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Greetings All,
I have the following array of data, column 1 is the date and column 2 is the amount in US$. I am trying to use regression to forecast three years or so hence. 3-Jan-06 750,000 3-Apr-06 1,000,000 27-Apr-06 1,009,778 14-Jul-06 1,139,739 26-Oct-06 1,109,484 5-Dec-06 1,169,886 9-Jan-07 1,244,587 10-Mar-07 1,421,393 28-Mar-07 1,419,597 14-Aug-07 1,642,892 If I use a linear regression trendline on this data, the R^2 is around 90% which is not bad. If I use the "Forecast Forward Periods" function, this gives me an idea of when the regression line passes 2,000,000 or 2,500,000. However, my question is this: what is the correct x-value input to use with the regression equation y = 1332.4x - 5E+07 since using the 1900-calendar number clearly does not work. Many thanks, Peter SS |
#2
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With your data in cols A & B, define x to be A1:A10
define y to be Y1:Y10 In A11, enter: =DATE(YEAR(A10)+3,MONTH(A10),DAY(A10)) the three years hence date In any two un-used cells, say C1 and C2: =SLOPE(y,x) in C1 and =INTERCEPT(y,x) in C2. They will display: 1332.387797 -50781315.47 Finally lets put the projection equation in B11: =C2+C1*A11 simple linear fit We end up seeing: 03-Jan-06 750,000 1332.387797 03-Apr-06 1,000,000 -50781315.47 27-Apr-06 1,009,778 14-Jul-06 1,139,739 26-Oct-06 1,109,484 05-Dec-06 1,169,886 09-Jan-07 1,244,587 10-Mar-07 1,421,393 28-Mar-07 1,419,597 14-Aug-07 1,642,892 14-Aug-10 3,052,481 For other models, see: http://j-walk.com/ss///excel/tips/tip101.htm -- Gary''s Student - gsnu200745 "petess" wrote: Greetings All, I have the following array of data, column 1 is the date and column 2 is the amount in US$. I am trying to use regression to forecast three years or so hence. 3-Jan-06 750,000 3-Apr-06 1,000,000 27-Apr-06 1,009,778 14-Jul-06 1,139,739 26-Oct-06 1,109,484 5-Dec-06 1,169,886 9-Jan-07 1,244,587 10-Mar-07 1,421,393 28-Mar-07 1,419,597 14-Aug-07 1,642,892 If I use a linear regression trendline on this data, the R^2 is around 90% which is not bad. If I use the "Forecast Forward Periods" function, this gives me an idea of when the regression line passes 2,000,000 or 2,500,000. However, my question is this: what is the correct x-value input to use with the regression equation y = 1332.4x - 5E+07 since using the 1900-calendar number clearly does not work. Many thanks, Peter SS |
#3
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Dear Gary''s Student,
Many thanks for your kind assistance. As a matter of interest, what "x-value" is being used? Is it the 1900-convention date or something else? My understanding of regression statistics is not complete. All the best. Peter SS "Gary''s Student" wrote: With your data in cols A & B, define x to be A1:A10 define y to be Y1:Y10 In A11, enter: =DATE(YEAR(A10)+3,MONTH(A10),DAY(A10)) the three years hence date In any two un-used cells, say C1 and C2: =SLOPE(y,x) in C1 and =INTERCEPT(y,x) in C2. They will display: 1332.387797 -50781315.47 Finally lets put the projection equation in B11: =C2+C1*A11 simple linear fit We end up seeing: 03-Jan-06 750,000 1332.387797 03-Apr-06 1,000,000 -50781315.47 27-Apr-06 1,009,778 14-Jul-06 1,139,739 26-Oct-06 1,109,484 05-Dec-06 1,169,886 09-Jan-07 1,244,587 10-Mar-07 1,421,393 28-Mar-07 1,419,597 14-Aug-07 1,642,892 14-Aug-10 3,052,481 For other models, see: http://j-walk.com/ss///excel/tips/tip101.htm -- Gary''s Student - gsnu200745 "petess" wrote: Greetings All, I have the following array of data, column 1 is the date and column 2 is the amount in US$. I am trying to use regression to forecast three years or so hence. 3-Jan-06 750,000 3-Apr-06 1,000,000 27-Apr-06 1,009,778 14-Jul-06 1,139,739 26-Oct-06 1,109,484 5-Dec-06 1,169,886 9-Jan-07 1,244,587 10-Mar-07 1,421,393 28-Mar-07 1,419,597 14-Aug-07 1,642,892 If I use a linear regression trendline on this data, the R^2 is around 90% which is not bad. If I use the "Forecast Forward Periods" function, this gives me an idea of when the regression line passes 2,000,000 or 2,500,000. However, my question is this: what is the correct x-value input to use with the regression equation y = 1332.4x - 5E+07 since using the 1900-calendar number clearly does not work. Many thanks, Peter SS |
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