hello
your answer is true but you should put z (.05/2) in two equation
z (.05/2) =1.96
now you can solve two equation and will get mean and sd
Is there anyway to calculate mean and standard deviation if we are given the 95% confidence interval?
I have the following interval 1.339-13.833
I know that the sample size is 128.
I know that the way to calculate a two-tailed confidence interval is
sample mean +/- Z(alpha/2) * s/sqrt(n)
If I put in
1.330=xbar - z (.05/2) * s/sqrt (128)
13.833 = xbar + z (.05/2) * s/sqrt (128)
How do I solve for mean and sd at the same time?
Is this problem even solvable?
Thanks!
hello
your answer is true but you should put z (.05/2) in two equation
z (.05/2) =1.96
now you can solve two equation and will get mean and sd
As elnaz said, you can solve this set of equations for s and xbar.
Solve one of the equations for xbar:
xbar = 13.833 - 1.96*s/sqrt(128)
substitute this xbar into the second equation:
1.330 = (13.833 - 1.96*s/sqrt(128)) - 1.96*s/sqrt(128)
solve this equation for s:
s = (13.833-1.330)*sqrt(128)/(2*1.96) = 36.0855
plug s into either equation and solve for xbar to get:
xbar = 13.833 - 1.96*(36.0855)/sqrt(128) = 7.583
~Matt
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