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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 Mean Damage Sample Std. Dev Repair Incidents Control Group X1 $1,078 X2 $1,786 S1 $ 652 S2 819 Test Group 15 n2 Source: Unpublished study by Thomas W. Lauer and Floyd G. Willoughby (a) Construct a 95 percent confidence interval for the true difference of the means assuming equal variances.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 95% confidence interval is from to (b) Repeat part (a), using the assumption of unequal variances with Welchs formula for d.f. (Round your intermediate terit Value to 3 decimal places. Round your final answers to 3 decimal places Negative values should be indicated by a minus sign.) The 95% confidence interval is from to c) Did the assumption about variances change the conclusion? O Yes O No (d) Construct separate 95% confidence intervals for each mean. (Round your intermediate trit value to 3 decimal places. Round your final answers to 2 decimal places.) Mean Damage 1 $1,078 2$1,786 Confidence Interval

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Answer #1

(a)

Here we have 5-1078,SI-652, n-15 T2-1786,S2 -819, n-12 Level of significance Confidence level α 0.05 CL 0.95 Since we can assume that variances are equal so degree of freedom of t test will be: The pooled standard deviation is (ni-)s, + (n--)s*2 =730.2007 Critical value: te 2.06 The standard error is: SE sp- 730.2007*sqrt((1/15)+(1/12) i ni -282.8055 Margon of error: ME=t,.SE -582.5793 Confidence interval is: (-52)t ME -(1078-1786) + 582.5793 (-1290.5793,-125.4207) Excel function for critical value:TINV(0.05,25)Answer: (-1290.579, -125.421)

(b)

Here we have -1078,s -652, n 15 T-1786, s2-819, n2 -12 Level of significance Confidence level α 0.05 CL 0.95 Since we cannot assume that variances are equal so degree of freedom of CI will be -21 Critical value: t-2 I2.08 The standard error is: SEsqrt((425104/15)+(670761/12)) = 290.2361 Margon of error: ME-t, SE -603.6911 Confidence interval is: (4-5)± ME =(1078-1786) ± 603.6911 1311.6911,-104.3089) Excel function for critical value:TINV(0.05,21) Hence, the required confidence interval is (-1311.6911,-104.3089)

Answer: (-1311.691, -104.309)

(c)

No since both confidence interval does not contain zero.

(d)

Here we have x =1078, s-652, n=15 Since populaiton SD is unknown so t critical value will be used. Level of significance: Critical value: Degree of freedom: α=0.05 te 2.145 df -n-1-14 The requried confidence interval is: ītr .|-1078 ± (652*2.145/sqrt(15)) 1078 +361.101 (716.899,1439.101) Excel function for critical value: TINV(0.05,14) Hence, the required confidence interval is (716.899,1439.101)

Answer: ($716.90, $1439.10)

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Here we have r1786, s-819, n-12 Since populaiton SD is unknown so t critical value will be used. Level of significance: Critical value: Degree of freedom: α=0.05 te -2.201 df -n-1-11 The requried confidence interval is: ītr .|-1786 ± (819*2.201/sqrt(12)) 1786+520.371 =(1265629,2306371) Excel function for critical value: TINV(0.05,11) Hence, the required confidence interval is (1265.629,2306.371)

Answer: ($1265.63, $2306.37)

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