0
$\begingroup$

This question already has an answer here:

I have two samples of $n=10$ with the following values

Sample 1:

  • Mean = $3$
  • $s_{d} = 0.4$

Sample 2:

  • Mean = $3.35$
  • $s_{d} = 0.3$

but the observations of both samples are unknown and I want to know the common $s_{d}$ to calculate $t_{0}$


Note: this is no homework is only an exercise from the tutorial, which has not been solved.

$\endgroup$

marked as duplicate by whuber Dec 1 '13 at 21:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ is this homework? $\endgroup$ – John Nov 30 '13 at 19:09
  • $\begingroup$ no it's normal exercise $\endgroup$ – Ben Ishak Nov 30 '13 at 19:33
  • $\begingroup$ What is this $t_0$? $\endgroup$ – Glen_b Dec 1 '13 at 0:56
  • $\begingroup$ $t_{0} = (d- µ_{0}) / (s_{d} / \sqrt{n})$ for paired t-test $\endgroup$ – Ben Ishak Dec 1 '13 at 14:21
0
$\begingroup$

i may suggest the following solution

$$ s_{d_{1}-d_{2}} = \sqrt{1/_n * (s_{d1}^2 + s_d{2}^2)} = \sqrt{1/_{10} * (0.4^2 + 0.3^2)} = 0.158 $$

$$ t = d_{1} - d_{2} / s_{d_{1}-d_{2}} = -0.35 / 0.158 = -2.215 $$

source: http://en.wikipedia.org/wiki/Student's_t-test#Equal_or_Unequal_sample_sizes.2C_unequal_variances

$\endgroup$
  • $\begingroup$ This unfortunately is incorrect. The duplicate question shows how to combine the SDs into a pooled SD, because their squares are variances and that thread concerns combining variances. $\endgroup$ – whuber Dec 1 '13 at 21:57
  • $\begingroup$ i still not sure if my solution was correct, but the thread that you sepposed is not the same as my problem, as he has 2 matrices with all values $(x_{i}, y_{i})$ but i have only the SDs and the means $\endgroup$ – Ben Ishak Dec 2 '13 at 17:10
  • $\begingroup$ Your problem is identical to that one, but happens to be much simpler: the squares of your SDs are one-by-one covariance matrices and your means are the means of vectors of dimension 1. $\endgroup$ – whuber Dec 2 '13 at 17:12
  • $\begingroup$ [this][1] what i was looking for [1] = en.wikipedia.org/wiki/… $\endgroup$ – Ben Ishak Jan 18 '14 at 14:44

Not the answer you're looking for? Browse other questions tagged or ask your own question.