Use the data to fit a simple linear regression model for the relationship betwee
ID: 3374982 • Letter: U
Question
Use the data to fit a simple linear regression model for the relationship between the diameter of the bearing and the friction power loss that occurs in the bearing of the automobile engine.Using Excel (a) Write the expression for the model of the relationship assumed between power loss and diameter of bearing (b) Is there enough evidence to say the fitted model is statistically significant (that is, the slope of the straight line is different from zero)? Explain and provide evidence (c) Report the coefficient of determination for this model. (keep 4 decimal places) (d) Interpret the coefficient of determination for this model Diameter Clearance Length Loss 0.075 0.082 0.081 0.075 0.064 0.074 0.057 0.06 0.053 0.081 0.086 0.054 0.086 0.081 0.065 0.093 0.079 0.068 0.069 0.054 0.076 0.055 0.055 0.094 0.076 0.081 23.9 24.4 25.6 27.3 28.5 26.1 23.6 28.4 26.9 25.7 28.2 26.1 23.2 25.6 28.7 14.2 32.19 12 32.17 13.9 32.28 14 32.52 15 32.68 32.3 32.1 11.8 32.56 14 32.54 12.3 32.21 11.7 32.35 12.7 32.49 11 32.03 13.2 32.38 32.8 10.6 32.18 14.3 32.44 14.1 32.29 13.1 32.16 10.6 32.12 32.2 10.8 32.41 12.7 32.25 13.6 32.44 11.8 32.25 15.1 32.41 14.8 12.5 25.7 24.9 23.2 23.2 10.6 25.9 23.2 27.9 25.1 26Explanation / Answer
Regression analyasis for diameter of bearing and friction power loss
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.87823
R Square
0.771288
Adjusted R Square
0.761758
Standard Error
0.867767
Observations
26
ANOVA
df
SS
MS
F
Significance F
Regression
1
60.946
60.946
80.93549
3.72E-09
Residual
24
18.07247
0.753019
Total
25
79.01846
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Lower 95.0%
Upper 95.0%
Intercept
-244.485
30.02273
-8.14333
2.3E-08
-306.449
-182.521
-306.449
-182.521
X Variable 1
8.352551
0.928431
8.996415
3.72E-09
6.436364
10.26874
6.436364
10.26874
a) From the above regression model
The regression equation is Y = -244.485 + 8.3525 X
Intercept of line = -2.44.485
Slope of line = 8.352551
b) This model is statistically significance because p value for slope of line is 0 which is less than 0.05 so we reject null hypothesis and we can conclude that slope of line is not equal to zero.
c) Coefficient of determination for this model is R-square = 0.7712
d) Interpretation of R square: It tells you how many points fall on the regression line. 77% of the variation of y-values around the mean are explained by the x-values. In other words, 77% of the values fit the model.
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.87823
R Square
0.771288
Adjusted R Square
0.761758
Standard Error
0.867767
Observations
26
ANOVA
df
SS
MS
F
Significance F
Regression
1
60.946
60.946
80.93549
3.72E-09
Residual
24
18.07247
0.753019
Total
25
79.01846
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Lower 95.0%
Upper 95.0%
Intercept
-244.485
30.02273
-8.14333
2.3E-08
-306.449
-182.521
-306.449
-182.521
X Variable 1
8.352551
0.928431
8.996415
3.72E-09
6.436364
10.26874
6.436364
10.26874
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