Question

(1 point) The amounts of 6 restaurant bills (r) and the corresponding amounts of the tips (y) are given in the table below Bill 70.29 64.30 49.72 43.58 52.44 88.01 Tip 10.00 7.70 5.28 5.50 7.00 10.00 Use a 0.05 confidence level to find the following: The correlation coefficient r The test statistic for a test of significant correlation t Is there a significant correlation? A. No B. Yes The regression equation is y . If the amount of the bill is $90, the best prediction for the amount of the tip is

Answer 1

	X	Y	X * Y	X2	Y2
	70.29	10	702.9	4940.684	100
	64.3	7.7	495.11	4134.49	59.29
	49.72	5.28	262.5216	2472.078	27.8784
	43.58	5.5	239.69	1899.216	30.25
	52.44	7	367.08	2749.954	49
	88.01	10	880.1	7745.76	100
Total	368.34	45.48	2947.402	23942.18	366.4184

$r = ( ( N * \Sigma XY) - \Sigma X \Sigma Y) / (\sqrt{ (N \Sigma (X^{2}) - (\Sigma X)^{2} )(N \Sigma (Y^{2}) - (\Sigma Y)^{2} })$
$r = ( ( 6 * 2947.4016 ) - 368.34 * 45.48 ) / ( \sqrt{ ( 6 * 23942.1826 ) - ( 368.34 )^{2} )( 6 * 366.4184 ) - ( 45.48 )^{2} })$
r = 0.9151

To Test :-

H0 :- ρ = 0
H1 :- ρ ≠ 0

Test Statistic :-
t = (r * √(n - 2) / (√(1 - r2))
t = ( 0.9151 * √(6 - 2) ) / (√(1 - 0.8375) )
t = 4.5402

Test Criteria :-
Reject null hypothesis if t > t(α/2,n-2)
t(α/2,n-2) = t(0.05/2 , 6 - 2 ) = 2.7764
t > t (α/2, n-2) = 4.5402 > 2.7764
Result :- Reject null hypothesis

Decision based on P value
P - value = P ( t > 4.5402 ) = 0.0105
Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.0105 < 0.05 ,hence we reject null hypothesis
Conclusion :- We reject H0

Yes, correlation is significant

Equation of regression line is Ŷ = a + bX
$b = ( n \Sigma XY - \Sigma X \Sigma Y) / ( n\Sigma X^{2} - (\Sigma X)^{2})$
$b = ( 6 * 2947.4016 - 368.34 * 45.48 ) / ( 6 * 23942.1826 - ( 368.34 )^{2})$
b = 0.117
a =( Σ Y - ( b * Σ X) ) / n
a =( 45.48 - ( 0.1168 * 368.34 ) ) / 6
a = 0.407
Equation of regression line becomes Ŷ = 0.4067 + 0.1168 X

When X = 90
Ŷ = 0.407 + 0.117 X
Ŷ = 0.407 + ( 0.117 * 90 )
Ŷ = 10.94