A manufacturer stores drums of chemicals - Edexcel - A-Level Maths Statistics - Question 3 - 2006 - Paper 1

The points in the scatter plot lie close to a straight line, which suggests a linear relationship between time in storage and evaporation loss. This warrants the use of a linear regression model.

Using the provided sums:

$S_y = 8354, ext{ } S_x = 1352, ext{ } S_{xx} = 704 \text{ (where } n = 10)$

The formula for b is given by:

$b = \frac{n \Sigma xy - \Sigma x \Sigma y}{n \Sigma x^2 - (\Sigma x)^2}$

Substituting in the values:

$b = \frac{10(8354) - (1352)(53112)}{10(704) - (1352)^2} = 3.90$

Next, we calculate a using:

$a = \frac{\Sigma y - b \Sigma x}{n}$

Substituting values:

$a = \frac{8354 - 3.90(1352)}{10} = -29.02...$

Thus, a = -29.02 and b = 3.90.

The value of b, approximately 3.90, indicates that for every additional week in storage, the evaporation loss increases by about 3.90 ml.

Using the regression equation:

$y = a + bx$ Substituting in the values from above:

$y = -29.02 + 3.90(19) \approx 48.88$

Thus, the predicted evaporation loss after 19 weeks is approximately 48.88 ml.

Again using the regression equation:

$y = -29.02 + 3.90(35) \approx 98.68$

Thus, the predicted evaporation loss after 35 weeks is approximately 98.68 ml.

The prediction at 19 weeks is close to the range of x values (3 to 18 weeks) used to create the model, making it reasonably reliable. However, the prediction at 35 weeks exceeds the maximum x value of the dataset, making it less reliable since it is well outside the observed range, and we cannot be sure that the linear relationship holds beyond the tested range.