3)Interpret the results of the tables below. Table 1: Model Summary Model R R Sq
ID: 3293219 • Letter: 3
Question
3)Interpret the results of the tables below.
Table 1: Model Summary
Model
R
R Square
Adjusted R Square
Std. Error of the Estimate
1
.716a
.512
.507
.8711
a. Predictors: (Constant), X17 - Price Flexibility, X18 - Delivery Speed
Table 2:ANOVAa
Model
Sum of Squares
df
Mean Square
F
Sig.
1
Regression
157.067
2
78.534
103.505
.000b
Residual
149.472
197
.759
Total
306.539
199
a. Dependent Variable: X19 - Satisfaction
b. Predictors: (Constant), X17 - Price Flexibility, X18 - Delivery Speed
Table 3: Coefficientsa
Model
Unstandardized Coefficients
Standardized Coefficients
t
Sig.
B
Std. Error
Beta
1
(Constant)
3.524
.331
10.661
.000
X18 - Delivery Speed
1.380
.096
.833
14.374
.000
X17 - Price Flexibility
-.412
.060
-.396
-6.824
.000
a. Dependent Variable: X19 - Satisfaction
4)3)
Table 1: Model Summary
Model
R
R Square
Adjusted R Square
Std. Error of the Estimate
1
.716a
.512
.507
.8711
a. Predictors: (Constant), X17 - Price Flexibility, X18 - Delivery Speed
Table 2:ANOVAa
Model
Sum of Squares
df
Mean Square
F
Sig.
1
Regression
157.067
2
78.534
103.505
.000b
Residual
149.472
197
.759
Total
306.539
199
a. Dependent Variable: X19 - Satisfaction
b. Predictors: (Constant), X17 - Price Flexibility, X18 - Delivery Speed
Table 3: Coefficientsa
Model
Unstandardized Coefficients
Standardized Coefficients
t
Sig.
B
Std. Error
Beta
1
(Constant)
3.524
.331
10.661
.000
X18 - Delivery Speed
1.380
.096
.833
14.374
.000
X17 - Price Flexibility
-.412
.060
-.396
-6.824
.000
a. Dependent Variable: X19 - Satisfaction
4)
Table 1: Model Summary
Model
R
R Square
Adjusted R Square
Std. Error of the Estimate
1
.716a
.512
.507
.8711
a. Predictors: (Constant), X17 - Price Flexibility, X18 - Delivery Speed
Explanation / Answer
From the 1st table, we see that the value of R2 is 0.512. This means that 51.2% of the variation is explained by the model and the rest of the variation is not explained by the model.
From the 2nd table, we see that the significant value is 0.000 which is less than 0.05 and so we have sufficient evidence to reject the null hypothesis H0.
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