Bivariate Data
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
The following data shows hours studied (\(x\)) and test score (\(y\)) for 6 students:
\[egin{array}{c|cccccc}x&1&2&3&4&5&6\\hline y&45&52&60&65&72&80\end{array}\]
(a)Plot a scatter graph and describe the correlation.
(b)Calculate \(ar{x}\) and \(ar{y}\).
(c)The regression line is \(y=6.74x+38.2\). Predict the score for 4.5 hours studied.
(a)
Answer
Strong positive correlation — as hours studied increase, test score increases.
(b)
Means
\(ar{x}= frac{21}{6}=3.5;\;ar{y}= frac{374}{6}pprox62.3\)
Answer
\(ar{x}=3.5,\;ar{y}pprox62.3\)
(c)
Substitute \(x=4.5\)
\(y=6.74(4.5)+38.2=30.33+38.2\)
Question 2
The PMCC for a data set is \(r=-0.92\).
(a)Describe the strength and direction of the correlation.
(b)Would it be appropriate to use the regression line to make predictions? Explain.
(a)
Answer
Strong negative correlation.
(b)
Answer
Yes — the strong correlation (\(|r|\) close to 1) suggests a linear model is a good fit, so interpolation within the data range is appropriate. Extrapolation beyond the data range should be used with caution.
Question 3
Data for temperature (\(x\), °C) and ice cream sales (\(y\), £100s):
\[egin{array}{c|ccccc}x&18&22&25&28&31\\hline y&3.2&4.8&6.1&7.5&9.0\end{array}\]
(a)Find the equation of the regression line \(y\) on \(x\).
(b)Estimate the sales when the temperature is 24°C.
(c)Comment on the reliability of using this model for a temperature of 40°C.
(a)
Summary stats: \(ar{x}=24.8,\;ar{y}=6.12\)
\(S_{xx}=\sum x^2-nar{x}^2=(18^2+\ldots+31^2)-5(24.8)^2=3218-3075.2=142.8\)
\(S_{xy}=\sum xy-nar{x}ar{y}\)
\(=(18 imes3.2+\ldots+31 imes9)-5(24.8)(6.12)=793-759.28=33.72\)
Gradient \(b=S_{xy}/S_{xx}\)
\(bpprox0.236\); \(a=6.12-0.236 imes24.8pprox0.26\)
Answer
\(ypprox0.236x+0.26\)
(b)
Substitute \(x=24\)
\(ypprox0.236(24)+0.26pprox5.93\)
(c)
Answer
40°C is well outside the data range (18–31°C) — this is extrapolation and the prediction may be unreliable.
Question 4
Distinguish between the regression line of \(y\) on \(x\) and the regression line of \(x\) on \(y\). When should each be used?
Difference
\(y\) on \(x\): minimises vertical distances — use to predict \(y\) from \(x\). \(x\) on \(y\): minimises horizontal distances — use to predict \(x\) from \(y\).
Answer
Use \(y\) on \(x\) when \(x\) is the independent variable and you want to predict \(y\). Use \(x\) on \(y\) when predicting \(x\) from a known \(y\). The two lines coincide only when \(|r|=1\).
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