Correlation (1) Flashcards
What is interpolation?
Use values of x within the data range given in the example. The predicted value should be reliable
Use values of x within the data range given in the example. The predicted value should be reliable
What is interpolation?
What is extrapolation?
Use values of x outside the data range given in the example. The predicted value can be unreliable
Use values of x outside the data range given in the example. The predicted value can be unreliable
What is extrapolation?
y = ax^n in logarithms
log y = n log x + log a
log y = n log x + log a
y = ax^n in logarithms
y = ab^x in logarithms
log y = x log b + log a
log y = x log b + log a
y = ab^x in logarithms
The sales per week, s (in thousands), of a new book, t weeks after its launch can be modelled by the equation s = at^b. Using values of s for t = 1,2,5,10 a scatter graph was drawn with the line of best fit: log s = 1.878 log t - 0.7122
Find the values of the constants a and b to 3 d.p.
Rearrange s = at^b
log s = b log t + log a
So log a = -0.7122, a = 10^-0.7122 = 0.194
b = 1.878
The sales per week, s (in thousands), of a new book, t weeks after its launch can be modelled by the equation s = at^b. Using values of s for t = 1,2,5,10 a scatter graph was drawn with the line of best fit: log s = 1.878 log t - 0.7122
Estimate the weekly sales of the book 8 weeks after its launch
s = at^b = 0.194b^(1.878)
So t = 8
s = 0.194 * 8^(1.878) = 9.63
So the model predicts 9630 copies 8 weeks after its launch
The sales per week, s (in thousands), of a new book, t weeks after its launch can be modelled by the equation s = at^b. Using values of s for t = 1,2,5,10 a scatter graph was drawn with the line of best fit: log s = 1.878 log t - 0.7122
Comment on the suitability of this model for large values of t
The model predicts that sales will grow exponentially, which isn’t realistic for a large value of t. For example, in the 300th week 15000000 copies of the book is sold, which is unrealistic
