moving Average (EMA), what is the effect of increasing the smoothing factor alpha from 0.1 to 0.9?

A The EMA becomes smoother and reacts slower to recent changes.

B The EMA becomes equivalent to a Simple Moving Average.

C The EMA ignores recent data and focuses on older data.

D The EMA becomes less smooth and reacts faster to recent changes.

💡 Hint

Alpha controls the weight of the newest data point.

Higher alpha means more weight on recent data, so EMA reacts faster but is less smooth.

❓ Metrics

expert

Comparing Moving Average Smoothing with MSE

You apply two smoothing methods on a noisy time series: SMA with window size 3 and EMA with alpha=0.3. Which metric best compares their smoothing quality against the original clean signal?

A Sum of absolute differences between noisy and smoothed signals

B Accuracy score of smoothed signal classification

C Mean Squared Error (MSE) between smoothed and clean signals

D Number of data points in the smoothed signal

💡 Hint

Think about measuring how close the smoothed signal is to the true clean signal.

MSE measures average squared difference between smoothed and clean signals, quantifying smoothing quality.

Practice

1. What is the main purpose of using a moving average in data analysis?

easy

A. To smooth out short-term fluctuations and highlight longer-term trends

B. To increase the number of data points in a dataset

C. To remove all noise from the data completely

D. To predict exact future values without error

Solution

1. Step 1: Understand the role of moving averages

   Moving averages smooth data by averaging nearby points, reducing short-term ups and downs.

2. Step 2: Identify the main goal

   The goal is to reveal longer-term trends by reducing noise, not to remove noise completely or predict exact values.

3. Final Answer:

   To smooth out short-term fluctuations and highlight longer-term trends -> Option A

4. Quick Check:

   Moving average = smoothing trends [OK]

Hint: Moving averages smooth data to show trends clearly [OK]

Common Mistakes:

• Thinking moving averages increase data points
• Believing moving averages remove all noise
• Assuming moving averages predict exact future values

2. Which of the following Python code snippets correctly computes a simple moving average with window size 3 for a list data?

easy

A. [data[i] / 3 for i in range(len(data))]

B. [sum(data[i:i+3]) for i in range(len(data)-3)]

C. [sum(data[i:i+3]) / 3 for i in range(len(data)-3)]

D. [(data[i] + data[i+1] + data[i+2]) / 3 for i in range(len(data)-2)]

Solution

1. Step 1: Understand moving average calculation

   A simple moving average with window 3 averages each group of 3 consecutive elements.

2. Step 2: Check each option's correctness

   [(data[i] + data[i+1] + data[i+2]) / 3 for i in range(len(data)-2)] correctly sums three consecutive elements and divides by 3, iterating till len(data)-2.
   [sum(data[i:i+3]) for i in range(len(data)-3)] sums but does not divide by 3.
   [sum(data[i:i+3]) / 3 for i in range(len(data)-3)] divides but uses range(len(data)-3), which is too short.
   [data[i] / 3 for i in range(len(data))] divides single elements by 3, not averaging groups.

3. Final Answer:

   [(data[i] + data[i+1] + data[i+2]) / 3 for i in range(len(data)-2)] -> Option D

4. Quick Check:

   Sum 3 elements / 3, range correct = [(data[i] + data[i+1] + data[i+2]) / 3 for i in range(len(data)-2)] [OK]

Hint: Sum 3 elements and divide by 3, loop till len-2 [OK]

Common Mistakes:

• Forgetting to divide by window size
• Using wrong range length causing index errors
• Averaging single elements instead of groups

3. Given the code below, what is the output?

data = [2, 4, 6, 8, 10]
window = 2
moving_avg = [sum(data[i:i+window]) / window for i in range(len(data) - window + 1)]
print(moving_avg)

medium

A. [2.0, 4.0, 6.0, 8.0, 10.0]

B. [3.0, 5.0, 7.0]

C. [3.0, 5.0, 7.0, 9.0]

D. [6.0, 8.0, 10.0]

Solution

1. Step 1: Calculate moving averages manually

   Window size is 2, so average pairs:
   (2+4)/2=3.0
   (4+6)/2=5.0
   (6+8)/2=7.0
   (8+10)/2=9.0

2. Step 2: Confirm output list length and values

   Length is len(data)-window+1 = 5-2+1=4, matching 4 values above.

3. Final Answer:

   [3.0, 5.0, 7.0, 9.0] -> Option C

4. Quick Check:

   Pairs averaged = [3.0, 5.0, 7.0, 9.0] [OK]

Hint: Average pairs sliding by one, length = len - window + 1 [OK]

Common Mistakes:

• Confusing window size with output length
• Calculating sums but forgetting to divide
• Off-by-one errors in range length

4. The following code is intended to compute a moving average with window size 3, but it misses the last window. What is the problem?

data = [1, 2, 3, 4, 5]
window = 3
moving_avg = [sum(data[i:i+window]) / window for i in range(len(data)-window)]
print(moving_avg)

medium

A. The range should be len(data) - window + 1 to include the last window

B. The window size is too large for the data list

C. sum() cannot be used on list slices

D. Division by window size should be outside the list comprehension

Solution

1. Step 1: Analyze the range length

   Range is len(data)-window = 5-3=2, but to cover all windows it should be len(data)-window+1 = 3.

2. Step 2: Understand impact of incorrect range

   Using len(data)-window misses the last valid window slice, causing incomplete results.

3. Final Answer:

   The range should be len(data) - window + 1 to include the last window -> Option A

4. Quick Check:

   Range length = len - window + 1 [OK]

Hint: Use range(len(data) - window + 1) for full coverage [OK]

Common Mistakes:

• Using len(data) - window instead of +1
• Thinking sum() can't handle slices
• Misplacing division outside comprehension

5. You have daily sales data for 10 days: [10, 12, 11, 14, 13, 15, 16, 14, 13, 12]. You want to smooth this data using a moving average with window size 4 but only want to keep averages where the window's average is greater than 13. Which Python code correctly computes this filtered moving average?

hard

A. [sum(data[i:i+4])/4 for i in range(len(data)-4) if sum(data[i:i+4])/4 > 13]

B. [avg for i in range(len(data)-3) if (avg := sum(data[i:i+4])/4) > 13]

C. [sum(data[i:i+4])/4 for i in range(len(data)-3) if sum(data[i:i+4]) > 13]

D. [sum(data[i:i+4])/4 for i in range(len(data)-3) if sum(data[i:i+4])/4 < 13]

Solution

1. Step 1: Understand window size and range

   Window size 4 means averaging groups of 4 elements, so range is len(data)-3 = 10-3=7.

2. Step 2: Filter averages greater than 13

   [avg for i in range(len(data)-3) if (avg := sum(data[i:i+4])/4) > 13] uses assignment expression to compute average once and filter if > 13.
   [sum(data[i:i+4])/4 for i in range(len(data)-4) if sum(data[i:i+4])/4 > 13] uses wrong range (len(data)-4=6), missing last window.
   [sum(data[i:i+4])/4 for i in range(len(data)-3) if sum(data[i:i+4]) > 13] filters sum > 13, not average > 13.
   [sum(data[i:i+4])/4 for i in range(len(data)-3) if sum(data[i:i+4])/4 < 13] filters averages less than 13, opposite condition.

3. Final Answer:

   [avg for i in range(len(data)-3) if (avg := sum(data[i:i+4])/4) > 13] -> Option B

4. Quick Check:

   Use assignment expression to filter averages > 13 [OK]

Hint: Use assignment expression (walrus) to filter averages [OK]

Common Mistakes:

• Using wrong range length missing last windows
• Filtering sum instead of average
• Using wrong comparison operator