Question

In: Statistics and Probability

The weights (in pounds) of 6 vehicles and the variability of their braking distances (in feet)...

Weight, x

5910

5350

6500

5100

5850

4800

Variability in braking distance, y

1.79

1.99

1.91

1.55

1.68

1.50

Setup the hypotheis for the test

Idenfity the critical values. Round to three decimal places

Calculate the test statistic

Solutions

Expert Solution

	Weight X	Braking Distance Y	X * Y
	5910	1.79	10578.9	34928100	3.2041
	5350	1.99	10646.5	28622500	3.9601
	6500	1.91	12415	42250000	3.6481
	5100	1.55	7905	26010000	2.4025
	5850	1.68	9828	34222500	2.8224
	4800	1.5	7200	23040000	2.25
Total	33510	10.42	58573.4	1.89E+08	18.2872

To Test :-

H0 :-

H1 :-

Test Statistic :-

t = 1.5951

Test Criteria :-
Reject null hypothesis if

-4.6041 < 1.5951 < 4.6041
Result :- We fail to Reject null hypothesis

Decision based on P value
P - value = P ( t > 1.5951 ) = 0.1859
Reject null hypothesis if P value < level of significance
P - value = 0.1859 > 0.01 ,hence we fail to reject null hypothesis
Conclusion :- We Accept H0

There is no linear correlation between two variables.

orchestra answered 1 year ago

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