Yes it does.
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n-1
Yes. The parameters of the t distribution are mean, variance and the degree of freedom. The degree of freedom is equal to n-1, where n is the sample size. As a rule of thumb, above a sample size of 100, the degrees of freedom will be insignificant and can be ignored, by using the normal distribution. Some textbooks state that above 30, the degrees of freedom can be ignored.
Given Z~N(0,1), Z^2 follows χ_1^2 Chi-square Probability Distribution with one degree of freedom Given Z_i~N(0,1), ∑_(i=1)^ν▒Z_i^2 follows χ_ν^2 Chi-square Probability Distribution with ν degree of freedom Given E_ij=n×p_ij=(r_i×c_j)/n, U=∑_(∀i,j)▒(O_ij-E_ij )^2/E_ij follows χ_((r-1)(c-1))^2 Chi-square Probability Distribution with ν=(r-1)(c-1) degree of freedom Given E_i=n×p_i, U=∑_(i=1)^m▒(O_i-E_j )^2/E_i follows χ_(m-1)^2 Chi-square Probability Distribution with ν=m-1 degree of freedom
The t-distribution and the normal distribution are not exactly the same. The t-distribution is approximately normal, but since the sample size is so small, it is not exact. But n increases (sample size), degrees of freedom also increase (remember, df = n - 1) and the distribution of t becomes closer and closer to a normal distribution. Check out this picture for a visual explanation: http://www.uwsp.edu/PSYCH/stat/10/Image87.gif
See: http://en.wikipedia.org/wiki/Confidence_interval Includes a worked out example for the confidence interval of the mean of a distribution. In general, confidence intervals are calculated from the sampling distribution of a statistic. If "n" independent random variables are summed (as in the calculation of a mean), then their sampling distribution will be the t distribution with n-1 degrees of freedom.