A full bivariate investigation practice questions | General Mathematics (Year 11, VCE Units 1 & 2, Bivariate data, Fitting linear models)

Question 1 of 8

Calculator allowed

Treat weekly training distance as the explanatory variable for fitness test total. Use technology to analyse the displayed bivariate data

fitness test total (points)	weekly training distance (km)
462	66
472	68
462	70
472	72
470	74
480	76
478	78
472	80

(a)1 mark

Identify the explanatory variable and the response variable

weekly training distance is the explanatory variable and fitness test total is the response variable

fitness test total is the explanatory variable and weekly training distance is the response variable

(b)1 mark

Select the statement that correctly gives direction, form and strength

The scatterplot shows a moderate positive linear association

The scatterplot shows a moderate negative linear association

The scatterplot shows a strong positive linear association

The scatterplot shows a weak positive linear association

Determine Pearson's r for the displayed data using technology. Give the answer to 2 decimal places

\(r\) Format: 2 d.p.

(d)1 mark

Select the contextual meaning of b = 0.9, including whether the intercept is sensible. The observed weekly training distance values are 66–80 km

For each 1-unit increase in fitness test total (points), predicted weekly training distance increases by about 0.5 on the displayed km scale; the intercept at EV = 0 is outside the observed range

Across the displayed bivariate data, predicted fitness test total increases by about 0.9 on the displayed points scale in total; the intercept at EV = 0 is outside the observed range

For each 1-unit increase in weekly training distance (km), predicted fitness test total increases by about 0.9 on the displayed points scale; the intercept at EV = 0 is outside the observed range

For each 1-unit increase in weekly training distance (km), predicted fitness test total increases by about 0.9 on the displayed points scale; the intercept at EV = 0 is a reliable prediction

For each 1-unit increase in weekly training distance (km), the slope 0.9 proves the association between the two plotted variables is strong; the intercept at EV = 0 is outside the observed range

(e)1 mark

Calculate the model prediction at weekly training distance = 15 km. Give the answer to 1 decimal place

\widehat{\text{fitness test total}}=\frac{8504}{21}+\frac{19}{21}\bigl(\text{weekly training distance}\bigr)

fitness test total (points) Format: 1 d.p.

(f)1 mark

Classify the prediction and select the reason supported by the observed explanatory range. The observed weekly training distance range is 66–80 km, and the query uses 15 km

This is extrapolation: the predicted response is outside its plotted range, so the response range decides whether the fitted data support it

This is extrapolation: the least-squares line returns a value at 15 km, so the fitted data support the prediction at every explanatory value

This is extrapolation: 15 km is outside the observed 66–80 km explanatory range, so the fitted data do not support the prediction

This is interpolation: 15 km is outside the observed 66–80 km explanatory range, so the fitted data support the prediction