With the growth of internet service providers, a researcher decides to examine w
ID: 3300557 • Letter: W
Question
With the growth of internet service providers, a researcher decides to examine whether there is a correlation between cost of internet service per month (rounded to the nearest dollar) and degree of customer satisfaction (on a scale of 1 - 10 with a 1 being not at all satisfied and a 10 being extremely satisfied). The researcher only includes programs with comparable types of services. A sample of the data is provided below:
a) Find the equation of the fitted regression line.
c) Compute the correlation coefficient
d) What is the R2 for this model?
g) Perform the regression using SAS, and give the p-value to the test in part f ). Verify that the p-value agrees with your conclusion in part f ).
DOLLARS Satisfaction 11 6 18 8 17 10 15 4 9 9 5 6 12 3 19 5 22 2 25 10Explanation / Answer
Solution:
Part a
First of all we have to find the regression equation for satisfaction. The formulas for the regression coefficients and calculation table are summarised below:
No.
Dollars (X)
Satisfaction (Y)
X^2
Y^2
X*Y
1
11
6
121
36
66
2
18
8
324
64
144
3
17
10
289
100
170
4
15
4
225
16
60
5
9
9
81
81
81
6
5
6
25
36
30
7
12
3
144
9
36
8
19
5
361
25
95
9
22
2
484
4
44
10
25
10
625
100
250
Total
153
63
2679
471
976
From above table, we have
X = 153
Y = 63
X^2 = 2679
Y^2 = 471
XY = 976
n = 10
Xbar = 153/10 = 15.3
Ybar = 63/10 = 6.3
b = (XY – n*Xbar*Ybar)/(X^2 – n*Xbar^2)
b = (976 – 10*15.3*6.3)/(2679 – 10*15.3^2)
b = 12.1/338.1
b = 0.035788228
a = Ybar – b*Xbar
a = 6.3 - 0.035788228*15.3
a = 5.752440112
Regression equation is given as below:
Y = a + b*X
Y = 5.752440112 + 0.035788228*X
Part b
Formula for standard deviation of the error is given as below:
S = sqrt[SSE/(n – k – 1)]
S = sqrt[73.67/(10 – 1 – 1)]
S = sqrt[73.67/8]
S = sqrt(9.20875)
S = 3.034592
Part c
Formula for correlation coefficient is given as below:
Correlation coefficient = r = [nxy - xy]/sqrt[(nx^2 – (x)^2)*(ny^2 – (y)^2)]
Correlation coefficient = r = [10*976 – 153*63]/sqrt[(10*2679 – 153^2)*(10*471 – 63^2)]
Correlation coefficient = r = 0.076445809
Part d
Formula for R^2 or coefficient of determination is given as below:
R^2 = r^2 * 0.076445809^2 = 0.076445809* 0.076445809 = 0.005843962
R^2 for this model is given as 0.58%.
Regression output for more reference is given as below:
Regression Statistics
Multiple R
0.076445809
R Square
0.005843962
Adjusted R Square
-0.118425543
Standard Error
3.034529668
Observations
10
ANOVA
df
SS
MS
F
P-value
Regression
1
0.43
0.433038
0.047027
0.83374863
Residual
8
73.67
9.20837
Total
9
74.1
Coefficients
std error
t Stat
P-value
Lower 95%
Upper 95%
Intercept
5.752440106
2.701190162
2.129595
0.065835
-0.47651557
11.98139578
Dollars (X)
0.035788228
0.165032243
0.216856
0.833749
-0.34477681
0.416353263
No.
Dollars (X)
Satisfaction (Y)
X^2
Y^2
X*Y
1
11
6
121
36
66
2
18
8
324
64
144
3
17
10
289
100
170
4
15
4
225
16
60
5
9
9
81
81
81
6
5
6
25
36
30
7
12
3
144
9
36
8
19
5
361
25
95
9
22
2
484
4
44
10
25
10
625
100
250
Total
153
63
2679
471
976
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