In: Statistics and Probability
People in the aerospace industry believe the cost of a space
project is a function of the mass of the major object being sent
into space. Use the following data to develop a regression model to
predict the cost of a space project by the mass of the space
object. Determine r2 and
se.
|
Weight (tons) |
Cost ($ millions) |
|---|---|
|
1.897 |
$ 53.6 |
|
3.019 |
184.9 |
|
0.453 |
6.4 |
|
0.971 |
23.5 |
|
1.058 |
33.1 |
|
2.100 |
110.4 |
|
2.385 |
104.6 |
*(Do not round the intermediate values. Round your
answers to 4 decimal places.)
**(Round the intermediate values to 4 decimal places. Round
your answer to 3 decimal places.)
2) For a least squares regression line, the sum of the residuals
is __________.
a) sometimes positive and sometimes negative
b) always zero
c) always positive
d) always negative
3) In the regression equation, y = 2.164 + 1.3657x, n = 6, the mean of x is 8.667, SSxx = 89.333 and Se = 3.44. A 95% confidence interval for the average of y when x = 8 is _________.
a) (3.56, 22.62)
b) (10.31, 15.86)
c) (9.13, 17.05)
d) (12.09, 14.09)
using excel>data >data analysis >regression
we have
| Simple Linear Regression Analysis | ||||||
| Regression Statistics | ||||||
| Multiple R | 0.9521 | |||||
| R Square | 0.9065 | |||||
| Adjusted R Square | 0.8878 | |||||
| Standard Error | 21.0780 | |||||
| Observations | 7 | |||||
| ANOVA | ||||||
| df | SS | MS | F | Significance F | ||
| Regression | 1 | 21547.3807 | 21547.3807 | 48.4994 | 0.0009 | |
| Residual | 5 | 2221.4078 | 444.2816 | |||
| Total | 6 | 23768.7886 | ||||
| Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
| Intercept | -38.7124 | 18.0116 | -2.1493 | 0.0843 | -85.0126 | 7.5878 |
| Weight (tons) | 66.2700 | 9.5159 | 6.9641 | 0.0009 | 41.8087 | 90.7314 |
Ans 1 )
| R Square | 0.9065 |
| Standard Error | 21.0780 |
2) For a least squares regression line, the sum of the residuals is b) always zero
3) In the regression equation, y = 2.164 + 1.3657x, n = 6, the mean of x is 8.667, SSxx = 89.333 and Se = 3.44. A 95% confidence interval for the average of y when x = 8 is 13.08+/- 2.78*3.44 = (3.56, 22.62)
option a is true
a) (3.56, 22.62)