how to compute p-value when the confidence interval is given? Please help on this problem...This is the statement:
"The immunological assay verified the presence of RuBisCO in non-treated (control) and Pb-induced leaves with average relative band intensities of 0.156 ± 0.012 and 0.128 ± 0.013 respectively" My professor told me that the p-value is 0.0518. He have not seen the raw data but he gives me the p-value base on the interval i have given.While I do the computation using the excel I get 0.00114 only.. I s it possible to compute the p- value using the given interval only? Thank you..
 A: If the sample sizes and confidence levels (95%?) are known, then the sample means and variances of the two separate groups can be deduced. From there a two-sample test can be performed and the p-value determined. 
Here is an outline of the procedure: The standard formula for a 95% confidence interval (CI) is
$\bar X \pm t^*S/\sqrt{n}.$ So the CI for the first group is based on sample mean $\bar X_1 = .0156$ and sample standard deviation 
$S_1 = .012\sqrt{n}/t^*,$ where $t^*$ cuts 2.5% from the upper
tail of Student's t distribution with $n-1$ degrees of
freedom, a number that can be found from a printed "t-table"
or from software.  Similarly, you could find $\bar X_2$ and $S_2$
for the second sample. Finally, use $\bar X_1,\, \bar X_2,\,S_1,\, S_2$ and the two sample sizes to compute then T-statistic for a
2-sample t test, and determine its p-value. 
If "band intensity" does not mean "confidence interval," then you'd need to know the formula for getting the band intensities to see if sample mean and variance can be deduced from them.
I do not see a reliable way just to look at two confidence levels and convert those to a p-value. You do not say how you (or Excel) found your
p-value of .00114. I have a hunch that may not be right, but can't
say for sure.
However, there is a $clue$ that your
professor's p-value may be approximately correct:
(a) $.156 - .012 = .144$ and $.129 + .013 = .141$ so the confidence
intervals do not quite overlap. The confidence intervals each have
a 5% 'error probability' so the error probability of this
inference may be more than 5%. It's a 'close call' whether
the two groups are significantly different. (b) By comparison, a p-value of 0.0518
is very near the 5% significance level; another 'close call'
as to significance. (c) Roughly speaking from either (a) or (b), I'd say there may be weak evidence that the two groups have different population means,  
