Question

#14 . Listed below are speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, CA. This simple random sample was obtained at 3:30 p.m. on a weekday. Use a 0.05 significance level to test the claim of the highway engineer that the standard deviation of speeds is equal to 5.0 mi/h. (With a 95% confidence) 62 61 61 57 61 54 59 58 59 69 60 67

A) Construct a confidence interval using the confidence level provided.

B) Test the hypothesis

C) Enter the results in excel as following

confidence level critical value confidence interval

   XL^2

   XR^2

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	62	1.7777
	61	0.1111
	61	0.1111
	57	13.4447
	61	0.1111
	54	44.4449
	59	2.7779
	58	7.1113
	59	2.7779
	69	69.4439
	60	0.4445
	67	40.1107
Total	728.0	182.6668

Mean X̅ = Σ Xi / n
X̅ = 728 / 12 = 60.6667
Sample Standard deviation S_X = √ ( (Xi - X̅ )2 / n - 1 )
S_X = √ ( 182.6668 / 12 -1 ) = 4.0751

Part a)

S² = 16.6064
α = 0.05
n = 12
((n-1)S² / χ² (0.05/2)) < σ² < ((n-1)S² / χ² (1 - 0.05/2) )
(( 12-1 ) 16.6064 / χ² (0.05/2) ) < σ² < ((12-1)16.6064 / χ² (1 - 0.05/2) )
χ2(0.05/2) = 21.92
χ² (1 - 0.05/2) ) = 3.8157
Lower Limit = (( 12-1 ) 16.6064 / χ² (0.05/2) ) = 8.3335
Upper Limit = (( 12-1 ) 16.6064 / χ² (0.05/2) ) = 47.8734
95% Confidence interval is ( 8.3335 , 47.8734 )
( 8.3335 < σ² < 47.8734 )
( 2.8868 < σ < 6.9191 )

Part b)

To Test :-

H0 :-

H1 :-  

Test Statistic :-

χ² = ( ( 12-1 ) * 16.6064 ) / 25
χ² = 7.3068

Test Criteria :-
Reject null hypothesis if
χ² (0.05/2,12 - 1) = 21.92
χ² (0.05/2,12 - 1) = 3.816
7.3068 lies between the value 21.92 and 3.816 , hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Decision based on P value
P value = 2 * P ( χ² > 7.3068 )
P value = 0.4526
Reject null hypothesis if P value < α = 0.05
Since P value = 0.4526 > 0.05, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

There is sufficient evidence to support the claim of the highway engineer that the standard deviation of speeds is equal to 5.0.

Part c)

XL²   =   χ² (1 - 0.05/2) ) = 3.8157

   XR² = χ2(0.05/2) = 21.92