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Can Someone Help Me With This Chart I Dont Know How To Find The R^2 Value Of A Datas Fit To A

How do I find the corresponding value for a given data and graph in Excel?

What you are really asking for is not to read data from the chart, but to interpolate values from the data you originally put into it. In the data you provided there simply is no data pair (x,50), so your chart will never be able to return x without doing some kind of interpolation between the two surrounding points (2.5,55) and (5,32).Of course Excel shows  lines between the points in your chart, and that suggest that there is data, but there really isn't. The lines are simply a linear or some kind of polynomal interpolation between the surrounding points.So the only right way to do this is by adding a trend line.Select your data points and choose insert trend line. It is up to you to decide whether it will be a linear trend line, a logarithmic or a 3rd order polynomal. Make sure you check the "show equation" box. Then Excel will add a trendline and show its equation. Using that equation you can simply calculate X from any given Y. As far as I know the only way to do this is type the equation manually in your workbook. Keep in mind however that any trend line is always an approximation, and that you better not use it to extrapolate.Below two screenshots (in Dutch).

WHen making a scatter plot in Excel, what is the r-squared (R2) value?

I still don't really understand... I had to make a scatter plot of a chart comparing concentration (x-axis) to absorbance (y-axis) and my lab report is asking what the R-squared value tells me about the relationship between the two. Would it just mean if it's positive then when one number increases, so does the other?

How can I judge if the linear or quadratic model fits best for my data?

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Thanks, your answer is very clear, you the problem is that the Prof. has a data for at least three years and he divided the data on 8 students and gave every student the data for about 81days!!!! So the data had been divided and that what made the problem.
He already know that I will have a problem and he told me to go on with these data.

Thank you very much, you helped me a lot.
I dont know if I have further question, Should I ask you or not!
Anyway thanks for your answers
Kind regards

How does one find the intersect point between 2 trendlines in Excel?

You can use solver to find such points, but firstly you need to setup a sheet for that and calculate coordinates of the two equation within excel cell. For that you will need the two formulas: INTERCEPT() and SLOPE(), The first one will give you the constant of equation while the slope gives you the coefficient of x in the equation.You can find the following formulas in the attached sheet in the cells B17, B18 and C17 and C18.After that you need to have process of trial and error where you need to find a point where both equations have same value of x. For that you need to setup a Solver Problem. What i have done is to that for an intersecting point, the value of x should be same for both of them!See the below setup of Solver, you can run for the values in cell D20 to get the desired result.Once you are done with the x coordinate you can find y-coordinate by substituting it in either of the equations. See this worksheet for your convinience:Trendline.xlsx

How to calculate standard deviation on TI-36X Pro?

I've checked a Youtube video on it but it wasn't very reliable.

For example, once I press data put my data in the first column of the table, I'm supposed to press 2nd and then "stat-reg/distr". Then I press "statvars" and press enter.

But some message says "Calculate 1-Var, 2-Var or a Regression" and I don't know what that means.

Can anyone help?

How can I get de X-Y data from a trendline graph in Excell using VBA?

As others have mentioned, I don’t think that VBA is necessary in this case.For a simple linear trendline, you can determine the value of any value Y based on a value X using the SLOPE and INTERCEPT formulas. The formula would take the form ofy = mx + bwhere “y” is the dependent variable, “m” is the slope of the line, “x” is the independent variable and “b” is the Y-intercept. Assuming X values in cells A2:A10, Y values in cells B2:B10 and your desired “X” variable in cell C1, you would enter this as :=SLOPE(B2:B10, A2:A10)*C1+INTERCEPT(B2:B10, A2:A10)However, your question indicates that you are interested in a Moving Average trendline. In this case, you can calculate any given data point by taking the average of the previous “n” observations, where “n” is the number of periods in your moving average.For example, to calculate a 2-Period Moving Average (“n” = 2), simply average the last two values. A simple example is copied below.If you are set on using VBA, then this code should do the trick with minor modifications as you see fit. As is, it should work for simply selecting a range of X/Y pairs and running the macro. The result will be a very basic XY Scatter chart with the chart title indicating the next XY pair in the sequence.Hope this helps out.BenSub AddXYScatter()

Dim r As Range
Dim c As Chart
Dim x As Long
Dim s As String

Set r = Selection

If r.Columns.Count <> 2 Then
MsgBox "Please select your X/Y pairs and try again"
Exit Sub
End If

x = Application.InputBox("Number of Periods?", "Periods?", 2, , , , , 1)

If r.Rows.Count <= x Then
MsgBox "Not enough observations, please select more rows or reduce periods."
Exit Sub
End If

ActiveSheet.Shapes.AddChart(xlXYScatter).Select
Set c = ActiveChart

With c
.SetSourceData Source:=r
.SeriesCollection(1).Trendlines.Add
With .SeriesCollection(1).Trendlines(1)
.Type = xlMovingAvg
.Period = x
End With
.SetElement (msoElementChartTitleAboveChart)
s = "Next X | Y Pair: "
s = s & Format(WorksheetFunction.Average(r.Offset(r.Rows.Count - x, 0).Resize(x, 1)), "0.00")
s = s & " | " & Format(WorksheetFunction.Average(r.Offset(r.Rows.Count - x, 1).Resize(x, 1)), "0.00")

.ChartTitle.Text = s
End With

Set r = Nothing
Set c = Nothing

End Sub

How can I add an exponential trendline on Microsoft Excel?

You can add an exponential trendline as follows:Select your data and graph it. You can either use a SCATTER plot Do not use the LINE plot because if you have two columns of data it will graph the individually instead of an input-output graph.Select the curve on the graph itself and then right-click. Click on Add Trendline. A dialog box will pop up giving you the option for plotting a curve based on different criteria. Choose EXPONENTIAL.You have an option here as well. You can also have Excel display the trendline equation on the chart as well which is also very useful.If you are statistically savvy, what you can also do is have Excel display the R-squared value on the chart. This value rates the trendline on how close or far away it is from the actual data values between 0 and 1.Click OK. There you have it.

How do I determine the correlation between 2 variables?

Hi,

I don't know anything about Excel, so I can't help you with it. If you have a graphing calculator, you can analyze data on it. On my TI-83, I would enter my data under STAT, Edit, in L1 for one set of data and in L2 for the second set of data.

Then I would press STAT, CALC, and choose the type of equation I think would best fit the points. You can turn on the "r" value that lets you know how well the equation it finds as a line of best fit actually represents the data. The closer r's value is to either ±1, the better the equation. The closer "r" is to zero, the weaker the equation is as a good predictor. If you don't like one result, go back to STAT, CALC, and try a different type of equation. You have the choice of linear, quadratic, cubic, quartic, ln, log, exp, power, or sinreg.

I hope that helps!! Sorry I don't know Excel... :-)

What does R^2 tell in a linear regression analysis?

R-square tells the percentage of variance in dependent variable that can be explained by the independent variable.Example:y=B0+B1*xIf R-square=.73, then we can say that 73% of variation in dependent variable(y-variable) can be explained by the dependent variable(x-variable).

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