Question

A special bumper was installed on selected vehicles in a large fleet. The dollar cost of body repairs was recorded for all vehicles that were involved in accidents over a 1-year period. Those with the special bumper are the test group and the other vehicles are the control group, shown below. Each "repair incident" is defined as an invoice (which might include more than one separate type of damage).

Statistic	Test Group					Control Group
Mean Damage	X¯¯¯1X¯1	=	$	1,245		X¯¯¯2X¯2	=	$	1,790
Sample Std. Dev.	s1	=	$	721		s2	=	$	835
Repair Incidents	n1	=		17		n2	=		14

Source: Unpublished study by Thomas W. Lauer and Floyd G. Willoughby.

(a) Construct a 99 percent confidence interval for the true difference of the means assuming equal variances. (Round your intermediate t_crit value to 3 decimal places. Round your final answers to 3 decimal places. Negative values should be indicated by a minus sign.)

The 99% confidence interval is from  to  

(b) Repeat part (a), using the assumption of unequal variances with Welch's formula for d.f. (Round your intermediate t_critvalue to 3 decimal places. Round your final answers to 3 decimal places. Negative values should be indicated by a minus sign.)

The 99% confidence interval is from  to  

(c) Did the assumption about variances change the conclusion?
  

• Yes

• No

(d) Construct separate 99% confidence intervals for each mean. (Round your intermediate t_crit value to 3 decimal places. Round your final answers to 2 decimal places.)

Mean Damage	Confidence Interval
x¯1=$1,245x¯1=$1,245	($  , $  )
x¯2=$1,790x¯2=$1,790	($  , $  )

Answer 1

Part a)

Confidence interval is :-
( X̅1 - X̅2 ) ± t( α/2 , n1+n2-2) SP √( (1/n1) + (1/n2))
t(α/2, n1 + n1 - 2) = t( 0.01/2, 17 + 14 - 2) = 2.756

( 1245 - 1790 ) ± t(0.01/2 , 17 + 14 -2) 774.1821 √ ( (1/17) + (1/14))
Lower Limit = ( 1245 - 1790 ) - t(0.01/2 , 17 + 14 -2) 774.1821 √( (1/17) + (1/14))
Lower Limit = -1315.1504
Upper Limit = ( 1245 - 1790 ) + t(0.01/2 , 17 + 14 -2) 774.1821 √( (1/17) + (1/14))
Upper Limit = 225.1504
99% Confidence Interval is ( -1315.150 , 225.150 )

Part b)

Confidence interval :-

t(α/2, DF) = t(0.01 /2, 25 ) = 2.787

Lower Limit =
Lower Limit = -1335.2794
Upper Limit =
Upper Limit = 245.2794
99% Confidence interval is ( -1335.279 , 245.279 )

Part c)

No, decision does not changed.

Part d)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 17- 1 ) = 2.921
1245 ± t(0.01/2, 17 -1) * 721/√(17)
Lower Limit = 1245 - t(0.01/2, 17 -1) 721/√(17)
Lower Limit = 734.2482
Upper Limit = 1245 + t(0.01/2, 17 -1) 721/√(17)
Upper Limit = 1755.7518
99% Confidence interval is ( 734.25 , 1755.75 )

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 14- 1 ) = 3.012
1790 ± t(0.01/2, 14 -1) * 835/√(14)
Lower Limit = 1790 - t(0.01/2, 14 -1) 835/√(14)
Lower Limit = 1117.7711
Upper Limit = 1790 + t(0.01/2, 14 -1) 835/√(14)
Upper Limit = 2462.2289
99% Confidence interval is ( 1117.77 , 2462.23 )