How to read this in easy to understand language

MissSuzy

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Jan 26, 2013
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I'm working on a homework problem and have trouble deciphering this statement. Can you help in easy to understand terms?

Beta = -2.48, 95% Confidence Interval: -3.68 to -1.29, p < .0.001

Thank you.;-)
 
I'm working on a homework problem and have trouble deciphering this statement. Can you help in easy to understand terms?

Beta = -2.48, 95% Confidence Interval: -3.68 to -1.29, p < .0.001

Thank you.;-)

Looks like you are taking an advanced level statistics class.

What does your text-book say about the definition of:

confidence level
Beta
p

What does your class-notes say?

Did you try to google those words in conjunction with statistics?

Are you taking an on-line course?
 
I'm working on a homework problem and have trouble deciphering this statement. Can you help in easy to understand terms?

Beta = -2.48, 95% Confidence Interval: -3.68 to -1.29, p < .0.001

Thank you.;-)
I would take that to mean that the mean of some distribution is -2.48, and 95% of the distribution lies between -3.68 and -1.29 .. but that leaves me without knowing what p is.

What is the context? what kinds of distributions have you been dealing with? We need to know what the sticking points are before we can help effectively.
 
I'm working on a homework problem and have trouble deciphering this statement. Can you help in easy to understand terms?

Beta = -2.48, 95% Confidence Interval: -3.68 to -1.29, p < .0.001

Thank you.;-)

This class is a college research class which was an intensive. I had statistics historically. I'm not confident in my analysis yet of statistical problems. The statement reads this exactly, "From multiple regression, aromatherapy was found to be associated with the changes in leg cramp levels. (first day: Beta = -2.48, 95% Confidence Interval: -3.68 to -1.29, p < .0.001). The problem goes on to continue by each day after application of aromatherapy. I did look up the terms but again felt unconfident. We don't have class notes for this advanced statistics question. Thank you.
 
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