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	=	$	1,078		X¯¯¯2X¯2	=	$	1,786
Sample Std. Dev.	s1	=	$	652		s2	=	$	819
Repair Incidents	n1	=		15		n2	=		12

(a) Construct a 95 percent confidence interval for the true difference of the means assuming equal variances. (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 Welch's formula for d.f. (Round the calculation for Welch's df to the nearest integer. 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?
  

• Yes

• No

(d) Construct separate 95% 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=$1,078x¯1=$1,078	($  , $  )
x¯2=$1,786x¯2=$1,786	($  , $  )

Answer 1

Solution

Let X = dollar cost of body repairs for test group and Y = dollar cost of body repairs for control group.

Let (μ₁, σ₁) and (μ₂, σ₂) be the mean and standard deviation of X and Y, σ₁ = σ₂ = σ, say but unknown

Part (a)

95 % Confidence Interval for true difference of the means, i.e., (μ₁ - μ₂) assuming σ₁ = σ₂ = σ, say but unknown is: [- 1290.449 to - 125.551] Answer 1

Details

100(1 - α) % Confidence Interval for (μ₁ - μ₂) assuming σ₁ = σ₂ = σ, say but unknown is:

(Xbar – Ybar) ± MoE,

where

MoE{(t_{2n – 2, α/2})(s)√{(1/n₁) + (1/n₂)}

with

Xbar and Ybar as sample means,

s = pooled sample estimate of σ given by

s = sqrt[{(n₁ – 1)s₁² + (n₂ – 1)s₂²)}/(n₁ + n₂ – 2)];

s₁, s₂ being respective the sample standard deviations;

t_{n1 + n2 – 2, α/2} = upper (α/2) percent point of t-distribution with n₁ + n₂ – 2 degrees of freedom

n₁ , n₂ being the sample sizes.

Calculations

Given	n1	15
	n2	12
	Xbar	1078.0000
	Ybar	1786
	s1	652.0000
	s2	819.0000
	s^2	533193.0800
	s	730.2007
	α	0.05
	tα/2	2.0595
	MoE	582.4489
Lower Bound		-1290.4489
Upper Bound		-125.5511

Part (b)

Under the assumption of unequal variances, only the degrees of freedom for t changes as per Welch’s formula..

95 % Confidence Interval for true difference of the means, i.e., (μ₁ - μ₂) assuming σ₁ = σ₂ = σ, say but unknown is: [- 1296.126 to - 119.874] Answer 2

Details

Welch’s Formula for degrees of freedom:

ν = {(s₁²/n₁) + (s₂²/n₂)}/{(s₁⁴/n₁²x ν₁) + (s₂⁴/n₂²x ν₂)}; ν₁ = n₁ - 1 and ν₂ = n₂ - 1

Calculations

ν-calculation
S1 s1^2/n1	28340.27
S2 s2^2/n2	55896.75
S = S1 + S2	84237.02
ν1	14
ν2	11
F1	3150
F2	1584
D1 s1^4/F1	57369337
D2 s2^4/F2	2.84E+08
D D1 + D2	3.41E+08
ν S^2/D	20.78403
[ν]	21

CI Calculations

Given	n1	15
	n2	12
	Xbar	1078.0000
	Ybar	1786
	s1	652.0000
	s2	819.0000
	s^2	533193.0800
	s	730.2007
	α	0.05
	tα/2	2.0796
	MoE	588.1263
Lower Bound		-1296.1263
Upper Bound		-119.8737

Part (c)

Assumption on population variance does impact the confidence interval leading to widening of the interval.

When the population variances cannot be assumed equal, the width of the confidence interval increases by 12.

So, first Option Answer 3

Part (d)

100(1 - α) % Confidence Interval for population mean μ, when σ is not known is: Xbar ± MoEwhere

MoE = (t_{n- 1, α /2})s/√n

with

Xbar = sample mean,

t_{n – 1, α /2} = upper (α/2)% point of t-distribution with (n - 1) degrees of freedom,

s = sample standard deviation and

n = sample size.

95% Confidence Interval for population mean μ₁: [716.934 to 1439.066] Answer 4

Calculations

Given	α	0.05	1 - (α/2) =	0.975
	n	15	SQRT(n)	3.87298335
	Xbar	1078	n - 1	14
	s	652.0000
tα/2		2.1448
95% CI for μ1		1078	±	361.0656
Lower Bound		716.9344
Upper Bound		1439.0656

95% Confidence Interval for population mean μ₂: [1265.632 to 2306.368] Answer 5

Calculations

Given	α	0.05	1 - (α/2) =	0.975
	n	12	SQRT(n)	3.46410162
	Xbar	1786	n - 1	11
	s	819.0000
tα/2		2.2010
99% CI for μ2		1786	±	520.3678
Lower Bound		1265.6322
Upper Bound		2306.3678

DONE