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 /
= 3.012 * (1863.5 /
= 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 )
orchestra answered 2 years agoRelated Solutions
he table below contains the overall miles per gallon (MPG) of a type of vehicle. Complete...
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...
The table below contains the overall miles per gallon (MPG) of a type of vehicle. 28...
The table below contains the overall miles per gallon (MPG) of
a type of vehicle.
28
29
24
22
29
32
23
25
35
32
35
23
26
34
Construct a 95% confidence interval estimate for the population
mean MPG for this type of vehicle, assuming a normal
distribution.The 95% confidence interval estimate is from ____ MPG
to ____ MPG.
(Round to one decimal place as needed.)
The file Sedans contains the overall miles per gallon (MPG) of 2013 midsized sedans: 38 26...
The file Sedans contains the overall miles per gallon
(MPG) of 2013 midsized sedans:
38 26 30 26 25 27 22 27 39 24 24 26 25
23 25 26 31 26 37 22 29 25 33 21 21
Source: Data extracted from “Ratings,” Consumer Reports,
April
2013, pp. 30–31.
a. Compute the mean, median, and mode.
b. Compute the variance, standard deviation, range, coefficient
of
variation, and Z scores.
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The file Sedans contains the overall miles per gallon (MPG) of 2013 midsized sedans: 38,26,30,26,25,27,22,27,39,24,24,26,25,23,25,26,31,26,37,22,29,25,33,21,21 Source:...
The file Sedans contains the overall miles per gallon (MPG) of
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38,26,30,26,25,27,22,27,39,24,24,26,25,23,25,26,31,26,37,22,29,25,33,21,21
Source: Data extracted from "ratings," Consumer
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MPG
Frequency
7 - 12
57
13 - 18
177
19 - 24
167
25 - 30
77
31 - 36
40
37 - 42
12
(a) Identify the following
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class boundaries (enter as a comma-separated list)
class width
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