In: Math
One late Friday afternoon your obnoxious boss comes into your office as you are about to leave, and shows you 26 observations that he believes to be related. Y, he believes, is the dependent variable and X1 is the independent variable. He also thinks there is a 2nd order polynomial relationship in the data ( Y = B1X1 +B2X12 + B0 ), and, as you casually view the data, you tend to agree. He insists that the determination of B1, B2 and B0 is far more important than your Friday afternoon gathering of young-urban-millennial-professionals (YUMPS) at a local watering hole. So, you perform the analysis using the scatter diagram and Trendline tool in Excel. Then you quietly exit for the YUMPS gathering.
Now, it is Monday morning. You want to use the assumption of a 2nd order polynomial to find the exact values of B1, B2 and B0 . (Hint: generate a new variable that fits the assumed polynomial model and then use the regression tool in Excel)
a) Use a regression tool in Excel to determine the exact values of B1, B2 and B0 .
b) Is the regression model significant? Use alpha 0.05.
Data:
X1 | X2 | Y |
68,067 | 4,633,116,489 | 1,598,278,294 |
70,103 | 4,914,430,609 | 1,695,432,796 |
76,370 | 5,832,376,900 | 2,011,698,722 |
86,686 | 7,514,462,596 | 2,592,259,645 |
86,759 | 7,527,124,081 | 2,597,646,596 |
91,805 | 8,428,158,025 | 2,907,666,308 |
92,306 | 8,520,397,636 | 2,940,058,192 |
93,731 | 8,785,500,361 | 3,030,806,599 |
100,913 | 10,183,433,569 | 3,512,923,429 |
102,199 | 10,444,635,601 | 3,603,809,834 |
109,399 | 11,968,141,201 | 4,129,345,241 |
113,430 | 12,866,364,900 | 4,439,623,027 |
118,133 | 13,955,405,689 | 4,814,663,900 |
122,820 | 15,084,752,400 | 5,203,806,895 |
123,417 | 15,231,755,889 | 5,255,000,229 |
123,054 | 15,142,286,916 | 5,224,742,756 |
127,860 | 16,348,179,600 | 5,640,704,277 |
132,868 | 17,653,905,424 | 6,091,148,092 |
131,160 | 17,202,945,600 | 5,935,115,092 |
132,132 | 17,458,865,424 | 6,022,925,591 |
132,583 | 17,578,251,889 | 6,064,201,852 |
136,859 | 18,730,385,881 | 6,462,049,499 |
140,562 | 19,757,675,844 | 6,816,147,332 |
144,594 | 20,907,424,836 | 7,212,915,055 |
141,493 | 20,020,269,049 | 6,906,705,120 |
139,465 | 19,450,486,225 | 6,710,038,170 |
Data > Data Analysis > Regression.
Output using excel:
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 1 | |||||
R Square | 1 | |||||
Adjusted R Square | 1 | |||||
Standard Error | 377471.72 | |||||
Observations | 26 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 2 | 7.816E+19 | 3.908E+19 | 274257835 | 1.4428E-85 | |
Residual | 23 | 3.277E+12 | 1.425E+11 | |||
Total | 25 | 7.816E+19 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | -3667666.87653 | 1821330.2 | -2.0137298 | 0.0558816 | -7435375.5 | 100041.8 |
X1 | 73.380775 | 34.876768 | 2.1040016 | 0.0465141 | 1.23268486 | 145.5289 |
X2 | 0.3446558 | 0.0001609 | 2142.2548 | 1.846E-62 | 0.34432302 | 0.344989 |
a) B1 = 73.3808
B2 = 0.3447
B0 = -3667666.8765
b) F test statistic using excel:
F = 274257835
p-value = 0.0000
As p-value < 0.05, we reject the null hypothesis.
And conclude that the regression model is significant.