In: Statistics and Probability
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)
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 )