For this assignment, I used a regression analysis to determine the missing rainfall data for Station A between 1931 and 1949. In order to accomplish this, I first had to determine the slope and the intercept coefficient for the relationship between the two sets of available data for the variables. Then I multiplied the the slope with the value of Station B for each year, and added the intercept coefficient to determine the rainfall that was missing for each year for Station A. While something like rainfall is impossible to precisely predict, the statistics used here could be very useful in similar scenarios. It's not very different from recent previous assignments, like surface interpolation. It may not be precise, but it gives you a good idea of what the reality probably is or was.
The results can be seen below:
| Year |
Station B |
Station A |
|
|
| 1931 |
1005.84 |
1013.45 |
|
|
| 1932 |
1148.08 |
1133.81 |
Slope: |
0.846171 |
| 1933 |
691.39 |
747.37 |
Intercept: |
162.3421 |
| 1934 |
1328.25 |
1286.27 |
|
|
| 1935 |
1042.42 |
1044.40 |
|
|
| 1936 |
1502.41 |
1433.64 |
|
|
| 1937 |
1027.18 |
1031.51 |
|
|
| 1938 |
995.93 |
1005.07 |
|
|
| 1939 |
1323.59 |
1282.33 |
|
|
| 1940 |
946.19 |
962.98 |
|
|
| 1941 |
989.58 |
999.70 |
|
|
| 1942 |
1124.60 |
1113.94 |
|
|
| 1943 |
955.04 |
970.47 |
|
|
| 1944 |
1215.64 |
1190.98 |
|
|
| 1945 |
1418.22 |
1362.40 |
|
|
| 1946 |
1323.34 |
1282.11 |
|
|
| 1947 |
1391.75 |
1340.00 |
|
|
| 1948 |
1338.97 |
1295.34 |
|
|
| 1949 |
1204.47 |
1181.53 |
|
|
No comments:
Post a Comment