Question 314394: Hello I have quite a lenghty problem. I think I figured it right but I am not sure. I appreciate all the help.
Vehicle speeds at a certain highway location are believed to have approx. a normal distribution with mean=60 mph and standard deviation=6 mph.
Speeds for random sample n=36. Determine the values in the sentence: For samples n=36 there is a 95% chance the vehicle will be between __ and ___.
I have figured it this way 60-2(6)=48 and 60+2(6)=72.
I also figured the standard error to be: 6/sqrt 36=6/6=1
I hoping I'm right because then I know what I am doing.
Thanks,
Donnie
Found 2 solutions by rapaljer, jrfrunner:
Answer by rapaljer(4671)   (Show Source):
Answer by jrfrunner(365)  (Show Source):
You can put this solution on YOUR website! Depends on what you are trying to estimate, the population average (mean) speed of an individual car or the populatin average (mean) speed for samples of size 36.
If we assume the speeds are normally distributed, then
for estimating the average speed of individual cars the formula would be
(Xbar - Z*S, Xbar + Z*S) where S=standard deviation
for 95% confidence interval Z=1.96
(60-1.96*6, 60+1.96*6) or 60-11.8=48.2, 60+11.2-71.2
For estimating the average speed for groups of size 36, in theory you should use the t distribution because your sample is small, and use the Standard Error not the standard deviation. Standard Error = Standard deviation/Sqrt(n)
with n=36 and 95% confidence
t at 0.05/2=0.025 and 35 degrees of freddom =2.03
SE = 6/Sqrt(36) =1
Xbar-t*SE, Xbar + t*SE which is (60-2.03*1, 60+2.03*1) or (57.97, 62.03)
Some people say that if n>30 you can use Z instead of t, but this is not correct since Z and t do not merge until n>200
if you did use Z, then for 95% confidence Z=1.96
(60-1.96*1, 60+1.96*1) or (58.04, 61.96)
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