Question

West Edmonton Mall, located in Edmonton, Canada, is the largest shopping mall in North America. The mall attracts thousands of shoppers daily. The actual daily visitor number depends on the day and season. Does the cold weather have any effect on the number of visitors, and eventually the retail business of the mall? In order to investigate this, suppose a sample of 6 weekdays was randomly selected from January 2019 to February 2019. The number of shoppers/visitors (in thousands) and the mean temperature (°C) were recorded for each of the selected days as shown in the following table. Day Mean Temperature Number of Visitors (°C) in thousands) ly) (x) February 8 January 25 February 29 January 4 January 17 January 10 -8.7 -1.3 -13.4 4.9 -29.4 -4.9 90 76 46 128 16 106

Answer 1

a)

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	-52.80	462.00	704.68	8294.00	2204.80
mean	-8.80	77.00	SSxx	SSyy	SSxy

Sample size,   n =   6      
here, x̅ = Σx / n=   -8.800          
ȳ = Σy/n =   77.000          
SSxx =    Σ(x-x̅)² =    704.6800      
SSxy=   Σ(x-x̅)(y-ȳ) =   2204.8      
              
estimated slope , ß1 = SSxy/SSxx =   2204.8/704.68=   3.1288      
intercept,ß0 = y̅-ß1* x̄ =   77- (3.1288 )*-8.8=   104.5334      
              
Regression line is, Ŷ=   104.53 + (   3.13 )*x

b)

Predicted Y at X=   -20   is          
Ŷ=   104.53341   +   3.12880   *-20=   42.0

α=   0.01              
t critical value=   t α/2 =    4.604   [excel function: =t.inv.2t(α/2,df) ]      
estimated std error of slope = Se/√Sxx =    18.6791/√704.68=   0.704          
                  
margin of error ,E= t*std error =    4.604   *   0.704   =   3.239690
estimated slope , ß^ =    3.1288              
                  
lower confidence limit = estimated slope - margin of error =   3.1288   -   3.240   =   -0.111
upper confidence limit=estimated slope + margin of error =   3.1288   +   3.240   =   6.369

c)

Ho:   β1=   0
H1:   β1 >   0
n=   6  
alpha =   0.01  
estimated std error of slope =Se(ß1) = Se/√Sxx =    18.6791/√704.68=   0.7037
t stat = estimated slope/std error =ß1 /Se(ß1) =    (3.1288-0)/0.7037=   4.45
Degree of freedom ,df = n-2=   4  
t-critical value=    3.7469   [Excel function: =T.INV(α,df) ]
      
p-value =    0.0056  
decison :    p-value<α , reject Ho  
Conclusion:   Reject Ho and conclude that slope is significantly different from zero  

d)

SSE=   (SSxx * SSyy - SS²xy)/SSxx =    1395.6305
std error ,Se =    √(SSE/(n-2)) =    18.6791
      
correlation coefficient ,    r = SSxy/√(SSx.SSy) =   0.91

e)

R² =    (SSxy)²/(SSx.SSy) =    0.8317      
Approximately    83.17%   of variation in observations of variable Y, is explained by variable x      

Please let me know in case of any doubt.

Thanks in advance!

Please upvote!