Question 1070240
it appears that the assumption is correct.
here's why i think so.


population mean = 40
sample mean = 44.5
sample standard deviation = 12
sample size = 36
significance level = .05 on a 2-tailed test.
degrees of freedom = 36 - 1 = 35.
critical t-factor with 35 degrees of freedom = 1 - .05/2 = .975
standard error = standard deviation of the distribution of sample means = 12/sqrt(36) = 12/6 = 2


the critical t-factor with 35 degrees of freedom will be at t = 2.03.


the t-score with 35 degrees of freedom will be at (sm-pm)/se = (44.5-40)/2 = 4.5/2 = 2.25.


sm = sample mean
pm = population mean
se = standard error = sddsm = standard deviation of the distribution of sample means.


the standard error is based on the fact that, the larger the sample size, the smaller the standard deviation of the distribution of sample means will be.


this can be seen from the formula.
se = ssd/sqrt(ss)


se = standard error
ssd = sample standard deviation.
ss = sample size.


assuming the sample standard deviation = 12, .....
with a sample size of 1, the standard error will be 12/sqrt(1) = 12
with a sample size of 36, the standard error will be 12/6 = 2.
with a sample size of 625, the standard error will be 12/25 = .48.


a t-score with 35 degrees of freedom greater than the critical t-score of 2.03 indicates that the results of this sample are statistically significant since the probability of getting a t-score greater than 2.03 is less than .025.