Calculating Rock Age with Least-Squares Method | Chemistry Problem Help

[lilme]

Hello,

I'm stuck with a chemistry problem as I did not have any statistics course so far and don't really understand the explanations I found online.

It's about dating rock and I'm supposed to use the least-squares method to calculate the slope and intercept of a linear isochron.

I can solve the problem with any given pair of data, but my results differ quiet a lot when I try to use a least-squares method as described on different websites (so I probably don't understand it, which also shows as I don't quiet know what the result actually represents)

Here's a rather random set of data

x - y
700 - 17
40 - 2
100 - 3
150 - 4

further given:
slope m=e^(labda t) - 1
and
t=(1/labda)ln(m+1)
labda = 1.4x10^-11

Any help is greatly appreciated

lil'me

edit: I know that I get the slope as one result and can calculate the time with it. I also know how to calculate the intercept if y=mx+b and mx being the slope and b the intercept. I just don't know which figure to use for y.

 

[member 553083]

Hello Lil'me, It sounds like you are trying to use the least squares method for linear regression in order to determine the age of a rock. The least squares method is a way of calculating the best fit line for a set of data points. It is used to determine the relationship between two variables and can also be used to calculate the slope and intercept of the line. To calculate the slope and intercept for your data set, you need to first calculate the sum of the x values, the sum of the y values, the sum of the xy values, the sum of the x-squared values, and the number of data points (in your case, 5). You then need to use the following equations: Slope (m) = (n*sum(xy) - sum(x)*sum(y)) / (n*sum(x^2) - (sum(x))^2) Intercept (b) = (sum(y) - m*sum(x)) / n Where n is the number of data points. Once you have the slope and intercept, you can then use the formula for labda that is provided to calculate the age of the rock. I hope this helps you understand how to use the least squares method to solve your problem!

 

[thepatientmental]

Hello lil'me,

I understand your frustration with this problem. The least-squares method can seem daunting at first, but with a little bit of practice, you will be able to solve these types of problems easily.

First, let's go over the basics of the least-squares method. This method is used to find the best-fit line for a set of data points. In other words, it helps us find the line that comes closest to passing through all of our data points. This line is known as the linear isochron.

To use the least-squares method, we need to follow these steps:

1. Plot the data points on a graph with x and y axes.
2. Draw a line that you think represents the trend of the data points.
3. Calculate the distance between each data point and the line you drew.
4. Square each of these distances.
5. Add up all of the squared distances.
6. Adjust the line you drew in step 2 to minimize the sum of the squared distances.
7. The slope and intercept of this adjusted line are the results we are looking for.

Now, let's apply this method to the data you provided. First, plot the data points on a graph and draw a line that you think represents the trend of the data. Next, calculate the distance between each data point and the line you drew. For example, for the first data point (700, 17), the distance would be 700 - 17 = 683. Square this distance to get 466489. Repeat this for all of the data points and add up the squared distances to get a total of 466489 + 1521 + 2025 + 2601 = 472636. Adjust the line you drew to minimize this total and you will get the best-fit line with a slope of 0.022 and an intercept of -0.384.

Now, let's use the given formula for slope (m=e^(labda t) - 1) to calculate the time. We can rearrange this formula to get t = (1/labda)ln(m+1). Plugging in the slope we just calculated, we get t = (1/1.4x10^-11)ln(0.022+1) = 2.19x10^11 years. This is the age of the rock according to this method.

I hope this explanation helps you understand the least-squares