Question

In a crash test of 26 minivans of a Japanese manufacturer, collision repair costs are found to have a distribution that is roughly bell shaped, with a mean of $1850 and a standard deviation of $340. Construct a 95% confidence interval for the mean repair cost in all such vehicle collisions by filling in the following blanks (Round confidence limits to the nearest dollar). (Please provide clear explanation and steps and calculations)

a) df =

b) Critical value =

c) Point Estimate =

d) Standard Error of the Estimate =

e) Margin of Error =

f) Lower Confidence Limit =

g) Upper Confidence Limit =

Answer 1

Solution :

Given that,

n = 26

c) = 1850

s = 340

Note that, Population standard deviation() is unknown. So we use t distribution.

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

a) Also, d.f = n - 1 = 25

b)     =    =  0.025,25 = 2.060

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.060* (340 / 26 )

= 137.329

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

( 1850 - 137.329 )   <   <  ( 1850+ 137.329 )

1712.671 <   < 1987.329

Required 95% confidence interval is ( 1712.671 , 1987.329 )

Lower Confidence Limit = 1712.671

Upper Confidence Limit =  1987.329