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I performed an actual scientific experiment and I was only able to collect 2 data points. I am wondering what kind of statistical method I can use to analyze the data.

I performed a similar experiment before where I was able to collect 4 data points. I calculated the mean, the standard deviation of the population and a confidence interval for this population of N = 4. Could I also calculate the same things for a population of N = 2 ?

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Well, sample mean and sample variance are well-defined for a sample of size $2$. But it would help if you describe a little bit more the context. – user10525 May 24 '12 at 18:03

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Yes you can calculate sample means and standard deviations for samples of size 2 or 4. But they are not going to be very accurate estimates for the population parameters. Estimating or constructing confidence intervals for population parameters is usually the reason to compute them Suppose the population variance is 36. So the standard devation is 6. Say the population mean is 3. For a sample of size two the variance of the sample mean is 36/2=18 and the standard deviation is approximately 4.24. The width of a 95% confidence interval fro the mean would be 4 times the standard deviation. which is about 16.97. so even if you were lucky enough to get a sample estimate very close to 3 the confidence interval would be awfully wide [-5.48, 11.48]. Increasing the sample size to 4 only halves the variance. So the standard deviation is only reduced by a factor of square root of 2 = 1.414. The width of the interval is reduced from 17 to 12. So even if the estimate of the mean is 3 the confidence interval would be [-3,9]. It is impossible to tell that the mean is greater than 0 when the sample size is this small! Also it is not the population that is small it is the sample. If you were sampling from a small finite population the story might be different.

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In this analysis it is unclear what you are assuming to be known about the population and what you are assuming the sample results are. This creates great potential for the argument to be misunderstood. I can follow for a while, being familiar with the ideas, but I'm afraid you lost even me with the $[-14,20]$ interval, whose width is twice the value of $16.97$ you had just claimed for it... – whuber♦ May 24 '12 at 21:03
@whuber I was trying to be very clear. I am especially surprised that this argument thatwas intended to be so simple confused you! I just intended to demonstrate how wide the confidence interval for the mean would be if it is based on just 2 samples. Then l also looked at what the interval might be like based on 4. Which is not much better. I start out with a population mean of 3 and a variance of 36 and hence std of 6. – Michael Chernick May 24 '12 at 22:12
For the first confidence interval I am assuming the best outcome an estimate close to 3 The variance for the mean is 18 and its std is 4.24 so the a 95% confidence interval centered at 3 would be [3-2(4.24), 3+2(4.24)] which is [-5.48, 11.48]. the width of is 16.97 as I stated. My mistake was to use that as the half width. The other result just cuts the width by dividing it by 1.414. I will fix the error. – Michael Chernick May 24 '12 at 22:12
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Maybe I can help by specifying some alternatives I had to consider when parsing your answer: (i) are you assuming a known or unknown population variance? (ii) Ditto for the population mean? (iii) Are you using a z-test, t-test, or something else? (iv) Are you making approximations (as suggested by "approximately") or precise calculations (as indicated by numbers like "16.97")? (v) When you contemplate taking more samples, in what sense can we say the SE of the mean will be halved, given the new sample results change? (vi) You seem to imply a 95% CI won't have 95% coverage. What gives? – whuber♦ May 24 '12 at 22:18
Read my reply and note the correction. It was a simple error. – Michael Chernick May 24 '12 at 22:19

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