Question 1115747: A career counselor claims that the mean annual salary of athletic trainers in New Mexico and Arizona is the same. The mean annual salary of 40 randomly selected athletic trainers in New Mexico is $35,630. Assume the population standard deviation is $4800. The mean annual salary of 35 randomly selected athletic trainers in Arizona is $39,440. Assume the population standard deviation is $6200. At α = 0.1, is there enough evidence to reject the counselor’s claim? (Adapted from U.S. Bureau of Labor Statistics)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you've got 2 populations.
the population of new mexico and the population of arizona.
you took a sample of 40 trainers in mexico and got a mean of 35,630 with a mexico population standard deviation of 4800.
you took a sample of 35 trainers in arizona and got a mean of 39,440 with an arizona population standard deviation of 6200.
the career counselor claims the salaries are the same.
the null hypothesis therefore assumes they are the same.
the alternate hypothesis therefore assumes they are different.
since the standard deviations are from each population, then the following two sample z-test is appropriate.
the formula will be:
z = ((x1-x2) - (m1-m2)) / sqrt(sd1^2/n1 + sd2^2/n2), where:
x1 = mean of mexico sample = 35630
x2 = mean of arizona sample 39440
m1 = assumed mean of mexico population = unknown
m2 = assumed mean of arizona population = unknown
sd1 = standard deviation of mexico population = 4800
sd2 = standard deviation of arizona population = 6200
n1 = sample size of mexico sample = 40
n2 = sample size of arizona sample = 35
since the mean of both populations is assumed to be the same, then m1 - m2 = 0 and you don't really need to know what m1 is or what m2 is.
the formula of:
z = (x1-x1) - (m1-m2) / sqrt([sd1^2/n1 + sd2^2/n2) becomes:
z = ((35630 - 39440) - 0) / sqrt(4800^2/40 + 6200^2/35)
simplify this to get:
z = -3810 / sqrt(1674285.714)
this result in z = -2.94449072
at alpha = .01, the 2 sided alpha will be .01/2 = .005, because the stated alpha needs to be split between the low side of the confidence interval and the high side of the confidence interval.
this results in a confidence interval of .005 to .995.
your low end critical z-score will have an area under the normal distribution curve of .005 to the left of it and your high end critical z-score will have an area under the normal distribution curve of .995 to the left of it.
this results in a low end z-score of -2.575829303 and end critical z-score of 2.575829303.
the z-score of your analysis is -2.94449072 which is outside of the confidence limits.
therefore you would reject the null hypothesis and conclude that it is more then likely that the average salary of mexico and arizona are not the same.
the reference for this test was found at https://www.cliffsnotes.com/study-guides/statistics/univariate-inferential-tests/two-sample-z-test-for-comparing-two-means
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