Question 922878: Dylan Jones kept careful records of the fuel efficiency of his new car. After the first thirteen times he filled up the tank, he found the mean was 25.6 miles per gallon (mpg) with a sample standard deviation of .9 mpg.
(a)
Compute the 95 percent confidence interval for his mpg. (Round your answers to 3 decimal places.)
The confidence interval is between ... and ....
(b)
How many times should he fill his gas tank to obtain a margin of error below 0.2 mpg? (Round your answer to the nearest whole number.)
Number of times
If you don't mind, please show your work. I will greatly appreciate it!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Dylan Jones kept careful records of the fuel efficiency of his new car. After the first thirteen times he filled up the tank, he found the mean was 25.6 miles per gallon (mpg) with a sample standard deviation of .9 mpg.
(a)
Compute the 95 percent confidence interval for his mpg. (Round your answers to 3 decimal places.)
----
CI:: x-bar - z(s/sqrt(n)) < u < x-bar + z(s/sqrt(n)
x-bar = 25.6
ME = z*s/sqrt(n)
ME = 1.96*0.9 = 1.276
Note:: s/sqrt(n) is the standard deviation of the sample means
You are told that the sample standard deviation is 0.9
---
Also, z is the z-value associated with a 2-tail test when alpha = 0.95
That value is invNorm(0.975) = 1.96
If you use a z-chart, find the z-value with 0.475 to the right of the mean.
---
95%CI:: 25.6-1.76 < u < 25.6+1.76
95%CI:: 23.84 < u < 27.36
-------------------------------------------
b)
How many times should he fill his gas tank to obtain a margin of error below 0.2 mpg? (Round your answer to the nearest whole number.)
s/sqrt(15) = 0.9
s = 3.486
----------------------
ME = z*s/sqrt(n)
Solve 1.96*3.486/sqrt(n) < 0.2
-----
sqrt(n) > 1.96*3.486/0.2
sqrt(n) > 34.16
n > 1167 times
--------------------
Cheers,
Stan H.
|
|
|
