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⎯⎯⎯1X1	= $	1,172	X⎯⎯⎯2X2	= $	1,740
Sample Std. Dev.	s1	= $	668	s2	= $	808
Repair Incidents	n1	=	15	n2	=	12

Source: Unpublished study by Thomas W. Lauer and Floyd G. Willoughby.

(a)	Construct a 90 percent confidence interval for the true difference of the means assuming equal variances. (Round your intermediate tcrit value to 3 decimal places. Round your final answers to 3 decimal places. Negative values should be indicated by a minus sign.)

The 90% 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 tcrit value to 3 decimal places. Round your final answers to 3 decimal places. Negative values should be indicated by a minus sign.)

The 90% confidence interval is from  to

(c)	Did the assumption about variances change the conclusion?
	Yes No

(d)	Construct separate 90% confidence intervals for each mean. (Round your intermediate tcrit value to 3 decimal places. Round your final answers to 2 decimal places.)

Mean Damage	Confidence Interval
x⎯⎯1=$1,172x1$1,172	($ , $ )
x⎯⎯2=$1,740x2$1,740	($ , $ )

Answer 1

(a)

Here we have =1172,s1-668, n-15 T2-1740,s2 -808, n2 -12 Level of significance: Confidence level α =0.1 CL -0.9 Since we can assume that variances are equal so degree of freedom of t test will be: df =n1 + n2-2 =25 The pooled standard deviation is --1)s2 2-732.902:2 Critical value: te1.708 The standard error is: SE -s-732.9022*sqrt((1/15)+(1/12)) i ni -283.8518 Margon of error: ME=..SE =484.8189 Confidence interval is: 4-5)± ME :(1172-1740) ± 484.8189 (-1052.8189, -83.1811) Excel function for critical value: =TINV(0.1,25) Hence, the required confidence interval is (-1052.8189,-83.1811)

Hence, the required confidence interval is (-1052.819, -83.181).

(b)

Here we have -1172,S -668, n 15 T-1740, s2-808, n2 -12 Level of significance Confidence level: -0.1 CL-0.9 Since we cannot assume that variances are equal so degree of freedom of CI will be =22 72 Critical value: te1.717 The standard error is: SEsqrt((446224/15)+(652864/12)) = 290.0924 Margon of error: ME=t.. SE -498.0887 Confidence interval is: (M-5) ± ME =(1172-1740) ± 498.0887 -(-1066.0887, -69.9113) Excel function for critical value:TINV(0.1,22) Hence, the required confidence interval is (-1066.0887,-69.9113)

Hence, the required confidence interval is (-1066.089, -69.911).

(c)

No because both confidence interval does not contain zero.

(d)

For X1:

Here we have T-1172, -668, -15 Since populaiton SD is unknown so t critical value will be used. Level of significance: α-0.1 Critical value: 1.7613 Degree of freedom df -n-1-14 The requried confidence interval is: ītr..--1172 ± (668" 1.7613/sqrt(15)) 7 -1172 303.783 -(868.217,1475.783) Excel function for critical value: TINV(0.1,14) Hence, the required confidence interval is (868.217,1475.783).

Hence, the required confidence interval is ($868.22, $1475.78).

For X2:

Here we have X-1740, s-808, n =12 Since populaiton SD is unknown so t critical value will be used. Level of significance: a α-0.1 Critical value: 1.7959 Degree of freedom df -n-1-11 The requried confidence interval is: ītr..--1740 ± (808" 1.7959/sqrt(12) Nm 1740+418.893 -(1321.107,2158.893) Excel function for critical value: TINV(0.1,11) Hence, the required confidence interval is (1321.107,2158.893).

Hence, the required confidence interval is ($1321.11, $2158.89).