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I am having problems in this course with following the professor. Can anyone help me with this problem Thanks

need to calculate 99% CI for true mean with the following data:
n=100, mean is 126 with sample variance of 402. I have to know the answer, theorems and interpret tne meaning...UGGGGG

need to calculate 99% CI for true mean with the following data:
n=100, mean is 126 with sample variance of 402. I have to know the answer, theorems and interpret tne meaning...UGGGGG

Using the normal distribution as your parameter, the confidence interval (CI) of 99.8% is 126 �} 3 standard deviations (SD). For 99% CI, it would be �} 2.575 SD. (These values can be found in the table in the back of most statistics texts labeled something areas under the normal distribution.)

SD^2 = variance = 402

However, in dealing with means, the Standard Error (SE) of the3 mean is used rather than the SD, but it cuts off the same percentages in the normal distribution.

SE = SD/(Sq.rt.of n)

You should be able to work the specific problem from this.

For the theorems, I searched Google under the key words "confidence intervals" to get these possible sources:

http://en.wikipedia.org/wiki/Confidence_interval
http://www.cas.lancs.ac.uk/glossary_v1.1/confint.html
http://davidmlane.com/hyperstat/confidence_intervals.html
http://www.tufts.edu/~gdallal/ci.htm
http://people.hofstra.edu/faculty/stefan_Waner/realworld/finitetopic1/confint.html

I hope this helps. Thanks for asking.

thank you so much, I may have to post some additional questions and I REALLY appreciate your helPp
Grace

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