How can i interpolate a data (y) on a graph to obtain the value of (x)?

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i want to do the same thing like in the image, but i can't find where is it.

Posted on Apr 11, 2013 4:20 AM

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10 replies

Apr 11, 2013 12:42 PM in response to IsaZhL

Well, this is my original graph User uploaded file

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it is suppose to be a linear regression, and those are the values of the graph. The "y" column are values of absorbance and "x" are concentration. So, i have another table of absorbances values to which i have to know what are their concentration:

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these values have to be interpolate on the graph to obtein x. This is what i don't know how to do on Numbers and thanks for all your help!

Apr 11, 2013 5:31 AM in response to IsaZhL

Here is a shot at it


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Scatter chart of columns B through E. Columns B and C are your data. Cell E2 is where you put the Y value of interest.


Formula in D2 =IF(AND(E$2>=C2,E$2<=C3),B2+(E$2-C2)*(B3-B2)/(C3-C2),"")

Fill down to complete the column


Formula in E3 =IF(D3="","",E$2)

Fill down to complete the column

You could use the more simple formula =E$2 . If you do, all rows will have your Y data point but only the rows with X data will plot on the chart.

Apr 11, 2013 5:32 AM in response to IsaZhL

Given two points you can calculate the slope (m), then you can use the first point to create a linear equation for the line between the two points. Then you can plug any X that is between the Xs of the two points.


Given:

(X1, Y1)

(X2, Y2)


m = (Y2 - Y1) / (X2 - X1)


equation for new line is:


y = mx + b


substitute a coordinate pair into above equation then solve for b:

Y1 = mX1 + b


no rearrange for b:

b = Y1 - mX1


m and b should be actual numbers


now use the formula:

y = mx + b where m and b are actual number and compute a new Y for a given X:


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Formulas above are:

D2=(B3-B2)/(A3-A2)

E2=B2-D2*A2

C4=D2*A4+E2


The chart is set up to graph X, Y and Y Interp by sharing x-values

Apr 11, 2013 5:16 AM in response to IsaZhL

Hi IsaZhL,


Your graph shows a straight line between the two black points. If the graph (response of Y to X) really is a straight line between those points, then the slope of that line and its intercept will enable you to interpolate from a Y value to an X value.


I assume that X (Time?) is the uncontrolled (and uncontrollable!) variable. Y is a measured value (such as height?)


Please post a screen shot of a small part of your data table. We need to see what you are trying to do!


Regards,

Ian.

Apr 11, 2013 3:02 PM in response to IsaZhL

The chart in your second post is completely different from the one you gave in your first post. Your first post showed a chart where straight lines were drawn from point to point and where there might be more than one X for a given Y. Your second chart is a linear-fit trendline of a set of data. I wasted my time answering the question you asked.

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How can i interpolate a data (y) on a graph to obtain the value of (x)?

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