Question 1143328: A simple random sample of size n=15 is drawn from a population that is normally distributed. The sample mean is found to be x overbar=28.2 and the sample standard deviation is found to be s=6.3. Determine if the population mean is different from 26 at the α=0.01 level of significance.
I need help with this next section
(b) Calculate the P-value.
P-value =
(Round to three decimal places as needed.)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 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.
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