Question

he table below contains the overall miles per gallon​ (MPG) of a type of vehicle. Complete parts a and b below.

29

27

23

35

28

20

28

30

29

27

35

29

34

33

a. Construct a 99​% confidence interval estimate for the population mean MPG for this type of​ vehicle, assuming a normal distribution.The 99​% confidence interval estimate is from

MPG to MPG. (MPG) Miles per gallon.

​(Round to one decimal place as​ needed.)

b. interpret the interval of (a)

Answer 1

Solution:

x	x2
29	841
27	729
23	529
35	1225
28	784
20	400
28	784
30	900
29	841
27	729
35	1225
29	841
34	1156
33	1089
x = 407	x2 = 12073

a ) The sample mean is

Mean    = (x / n) )

= ( 29+27+,23,+35,+28,+20,+28,+30,+29,+27,+35,+29,+34,+33/ 14 )

= 407 / 14

= 29.0714

Mean    = 29.1

The sample standard is S

  S = ( x² ) - (( x)² / n ) n -1

= (12073(407 )² / 14 ) 13

   = ( 12073 - 11832.0714 / 13)

= (240.9286 / 13)

= 18.533

= 4.305

The sample standard is 4.3

b ) At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,13 = 3.012

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

= 3.012 * (4.3 / 14)

= 3.5

Margin of error = 3.5

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

- E < < + E

29.1 - 3.5 < < 29.1 + 3.5

25.6< <32.6

(25.6, 32.6 )