Question 1143328
sample mean is 28.2
sample standard deviation is 6.3
population mean is assumed to be 26.


the standard error is equal to the standard deviation of the sample divided by the square root of the sample size.


that makes it equal to 6.3 / sqrt(15) = 1.600833116.


you would probably use a t-score rather than a z-score.


the t-score is equal to (28.2 - 26) / 1.600833116.


that makes it equal to 1.374284414.


the degrees of freedom is equal to the sample size minus 1 = 14.


the area to the left of a t-score of 1.374284414 with 14 degrees of freedom is equal to .9045


at the two tail .01 level of significance, the critical t-score with 14 degrees of freedom is equal to plus or minus 2.977


since the sample t-score is well within these limits, it can be concluded that the sample mean more then likely difference from the assumed population mean due to random variations in the mean of samples of size 15 taken from that population.


alternatively, the p-value of the sample t-score would be (1 - .9045) / 2 = 0.04775.


this is well above the critical p-value of (1 - .99) / 2 = .005, leading to the same conclusion.


the conclusion is is that the population mean cannot be assumed to be different from 26 at the .01 level of significance.


i believe this is correct.
you can test with the critical t-score, or you can test with the critical p-value.
either one will lead you to the same conclusion.