Question 1061792: The average cost of owning and operating an automobile is $8121 per 15,000 miles including fixed and variable costs. A random survey of 40 automobile owners revealed an average cost of $8350 with a population standard deviation of $750. Is there sufficient evidence to conclude that the average is greater than $8121? use a=.01
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! pm = 8121
sm = 8350
sd = 750
ca = .01
ss = 40
pm is the population mean.
sd is the population standard deviation.
sm is the sample mean.
ss is the sample size.
ca is the critical alpha
if you are looking at a double ended test, then the critical alpha will be .005 on each end.
if you are looking at a single ended test, than the critical alpha will be .01.
since they are looking to see if the sample average is greater than the population average, a single ended test is indicated.
you would look up the z-score for a single ended alpha of .01.
that z-score would be the z-score that give you an area of .99 to the left of it in the z-score table that i used (http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf)
that z-score would be somewhere between a z-score of 2.32 and 2.33.
i got .9901 for a z-score of 2.33 and i got .9898 for a z-score of 2.32.
it's a little bit closer to 2.33.
about two thirds of the way actually.
if you used a calculator to find the z-score, it would show up as something like 2.326347877.
i rounded it off to 2.327.
so my critical z-factor that i settled on was cz = 2.327.
i then found the standard error of the distribution of sample means.
the formula for that is:
se = sd / sqrt(ss) = 750/sqrt(40) = 118.5854
now i want to find the z-score of the sample.
the formula to use for that sz = (sm - pm) / se
replace variables with known values to get:
sz = (8350 - 8121) / 118.5854.
this results in sz = 1.931097757
since sz is less than cz, the results of the test are not statistically significant and you can conclude that the difference between the sample mean and the population mean is more than likely due to normal variations that you would expect to see in sample means of that size.
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