To read space-technology data, calculate the difference or change for each row, then write a conclusion that stays inside the numbers. Say what the data shows, not why it happened.
This lesson is part of space technology within the SPM Science guide. It follows comparing orbit and launch concepts.
What do I compute first?
Work out the gap between each pair of readings, then a summary such as the mean. Keep the unit on every number.
A single summary can hide large gaps in individual rows, so look at both the mean and the biggest gap.
Worked example: rainfall readings
The table is original. It compares rainfall in mm at five stations, measured by a ground gauge and estimated from a satellite.
| Station | Gauge (mm) | Satellite (mm) | Satellite minus gauge |
|---|---|---|---|
| A | 42 | 40 | −2 |
| B | 55 | 61 | +6 |
| C | 30 | 28 | −2 |
| D | 70 | 64 | −6 |
| E | 18 | 20 | +2 |
Means. Gauge mean = (42 + 55 + 30 + 70 + 18) ÷ 5 = 215 ÷ 5 = 43 mm. Satellite mean = (40 + 61 + 28 + 64 + 20) ÷ 5 = 213 ÷ 5 = 42.6 mm.
Largest gap. At stations B and D the gap is 6 mm. At B, 6 ÷ 55 × 100 = 10.9% of the gauge reading.
Conclusion. The means are close (43 mm and 42.6 mm), but individual stations differ by up to 6 mm. Assuming the gauges are accurate, the satellite gives a good overall picture but is less reliable at a single station.
The mistake that costs marks
The common slip is to compare the two means only and state that the satellite is accurate. The means hide the +6 and −6 gaps because positive and negative differences cancel.
| Step | Wrong | Right |
|---|---|---|
| Compare | Means 43 and 42.6 are the same | Means are close, rows differ by up to 6 mm |
| Differences | Ignored | Computed for each station |
| Claim | The satellite is accurate | Close for the mean, less so per station |
| Assumption | (none) | Gauges are accurate |
The fix is to check the largest gap before writing the verdict.
What can the data not show?
A table of readings shows a change. It does not name the cause.
For example, a satellite records forest cover of 4 200 km² in one year and 4 032 km² in the next. The change is 4 032 − 4 200 = −168 km², and 168 ÷ 4 200 × 100 = 4%. The forest area fell by 4%, but the table does not say whether logging, fire or farming caused it.
Check yourself
A satellite measures a lake area of 250 km² in March and 225 km² in May. Find the percentage change. Can you say the lake dried up because of less rain?
Answer
Change = 225 − 250 = −25 km². Percentage change = −25 ÷ 250 × 100 = −10%, which is a 10% fall.
No. The data shows that the area fell, but it does not show the cause. Less rain could contribute, though water use or evaporation could also be involved. The claim needs separate evidence.
What to study next
Continue with evaluating applications using specified evidence, then test the cluster with the space technology practice set.
If you want a teacher to go through your data conclusions with you, see online one-to-one Science tuition.