6.05 Making predictions

Making predictions

The line of best fit is a linear representation of the general trend of our data.

Once we have determined the equation of the line of best fit, we can use it as a model to predict the likely value of the response variable based on a given value of the explanatory variable.

The process of predicting has two parts:

• Substitute the $x$x value into the rule for the line of best fit to get the predicted $y$y value
• Then we need to consider whether our prediction is reliable or not

 

We can make predictions 'by hand' either using a graph of the line of best fit or by substituting into the equation of the line.

 

Predictions from a graph

The example below shows a scatter plot for the relationship of daily ice-cream sales versus the maximum daily temperature, along with the line of best fit.

Although there is a clear trend of increasing sales as the temperature increases, it would be difficult to predict the sales for a given day from the raw data.  However, we can estimate this by using the line of best fit.

If we want to predict the sales on a day when the temperature reaches $30$30 degrees, we could do so with these steps that follow the red arrows on the graph:

1. From $30$30 degrees on the horizontal axis, draw a vertical line to intersect with the line of best fit.
2. From the point of intersection, draw a horizontal line to the vertical axis
3. Read the predicted sales, approximately $245$245 ice-creams, form the vertical axis.

 

When we are making predictions from a graph we should take care to work accurately, but still expect a small amount of variation due to the limited precision of working with graphs.

 

Predictions from an equation

For the example above, the equation of the line of best fit is: 

$S$S

$=$=

$10\times T-45$10×T−45

Where $S$S is the number of ice cream sales and $T$T is the maximum daily temperature.

To predict the sales on a day where the maximum temperature is $30$30 degrees, we simply substitute the temperature into the equation:

$S$S	$=$=	$10\times30-45$10×30−45
	$=$=	$245$245

So, we can predict that $245$245 ice creams will be sold on that day.

 

Practice question

Question 1

Question 2

 

Interpolation and extrapolation

An important consideration when we are making predictions is recognising if the prediction is within the range of data values for which we have actual measurements.  If it is, we refer to the prediction as interpolation.  If not, we refer to the predication as extrapolation.

The bivariate data set on the right has generated a line of best fit, and the range of the $x$x-values has been highlighted. Making predictions within this range is interpolation, and making predictions outside this range is extrapolation.

 

 

The scatter plots below are annotated to show examples of interpolation (left) and extrapolation (right) from the line of best fit.

 

Reliability of predictions

To judge the reliability of the prediction we need to consider three things:

• How strong is the correlation?
• Is the data interpolated or extrapolated?
• How many points are contained in the data set?

We have already seen how to calculate the correlation coefficient and interpret the strength of a relationship, using this chart.

 

If the correlation is weak, then the data values are more widely scattered, so there will be greater uncertainty in the prediction than for data with a strong correlation.

An interpolated prediction is made within the range of $x$x-values upon which the model was based. Since the model is designed to align closely to the data in this range, interpolated predictions will be more reliable than extrapolated predictions.   

When we are extrapolating, we are making the assumption that the linear model continues. This assumption is generally valid when making a prediction relatively close to the given data range. A sudden change in trend is unlikely for most of the phenomena we will model, so we will consider such predictions made to be reliable when using a strong model.

When we extrapolate well beyond the data range, even if the linear trend continues, we will consider such predictions unreliable. As we are making a prediction for values that we have not made any similar measurements, it is possible that the underlying linear relationship is slightly different from the one generated by the given data. This will lead to predictions fairing worse as we predict further beyond the given data range. There are also many cases where it is clear the linear trend cannot continue well beyond the data range. In such cases, the predictions will be unreliable and may also be unreasonable - that is give values that are not possible. For example, a model may show a negative linear trend between a chicken's age and the number of eggs it lays per week, however, the linear model could not continue indefinitely as it would suggest the chicken would lay a negative number of eggs at some age.

 

 

Worked example

A farmer sprays his wheat fields with a fertiliser. The data below gives the yield of wheat per hectare for various spray concentrations:

Spray concentration ($x$x ml/l)	$1$1	$2$2	$4$4	$7$7	$8$8	$10$10
Yield of wheat, ($y$y tonnes per hectare)	$1.19$1.19	$1.50$1.50	$1.62$1.62	$1.65$1.65	$2.05$2.05	$2.07$2.07

The line of best fit for this data is $\hat{y}=0.088x+1.212$^y=0.088x+1.212 and $r=0.929$r=0.929 (given to $3$3 decimal places),

(a) Use the line to make predictions for the yield when using the following concentrations of fertiliser: $5$5 ml/l, $10.8$10.8 ml/l and $35$35 ml/l. (Give answers to $2$2 decimal places)

Think: Substitute the concentration for the value of $x$x in the equation of the line and evaluate.

Do:

Spray concentration ($x$x ml/l)	Predicted yield of wheat ($\hat{y}$^y tonnes/hectare)
$5$5	$1.65$1.65
$10.8$10.8	$2.16$2.16
$35$35	$4.29$4.29

(b) Comment on the reliability of each prediction made in part (a).

Think: Consider the strength of the model and whether the prediction is interpolation, extrapolation just beyond the data range or extrapolation well beyond the data range.

Do:

The correlation coefficient tells us we have a strong model.

The prediction for $5$5 ml/l is interpolation on a strong model and hence, should be reliable.

The prediction for $10.8$10.8 ml/l is extrapolation just beyond the data range on a strong model and hence, should be reasonably reliable. 

The prediction for $35$35 ml/l is extrapolation well beyond the data range and we know the linear trend cannot continue indefinitely since a crop of wheat must have a maximum possible yield and at some point the fertiliser will no longer be effective and may even be toxic. Hence, despite the strong model this prediction is not reliable and may not even be reasonable - may be beyond the maximum yield possible.

 

Practice questions

Question 3

Question 4

Question 5

Question 6