Question

In: Statistics and Probability

An online used car company sells second-hand cars. For 40 randomly selected transactions, the mean price...

An online used car company sells second-hand cars. For 40 randomly selected transactions, the mean price is 2800 dollars. Assuming a population standard deviation transaction prices of 140 dollars, obtain a 99% confidence interval for the mean price of all transactions. Please carry at least three decimal places in intermediate steps. Give your final answer to the nearest two decimal places.

Confidence interval: ( , ).

Solutions

Expert Solution

Solution =

Given,

= 2800

= 140

n = 40  

Note that, Population standard deviation() is known..So we use z distribution. Our aim is to construct 99% confidence interval.

c = 0.99

= 1- c = 1- 0.99 = 0.01

  /2 = 0.01 2 = 0.005 and 1- /2 = 0.995

Search the probability 0.995 in the Z table and see corresponding z value

= 2.576   

The margin of error is given by

E =  /2 * ( / n )

= 2.576 * (140 / 40 )

= 57.022

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

( - E ) <   <  ( + E)

( 2800 - 57.022 )   <   <  ( 2800 + 57.022 )

2742.98 <   < 2857.02

Required 99% confidence interval is ( 2742.98 , 2857.02 )

orchestra answered 2 years ago
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