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:
(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 ($\sigma^2_x=\sigma^2_y$ )
(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 :(