I understand that in the one sample case, z is employed where population variance is known and t for where it is unknown.

In the two sample case: we are trying to test $H_0: \mu_x=\mu_y$ against $H_1: \mu_x\not= \mu_y$, for finite variance; $\sigma^2_x$ and $\sigma^2_y$; while $n_x$ and $n_y$ are respective sample sizes

What test would we use if:

  1. (both variance are equal) AND either

    • (i) both population variance unknown (i.e. we use sample variance $S_x=S_y$)
    • (ii) both population variance known ($\sigma^2_x=\sigma^2_y$ )
  2. (variance not equal) AND either

    • (i) both population variance unknown (i.e. we use sample variance, $S^2_x\not=S^2_y$)
    • (ii) both population variance known ($\sigma^2_x \not=\sigma^2_y$ )
    • (iii) one population variance known other unknown (i.e. given $\sigma^2_x$ and $ S^2_y$)

My answers:
1. (i) use 2 sample t-test with $n_x+n_y-2$ degrees of freedom and variance: $S_p*\frac{n_x+n_y}{n_xn_y}$ (where $S_p$ is the pooled variance )
1. (ii) use 2 sample z-test with variance: $(\frac{\sigma^2_x}{n_x}+\frac{\sigma^2_y}{n_y})$
2. (i) no idea :(
2. (ii) same as 1. (ii)
2. (iii) no idea :(


1 Answer 1


In the unequal variance case with both variance unknown (normal distributions assumed throughout) Welch's test is used. This is called the Behrens-Fisher problem. The test statistic under the null hypothesis has an approximate t distribution with degrees's of freedom given by a formula where the df does not have to be integer. It is called the Satterthwaite approximation. In the case where both population variances are known you can normalize by the formula for the variance of the difference of two sample means and the use the z test.

I have never seen the case where one variance is known and the other is not. You could be conservative and use Welch's test but it may be better to find the exact null distribution when one variance is estimated and the other is a known constant.

  • $\begingroup$ Thank you @chl for enhancing my answer with two well-chosen links. $\endgroup$ Oct 1, 2012 at 16:26

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