Probability Q: find margin of error, stand. dev., mean, etc.

[srt927]

A confidence interval estimate for the mean is (3 , 9). If the level of confidence is 0.95 (95%), the sample size is 36, then determine:

a) the maximum margin of error

b) the standard deviation

c) the sample mean

I believe that I should use the Z = (X-bar - mu) over (stand. dev.) over (sqrt[n])

How would I plug the numbers into this equation for this problem and/or am I even using the right formula? I just can't seem to figure this one out.

Thanks in advance

 

[tkhunny]

A few hints?

9-3 = 6

95% CI implies 6 = (1.96*s)*2 -- z-standard deviations both directions.

Solving gives s = 1.531

Sample Size considerations suggest \(\displaystyle \L\,36 = \frac{1.96^{2}*s^{2}}{d^{2}}\)

Where does that leave us?

Note: I really don't like this problem, not that it is difficult or unfair, but that it does not provide sufficient information. In the solution I have suggested, I was required to assume the confidence interval was symmetric. If this is so, the problem statement should tell us. I am not at all comfortable assuming symmetry with such a small sample and no other indications. Some definitions of "robust" will suggest that it is okay to proceed, anyway, but again, the problem statement did not use that term. Did you supply the ENTIRE problem statement?

 

[srt927]

That is the whole question. I was just confused on the formula to use but now I was trying to decipher the formula that you had given and that is not even on the formula samples the book had given me the only one close is

means n= ( Z confidence* Standard Deviation/ E) squared

Is the formula you gave the same or different because you used a "d" or is the letters used irrelevant.

I am just confused

 

[tkhunny]

'd' is the maximum margin of error. It does not matter what it is called. A recent text book I saw called it "m". Look up "sample Size" in your index. The derivation of this formula will show you the relationship to the Confidence Interval.