Question

In: Statistics and Probability

You randomly select 20 cars of the same model that were sold at a car dealership...

You randomly select 20 cars of the same model that were sold at a car dealership and determine the number of days each car sat on the dealership’s lot before it was sold. The sample mean is 9.75 days, with the sample standard deviation of 2.39 days. Construct a 99% confidence interval for the population mean number of days the car model sits on the dealership’s lot. Assume the days on the lot is normally distributed.

Expert Solution

Solution :

Given that,

=9.75

s = 2.39

n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t_/2,df = t_0.005,19 =2.861

Margin of error = E = t_/2,df * (s / $\sqrt$ n)

= 2.861* ( 2.39 / $\sqrt$ 20)

= 1.53

Margin of error = 1.53

The 99% confidence interval estimate of the population mean is,

- E < < + E

9.75 - 1.53 < < 9.75 + 1.53

8.22 < < 11.28

(8.22,11.28)

orchestra answered 1 year ago

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