Question

The following data on price ($) and the overall score for 6 stereo headphones that were tested by Consumer Reports were as follows. Brand Score Price ($) 186 Bose 72 Scullcandy 162 Koss 84 72 66 54 Phillips/O'Neill Denon 78 78 48 30 JVC 48 a. Does the t test indicate a significant relationship between price and the overall score? The test t-Conclusion at a = .05 t = (to 2 decimal places.) p-value is between .025 and .05 What is your conclusion? Use a = .05. There is a significant relationship between price and overall score b. Test for a significant relationship using the F test. p-value is between .025 and .05 What is your conclusion? Use a = .05. Because p-value is less than or equal to .05, we reject HO: B 1 is equal to zero

Answer 1

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	636.00	342.00	14952.00	1350.00	3744.00
mean	106.00	57.00	SSxx	SSyy	SSxy

Sample size,   n =   6      
here, x̅ = Σx / n=   106.000          
ȳ = Σy/n =   57.000          
SSxx =    Σ(x-x̅)² =    14952.0000      
SSxy=   Σ(x-x̅)(y-ȳ) =   3744.0      
              
estimated slope , ß1 = SSxy/SSxx =   3744/14952=   0.2504      
intercept,ß0 = y̅-ß1* x̄ =   57- (0.2504 )*106=   30.4575      
              
Regression line is, Ŷ=   30.457   + (   0.250   )*x
              
SSE=   (SSxx * SSyy - SS²xy)/SSxx =    412.4976      
std error ,Se =    √(SSE/(n-2)) =    10.1550      

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

a)

Ho:   β1=   0
H1:   β1╪   0
n=   6  
alpha =   0.05  
estimated std error of slope =Se(ß1) = Se/√Sxx =    10.155/√14952=   0.0830

t stat = estimated slope/std error =ß1 /Se(ß1) =    (0.2504-0)/0.083=   3.02
Degree of freedom ,df = n-2=   4  

      
p-value =    0.0394  

p-value is between .025 and .05 What is your conclusion? Use a = .05. There is a significant relationship between price and overall score b. Test for a significant relationship using the Ftest. p-value is between .025 and .05 What is your conclusion? Use a = .05. Because p-value is less than or equal to .05, we reject HO: B 1 is equal to zero

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

Anova table
variation	SS	df	MS	F-stat	p-value
regression	937.502	1	937.502	9.091	0.039
error,	412.498	4	103.124
total	1350.000	5

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

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