Question

In: Statistics and Probability

The table below contains the overall miles per gallon​ (MPG) of a type of vehicle. Complete...

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

2929

2727

2323

3535

2828

2020

2828

3030

2929

2727

3535

2929

3434

3333

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.

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

b. Interpret the interval constructed in​ (a)

Solutions

Expert Solution

Solution:

x x2
2929 8579041
2727 7436529
2323 5396329
3535 12496225
2828 7997584
2020 4080400
2828 7997584
3030 9180900
2929 8579041
2273 5166529
5352 28643904
9293 86359849
4343 18861649
3333 11108889
x = 49753 x2 = 221884453

a ) The sample mean is

Mean   = (x / n) )

= ( 2929 + 2727 + 2323 + 3535 + 2828 + 2020 + 2828 + 3030 + 2929 + 2727 + 3535 + 2929 + 3434 + 3333 / 14 )

= 49753/ 14

= 3553.0714

Mean   = 3553.1

The sample standard is S

  S = ( x2 ) - (( x)2 / n ) n -1

= (221884453 ( 49753 )2 / 14 ) 13

   = ( 221884453 - 176740432.0714 / 13)

= (45144020.9286 / 913)

= 3472616.9945

= 1863.4959

The sample standard is 1863.5

b ) At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,13 = 3.012

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

= 3.012 * (1863.5 / 14)

= 1500.1

Margin of error = 1500.1

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

- E < < + E

3553.1 - 1500.1 < < 3553.1 + 1500.1

20523.0< < 5053.2

(20523.0, 5053.2 )


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