Listed below are the costs (in dollars) of repairing the front ends and rear end

Question

Listed below are the costs (in dollars) of repairing the front ends and rear ends of different cars when they were damaged in controlled low-speed crash tests (based on data from the Insurance Institute for Highway Safety).

Is there sufficient evidence to conclude that there is a difference? Use a 0.05 significance level.

What test would you use in STATDISK to make this decision?

Which distribution should be used for the test statistic and critical value?

What is the value of the test statistic?

Round your answer to three decimal places.

Will the critical value(s) be positive, negative, or one of each?

What is the p-value?

Round your answer to four decimal places.

What is the critical value(s)?

Enter the answer rounded to three decimal places. If the critical value is plus/minus, just enter the positive value. You indicated in the last question that it should be plus/minus.

What is your conclusion?

Construct a 95% confidence interval estimate of the mean difference between the front repair costs and the rear repair costs.

Round your limits to the nearest whole number.

Write a statement that interprets the resulting confidence interval. Make sure your statement addresses the claim and uses the confidence interval.

Toyota Mazda Volvo Saturn Subaru Hyundai Honda VW Nissan Front Repair Cost 936 978 2252 1032 3911 4312 3469 2598 4535 Rear Repair Cost 1480 1202 802 3191 1122 739 2769 3375 1787

Explanation / Answer

PART A. Hypothesis
Given that,
mean(x)=2669.2222
standard deviation , s.d1=1461.3965
number(n1)=9
y(mean)=1829.6667
standard deviation, s.d2 =1023.9707
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.306
since our test is two-tailed
reject Ho, if to < -2.306 OR if to > 2.306
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =2669.2222-1829.6667/sqrt((2135679.73021/9)+(1048515.99446/9))
to =1.4115
| to | =1.4115
critical value
the value of |t | with min (n1-1, n2-1) i.e 8 d.f is 2.306
we got |to| = 1.41147 & | t | = 2.306
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.4115 ) = 0.196
hence value of p0.05 < 0.196,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.4115
critical value: -2.306 , 2.306
decision: do not reject Ho
p-value: 0.196

PART B. Confidence
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 2669.2222-1829.6667) ± t a/2 * sqrt((2135679.73/9)+(1048515.994/9)]
= [ (839.556) ± t a/2 * 594.81]
= [-532.078 , 2211.189]

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