Question

Consumer price indices in 2018 in RK were:

Feb. 722.7

Mar. 726.5

Apr. 729.3

May 730.8

Jun. 732.2

Jul. 733.1

Aug. 734.2

Sep. 736.9

Oct. 739.9

Nov. 746.4

Dec. 751.5

a)      Produce the simple linear regression analysis. Present the fitted line analytically. Is there a significant linear dependence of СPI on time? Explain.

b)      Construct the 95% confidence predicted interval for the expected in the mean CPI for Jan. 2019 .

c)      How much is r squared and what is its meaning?

d)      What assumptions do you need for that prediction?

e)      Draw the corresponding graph and show on it your predicted interval.

Answer 1

A)

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
1	722.7	25.00	147.95	60.82
2	726.5	16.00	69.95	33.45
3	729.3	9.00	30.95	16.69
4	730.8	4.00	16.51	8.13
5	732.2	1.00	7.09	2.66
6	733.1	0.00	3.11	0.00
7	734.2	1.00	0.44	-0.66
8	736.9	4.00	4.15	4.07
9	739.9	9.00	25.36	15.11
10	746.4	16.00	133.09	46.15
11	751.5	25.00	276.77	83.18

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	66	8083.5	110	715.4	269.60
mean	6.00	734.86	SSxx	SSyy	SSxy

sample size ,   n =   11          
here, x̅ = Σx / n=   6.00   ,     ȳ = Σy/n =   734.86  
                  
SSxx =    Σ(x-x̅)² =    110.0000          
SSxy=   Σ(x-x̅)(y-ȳ) =   269.6          
                  
estimated slope , ß1 = SSxy/SSxx =   269.6   /   110.000   =   2.4509
                  
intercept,   ß0 = y̅-ß1* x̄ =   720.1582          
                  
so, regression line is   Ŷ =   720.1582   +   2.4509   *x

............

slope hypothesis test               tail=   2
Ho:   ß1=   0          
H1:   ß1╪   0          
n=   11              
alpha =   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    2.464   /√   110.00   =   0.2349
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    2.4509   /   0.2349   =   10.4344
                  
t-critical value=    2.262   [excel function: =T.INV.2T(α,df) ]          
Degree of freedom ,df = n-2=   9              
p-value =    0.0000              
decison :    p-value<α , reject Ho             
Conclusion:  reject Ho and conclude  that linear relations exists between X and y              
..............

B)

Predicted Y at X=   12   is                  
Ŷ =   720.15818   +   2.450909   *   12   =   749.569
For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   2.9337              
margin of error,E=t*std error=t*S(ŷ)=    2.2622   *   2.93   =   6.6366
                  
Prediction Interval Lower Limit=Ŷ -E =   749.569   -   6.64   =   742.933
Prediction Interval Upper Limit=Ŷ +E =   749.569   +   6.64   =   756.206

............

c)

R² =    (Sxy)²/(Sx.Sy) =    0.9236

there is 92.36% data is explained by time of CPI (y)

..........

d)

Linear relationship
Homoscedasticity
Multivariate normality
........

Please revert back in case of any doubt.

Please upvote. Thanks in advance.