Question

Question 1:

A random sample of   n = 16 mid-sized cars tested for fuel consumption gave a mean of   x ¯ = 22.81 miles per gallon with a standard deviation of   s = 2.08 miles per gallon.

Assuming that the miles per gallon given by all mid-sized cars have a normal distribution, find a 99 % confidence interval for the population mean,   μ .

Round your answers to two decimal places.

Answer 1

Solution :

Given that,

n = 16

   = 22.81

s = 2.08  

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

Also, d.f = n - 1 = 16 - 1 = 15

    =    =  _0.005,15 = 2.947

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

The margin of error is given by

E =  /2,d.f. * ( / n )

   = 2.947* ( 2.08/ 16 )

= 1.532

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

( - E ) <   <  ( + E)

( 22.81 - 1.532 )   <   <  ( 22.81 + 1.532 )

21.28 <   < 24.34

Required. 99% confidence interval is ( 21.28 , 24.34 )