Question

Question 2

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.

dollars	satisfaction
11	6
18	8
17	10
15	4
9	9
5	6
12	3
19	5
22	2
25	10

a) Find the equation of the fitted regression line.

b) Estimate the standard deviation of the error term e_i

c)   Compute the correlation coefficient

c)   What is the R² for this model?

e)   Calculate a 90% confidence interval for b₁

f)    Test the null hypothesis that b₁ = 0 (perform a two-sided test), using α = 0.1. Is the model useful?

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 ).

Answer 1

Soln

a)

	dollars (X)	satisfaction (Y)	X*Y	X2	Y2
	11	6	66	121	36
	18	8	144	324	64
	17	10	170	289	100
	15	4	60	225	16
	9	9	81	81	81
	5	6	30	25	36
	12	3	36	144	9
	19	5	95	361	25
	22	2	44	484	4
	25	10	250	625	100
Total	153	63	976	2679	471

Let the regression equation be: Y = a + bX

Where

Slope(b) = {n*∑XY - ∑X *∑Y}/{n*∑X² – (∑X)² } = 0.-36

and a = ∑Y/n – b*∑X/n = 5.75

Hence,

Regression Equation

Satisfaction = 5.75 + 0.036 * dollars

b)

satisfaction (Y)	Y Predicted	Residual
6	6.15	-0.15
8	6.40	1.60
10	6.36	3.64
4	6.29	-2.29
9	6.07	2.93
6	5.93	0.07
3	6.18	-3.18
5	6.43	-1.43
2	6.54	-4.54
10	6.65	3.35

Standard Deviation of the Error (Residual) is calculated using the below formula and above values:

Standard Deviation = 2.86

c)

Correlation Coefficient using the above formula and values calculated on part a = 0.0764

d)

R Square = Correlation Coefficient ² = 0.0058

e)

90% CI for b₁ = {-0.271, 0.343}

f)

alpha = 0.1

Null and Alternate Hypothesis

H₀: b₁ = 0

H_a: b₁ <> 0

Test Statistic’

t = 0.22

p-value = TDIST(0.22, 10-2,2) = 0.8337

Result

Since the p-value is greater than 0.1, we fail to reject the null hypothesis

Conclusion

The model is not useful as the slope is not significant. Also, the R square of the model is too less (0.58%)

g)

P-value using SAS = 0.8337

It is inline with the value calculated in part f

SAS Code

%web_drop_table(WORK.IMPORT);

FILENAME REFFILE '/home/rkvimal0/sasuser.v94/Data_5Feb20.csv';

PROC IMPORT DATAFILE=REFFILE

                DBMS=CSV

                OUT=WORK.IMPORT;

                GETNAMES=YES;

RUN;

PROC CONTENTS DATA=WORK.IMPORT; RUN;

ods noproctitle;

ods graphics / imagemap=on;

proc reg data=WORK.IMPORT alpha=0.05 plots(only)=(diagnostics residuals

                                fitplot observedbypredicted);

                model satisfaction=dollars/;

                run;

quit;

Output