SOLUTION: Use the standard normal distribution or the​ t-distribution to construct a
98% confidence interval for the population mean.
The gas mileages​ (in miles per​ g
Algebra.Com
Question 980413: Use the standard normal distribution or the t-distribution to construct a
98% confidence interval for the population mean.
The gas mileages (in miles per gallon) of 45 randomly selected cars are listed below.
2121
2424
1616
2727
1313
2323
1818
1919
1919
1515
1616
1414
2424
1717
2727
2222
2222
1818
1414
1515
1313
1919
1818
2727
2222
2222
1515
1515
2626
1414
2020
1919
2525
2626
1414
2222
2626
1818
1616
1919
2121
2727
2424
2424
2020
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
The mpg data values are in red. The other numbers in black are row numbers to keep track of all of the data in a neater form (compared to one long column of 45 values).
| Mileage (mpg) |
|---|
| 1 | 21 | 22 | 20 |
| 2 | 24 | 22 | 19 |
| 3 | 16 | 18 | 25 |
| 4 | 27 | 14 | 26 |
| 5 | 13 | 15 | 14 |
| 6 | 23 | 13 | 22 |
| 7 | 18 | 19 | 26 |
| 8 | 19 | 18 | 18 |
| 9 | 19 | 27 | 16 |
| 10 | 15 | 22 | 19 |
| 11 | 16 | 22 | 21 |
| 12 | 14 | 15 | 27 |
| 13 | 24 | 15 | 24 |
| 14 | 17 | 26 | 24 |
| 15 | 27 | 14 | 20 |
n = 45 is the sample size. Since n > 30, we can use a standard normal distribution instead of the T distribution.
Use a calculator to find
sample mean: xbar = 19.911111
sample standard deviation: s = 4.378783
critical value: z = 2.326 (at 98% confidence)
Optionally, you can use a table like this one to find the critical z value. Look in the bottom "Z" row and above the "98%" confidence level to find z = 2.326
----------------------------------------------------------------------
The confidence interval is of the form (L,U) where L is the lower limit and U is the upper limit.
Lower Limit (L):
L = xbar - z*s/sqrt(n)
L = 19.911111 - 2.326*4.378783/sqrt(45)
L = 19.911111 - 2.326*0.652750429781364
L = 19.911111 - 1.51829749967145
L = 18.3928135003285
L = 18.39
Upper Limit (U):
U = xbar + z*s/sqrt(n)
U = 19.911111 + 2.326*4.378783/sqrt(45)
U = 19.911111 + 2.326*0.652750429781364
U = 19.911111 + 1.51829749967145
U = 21.4294084996715
U = 21.43
The 98% confidence interval for mu is (L,U) = (18.39, 21.43) (rounded to 2 decimal places)
Side Note: the margin of error is approximately 1.51829749967145
RELATED QUESTIONS
another questions i need help with:
In a random sample of 42 bolts, the mean length... (answered by stanbon)
use a normal distribution
or a t-distribution to construct a 95% confidence interval... (answered by Boreal)
Construct the indicated confidence interval for the population mean μ using a (answered by Boreal)
In a random sample of 23 people, the mean commute time to work was 34.4 minutes... (answered by Boreal)
In a random sample of 2727 people, the mean commute time to work was 33.233.2... (answered by ikleyn)
In a random sample of 29 people, the mean commute time to work was 34.2 minutes... (answered by Boreal)
Use the given degree of confidence and sample data to find a confidence interval for the... (answered by ewatrrr)
In a random sample of 28 people, the mean commute time to work was 31.2 minutes and the... (answered by Boreal)
a) determine which distribution you would use to determine the critical value (normal... (answered by Boreal)