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New member
Oct 24, 2009
Okay I administered a certain aptitude test to a random sample of 9 students in my high school, and the average score is 105. I want to determine the mean ? of the population of all students in the school. Assume a standard deviation of ? = 15 for the test.

What is the margin of error?
What would be the interval for a 98% confidence interval?
Write a sentence explaining the 98% confidence interval in context of the question.

I have made no progress toward the problem i got asked this question and was totally in the dark but i usually help my friend with his class so i felt inclined to try and find him an answer can you help. Dont worry about me cheating and giving him the answer i never do just explain whats happing here what is ? and ? and what are the answer to those three question i asked. If i get all that I think i can help him.


Super Moderator
Staff member
Sep 28, 2005
Since you have a small sample, use the t-distribution.

For the margin of error, look up a 98% confidence interval in the t table with degrees of freedom equal to 8.

\(\displaystyle 2.896\cdot \frac{15}{\sqrt{9}}=14.48\)

Since the sample mean is 105, we add and subtract this result from 105 to get the C.I.

\(\displaystyle \text{Confidence Interval}=90.52<{\mu}<119.48\)

This means we can say with 98% confidence that the population mean test scores will fall between 90.52 and 119.48.


New member
Nov 9, 2009
I need toknow how to make a frequency ploygon, please help and explain.