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Scientific Enquiry - Graphing and Analysis

Grade 6IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Independent Variable: The factor that is changed or manipulated in an experiment, plotted on the xx-axis.

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Dependent Variable: The factor that is measured or observed, plotted on the yy-axis.

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Controlled Variables: Factors kept constant to ensure a 'fair test' so that only the independent variable affects the results.

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Line of Best Fit: A smooth line or straight line drawn through the center of data points to show the general trend, often ignoring anomaliesanomalies.

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Anomalies: Data points that do not fit the overall pattern or trend, often caused by measurement errors.

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Correlation: A relationship between variables. A positive correlation means as xx increases, yy increases (y∝xy \propto x).

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Scale: Choosing consistent intervals for axes so the data fills at least 50%50\% of the graph area.

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Units: Every axis label must include the unit of measurement in brackets, e.g., Time (s)Time \ (s) or Temperature (∘C)Temperature \ (^{\circ}C).

📐Formulae

Mean=∑data valuesn\text{Mean} = \frac{\sum \text{data values}}{n}

Range=Maximum value−Minimum value\text{Range} = \text{Maximum value} - \text{Minimum value}

Gradient (Slope)=ΔyΔx=y2−y1x2−x1\text{Gradient (Slope)} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}

% Change=New Value−Old ValueOld Value×100\% \text{ Change} = \frac{\text{New Value} - \text{Old Value}}{\text{Old Value}} \times 100

💡Examples

Problem 1:

A student measures the length of a shadow every hour. At 1:00 PM1:00 \ PM it is 15 cm15 \ cm, at 2:00 PM2:00 \ PM it is 30 cm30 \ cm, and at 3:00 PM3:00 \ PM it is 45 cm45 \ cm. Identify the variables and the trend.

Solution:

Independent Variable: Time (hh); Dependent Variable: Shadow Length (cmcm). Trend: Positive linear correlation.

Explanation:

The student decides when to take the measurement (Time is independent). The length changes because of the time (Length is dependent). Since the length increases by 15 cm15 \ cm every hour, it shows a constant positive trend: Length=15×TimeLength = 15 \times Time.

Problem 2:

Calculate the mean temperature from the following dataset: 22∘C22^{\circ}C, 25∘C25^{\circ}C, 21∘C21^{\circ}C, and an anomalous reading of 40∘C40^{\circ}C.

Solution:

Mean =22+25+213=22.67∘C= \frac{22 + 25 + 21}{3} = 22.67^{\circ}C.

Explanation:

When calculating a mean in scientific enquiry, anomaliesanomalies (like the 40∘C40^{\circ}C reading which is much higher than the rest) should be excluded to ensure the average represents the true trend.

Graphing and Analysis - Revision Notes & Key Formulas | IGCSE Grade 6 Science