Question 1173009: it is claimed that people drive on the average 18.000 km per year. A consumer form believes that the average mileage is probably lower. to check, the consumer firm abtained information from 40 randomly selected car owners that resulted in a sample mean of 17, 463 km with a sample standard deviation of 1348 km what can we conclude about this claim? use alpha
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! claim is 18,000 km per year.
sample of 40 randomly selected car owners results in a sample mean of 17,463 km with a sample standard deviation of 1348 km.
standard error = standard deviation of sample divided by square root of sample size = 1348 / sqrt(40) = 213.375143
since the standard deviation is taken from the sample, then you use t-score.
t-score = (x - m) / s
x is the raw score
m is the mean
s is the standard error.
you get:
t-score = (17,463 - 18,000) / 213.375143 = -2.515584261.
area to the left of that under the normal distribution curve is equal to .0080345273.
at 99% confidence level, the area to the left is less than .01.
this means the results are significant.
this means that the claim that the average mileage is probably lower is valid at the 99% confidence level.
with t-score, you need to use degrees of freedom.
with a sample size of 40, the degrees of freedom is 39.
i used the ti-84 plus to solve for the area to the left of the indicated z-score.
an online calculator that does the same can be found at https://stattrek.com/online-calculator/t-distribution.aspx
it provided the same answer rounded to 4 decimal places.
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