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S6 ICT: Artificial Intelligence and Machine Learning – Practice Paper 4 (Error Review, 10 Questions)
S6 ICT: Artificial Intelligence and Machine Learning – Practice Paper 4 (Error Review, 10 Questions)
Paper four targets the errors: unpicking the denominators of accuracy, precision and recall on a fresh 45/5/15/185 matrix to reach 92.0%, 90.0%, 75.0%, F1 0.818 and 97.37% specificity, with batching, five-fold validation and learning rate halving revisited.
Quick answer Most of these errors are denominator errors: accuracy uses all 250, precision uses the 50 positive predictions and recall uses the 60 actual positives, and they never swap.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | 25 minutes |
| Difficulty mix | Foundation 2 · Intermediate 2 · Challenge 2 · Error redo 4 |
| Chapter | Chapter: S6 ICT | Artificial Intelligence and Machine Learning |
| Level | Error Review |
02 Questions (10 items)
- FoundationA classifier gives a confusion matrix: 45 true positives, 5 false positives, 15 false negatives and 185 true negatives. How many samples are in the test set?
- FoundationA student reports precision as 45 divided by (45+15), which is 75.0%. What is wrong?
- IntermediateWhat is the accuracy of this classifier?
- IntermediateWhat is the precision of this classifier?
- ChallengeWhat is the recall of this classifier?
- ChallengeUsing a precision of 90.0% and a recall of 75.0%, find the F1 score to three decimal places.
- Error redo250 images are split 80/20, and the training batch size is 16. How many batches are there in one epoch?
- Error redoThe same 250 images are used for five-fold cross validation. How large are the test and training sets per fold, and how many runs are there?
- Error redoSpecificity is the share of actual negatives that are correctly labelled. What is the specificity of this classifier?
- Error redoThe learning rate is 0.008 and is halved every 10 epochs. What is it after 40 epochs?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 250 samples. | 45 + 5 + 15 + 185 = 250, so the four cells add up to the whole test set. |
| 2 | The denominator should be the 50 positive predictions; 75.0% is the recall. | Precision is measured over all positive predictions, while recall is measured over the actual positives. |
| 3 | 92.0%. | (45+185) divided by 250 is 230 over 250, which is 0.92. |
| 4 | 90.0%. | 45 divided by (45+5) is 45 over 50, which is 0.90. |
| 5 | 75.0%. | 45 divided by (45+15) is 45 over 60, which is 0.75. |
| 6 | 0.818. | Two times 0.90 times 0.75 is 1.35, and dividing by 1.65 gives about 0.818. |
| 7 | 13 batches. | The training set holds 200 images, and 200 divided by 16 gives 12 full batches with 8 left, which still counts, so 13 in total. |
| 8 | 50 for testing and 200 for training, five runs. | 250 divided by 5 is 50, and 250 minus 50 is 200, with the five folds taking turns as the test set. |
| 9 | 97.37%. | 185 divided by (5+185) is 185 over 190, which is about 0.9737. |
| 10 | 0.0005. | Halving 0.008 four times gives 0.004, 0.002, 0.001 and finally 0.0005. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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