For a set of data: x = (0,1,2,3,4,5,6) and y=(36, 28, 25, 24, 23, 21, 19), is it wise to use a linear regression to extrapolate
data for x = 50? Solution: Since the coefficient of determination is 0.8582, the linear model is a reasonably good fit for the data, so extrapolation for any x-value is acceptable. What is wrong with this solution?
The problem with this solution is that a regression model is not recommended to extrapolate because we do not know if the linear relation that we calculated for a specific range of x values still holds outside this range.
Step-by-step explanation:
We have a linear regression model, with a range of the independent variable "x" that goes from 0 to 6.
The regression model finds a good fit (r=0.8582).
As it has a good fit, it is proposed to use this model to extrapolate and calculate the value of y for x=50.
It is not recommended to extrapolate a regression model unless we are really sure that the model is still valid within the range within we are extrapolating.
This means that if we have no proof that y has a linear relation in a range of x that includes x, the extrapolation has no validity and can lead to serious errors.
A linear regression model is only suitable for interpolation or extrapolating within the range we are sure that the relation between y and x is linear within a certain acceptable error.
We have been given that an amount of $53,000 is placed in an investment account that grows at a fixed rate of 2% (compound growth) per year. We are asked to find the amount in the account after 4 years.
To solve our given problem we will use compound interest formula.\
, where,
A = Final amount after t years,
P = Principal amount,
r = Annual interest rate in decimal form,
n = Number of times interest is compounded per year,
t = Time in years.
Let us convert our given rate in decimal form.
Upon substituting our given values in compound interest formula we will get,
Therefore, an amount of $57,369 will be in the account after 4 years.