Question

Listed below are # of inches of snow in the storm and # of the kids absent from the school next day. a) Is there a linear correlation between Inches of snow and the absences the day after the snow ? use α= 10 % ? b) Use CI 90% to find the best estimated predicted inches of snow if 16 kids are absence next day.

2. Listed below are # of inches of snow in the storm and # of the kids absent from the school next day. a) Is there a linear correlation between Inches of snow and the absences the day after the snow ? use a= 10% ? b) Use Ci 90% to find the best estimated predicted inches of snow if 16 kids are absence next day. 2 4 5 7 12 12 12 12 16 18 Snow 1 111 4 10 7 3 6 11 20 Absence S

Answer 1

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	100.00	74.00	250.00	306.40	187.00
mean	10.00	7.40	SSxx	SSyy	SSxy

Sample size,   n =   10      
here, x̅ = Σx / n=   10.000          
ȳ = Σy/n =   7.400          
SSxx =    Σ(x-x̅)² =    250.0000      
SSxy=   Σ(x-x̅)(y-ȳ) =   187.0      
              
estimated slope , ß1 = SSxy/SSxx =   187/250=   0.7480      
intercept,ß0 = y̅-ß1* x̄ =   7.4- (0.748 )*10=   -0.0800      
              
Regression line is, Ŷ=   -0.080   + (   0.748   )*x
              
SSE=   (SSxx * SSyy - SS²xy)/SSxx =    166.5240      
std error ,Se =    √(SSE/(n-2)) =    4.5624      
              
correlation coefficient ,    r = SSxy/√(SSx.SSy) =   0.67566      

Ho:   β1=   0
H1:   β1╪   0
n=   10  
alpha =   0.1  
estimated std error of slope =Se(ß1) = Se/√Sxx =    4.5624/√250=   0.2886
t stat = estimated slope/std error =ß1 /Se(ß1) =    (0.748-0)/0.2886=   2.59
Degree of freedom ,df = n-2=   8  
t-critical value=    1.8595   [excel function: =T.INV.2T(α,df) ]
      
p-value =    0.0320  
decison :    p-value<α , reject Ho  
Conclusion:   Reject Ho and conclude that slope is significantly different from zero  

Hence there is correlation

b)

X Value=   16              
Confidence Level=   90%              
                  
                  
Sample Size , n=   10              
Degrees of Freedom,df=n-2 =   8              
critical t Value=tα/2 =   1.860   [excel function: =t.inv.2t(α/2,df) ]          
                  
X̅ =    10.00              
Σ(x-x̅)² =Sxx   250.00              
Standard Error of the Estimate,Se=   4.5624              
                  
Predicted Y at X=   16   is          
Ŷ=   -0.08000   +   0.74800   *16=   11.888

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    2.254              
margin of error,E=t*Std error=t* S(ŷ) =   1.8595   *   2.254   =   4.1908
                  
Confidence Lower Limit=Ŷ +E =    11.888   -   4.191   =   7.6972
Confidence Upper Limit=Ŷ +E =   11.888   +   4.191   =   16.0788

Please let me know in case of any doubt.

Thanks in advance!

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