For the data set shown below
x y
20 98
30 95
40 91
50 83
60 70
(a) Use technology to find the estimates of β0 and β1.
β0≈b0=114.60
(Round to two decimal places as needed.)
β1≈b1=−0.68
(Round to two decimal places as needed.)
(b) Use technology to compute the standard error, the point estimate for σ.
Se=3.7771
(Round to four decimal places as needed.)
(c) Assuming the residuals are normally distributed, use technology to determine sb1.
sb1=__?__
(Round to four decimal places as needed.)
Following is the output of regression line generated by excel:
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 0.956703233 | |||||
R Square | 0.915281077 | |||||
Adjusted R Square | 0.887041436 | |||||
Standard Error | 3.777124126 | |||||
Observations | 5 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 1 | 462.4 | 462.4 | 32.41121495 | 0.010744254 | |
Residual | 3 | 42.8 | 14.26666667 | |||
Total | 4 | 505.2 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | 114.6 | 5.067543784 | 22.61450614 | 0.000189348 | 98.47281401 | 130.727186 |
X | -0.68 | 0.119443152 | -5.693084836 | 0.010744254 | -1.06012142 | -0.299878581 |
-------------------------
(a)
β0≈b0=114.60
β1≈b1=−0.68
(b)
Se=3.7771
(c)
sb1=0.1194
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