SOLUTION: 2. The medical Rehabilitation Education Foundation reports that the average cost of rehabilitation forstroke victims is $24,672. To see if the average cost of rehab is different at

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Question 1185582: 2. The medical Rehabilitation Education Foundation reports that the average cost of rehabilitation forstroke victims is $24,672. To see if the average cost of rehab is different at a particular hospital, a researcher selects a random sample of 35 stroke victims at the hospital and find the average cost of their rehab is $25,250. The standard deviation of the population is $3,251. At α= 0.01, can it be concluded that the average cost of stroke rehabilitation at the particular hospital is different from $24,672?a) State thenull and alternativehypotheses.b) Find the critical value from the appropriate table.c) Compute the test value.d) Make the decision to reject or not reject the null hypothesis.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
population average cost of rehabilitation is 24672.
population standard deviation is 3251.

sample size is 35
sample mean is 25250

standard error = standard deviation / square root of sample size = 3251 / sqrt(35) = 549.5193 rounded to 4 decimal places.

sample size is over 30 and population standard deviation is known.
use of z-score is indicated.

here's a guideline on when to use the z-score versus when to use the t-score.

https://math.stackexchange.com/questions/1817980/how-to-know-when-to-use-t-value-or-z-value

use the z-score formula to find the z-score.
the formula is:

z = (x - m) / s

z is the z-score
x is the mean of the sample
m is the mean of the population
s is the standard error

the formula becomes:

z = (25250 - 24672) / 549.5193 = 1.05 rounded to 2 decimal places.

since you want to find out if the mean of the sample is significantly different from the mean of the population, you will use a two-tail significance level.

the two-tail significance level is .01/2 = .005.

the two-tail confidence interval is equal to .99.
the significance level is the tail that is equal to .01 divided between the lower and upper limit of the confidence interval.
that makes is .005 above and .005 below.

the critical z-score would be plus or minus 2.5758.

this was found using the z-score calculator at https://davidmlane.com/hyperstat/z_table.html

a display of the results of using that calculator is shown below.



the area was entered as .01 with the results of finding the z-score that had half of that above the confidence interval and half below.

since the absolute value of the test z-score is significantly less than the absolute value of the critical z-scores, the results of the test are not significant, indicating there is not enough evidence to say that the mean at the hospital is different than the mean at the general population of all hospitals.