Formula for confidence intervals for small samples and unknown population standard deviation

Question

I have consulted two texts on how to calculate to calculate confidence intervals when N is small and the population standard deviation is unknown. There are some differences in the formulas they give and the end result varies depending on which text I follow (although not by a large amount). Text one says:

1. Calculate the mean
2. Calculate the standard deviation using the formula: s= √ ((∑ X(squared)/N)–X-bar)
3. Calculate the standard error of the mean using the formula: s/√ N-1
4. Determine the value of T from the t-table
5. Obtain the margin of error by multiplying the standard error of the mean by multiplying it by the value obtained in step 4.
6. Add and subtract this product from the sample mean to obtain the C.I.

Steps, 1,4,5& 6 are exactly the same in the second text. However it gives different formulas for steps 2 & 3. It says:

1. Calculate the standard deviation using the formula: s= √ ((∑ X(squared) /N-1) –X-bar). The difference is that they reduce N by one.
2. Calculate the standard error of the mean using the formula: s/√ N The difference is that N is not reduced by 1.

Can anyone explain why the different formulas are used and why?

Thanks. Anne S

Answer 1

Here are some good notes on standard deviation and the standard error of the mean here.

The Wackerly et al text computes small sample confidence intervals in section 8.8 (page 430) you can see their formula here.

Confidence interval: $\bar{Y} \pm t_{\alpha/2} * \frac{S}{\sqrt{n}}$

Where $\bar{y}$ = $\frac{1}{n}$$\sum{\textstyle y_{i}}$ (the sample mean)

and $S = \sqrt{\frac{1}{n-1}\sum(y_{i}-\bar{y})^2}$ (the sample standard deviation)

t$_{\alpha/2}$ is the critical value for a given value of $\alpha$ (e.g., .1, .05, etc.) and has n-1 degrees of freedom, where n is the sample size, that you'd find in a table.

Now if your sample is a large proportion of a known finite population size there is something called a population correction factor, but for basic needs you probably don't have to worry about this.