When you use a paired T-test, you are essentially doing a one-sample test, where your one sample consists of the paired differences between outcomes in two groups. If you create a new
A couple decides to keep having children until they have the same number of boys and girls, and then stop. Assume they never have twins, that the "trials" are independent with probability
The information that at least one is a boy, however that has been decided to make that statement, does certainly exclude the probability of two girls. The information about the day is seemingly not important)
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Probability of having 2 girls and probability of having at least one girl Ask Question Asked 8 years, 7 months ago Modified 8 years, 7 months ago
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Considering the population of girls with tastes disorders, I do a binomial test with number of success k = 7, number of trials n = 8, and probability of success p = 0.5, to test my null hypothesis
Expected girls from one couple$ {}=0.5cdot1 + 0.25cdot1 =0.75$ Expected boys from one couple$ {}=0.25cdot1 + 0.25cdot2 =0.75$ 1 As I said this works for any reasonable rule that
Use standard type for Greek letters, subscripts and superscripts that function as identifiers (i.e., are not variables, as in the subscript “girls” in the example that follows), and
3 "Given that boys'' heights are distributed normally $mathcal {N} (68$ inches, $4.5$ inches$)$ and girls are distributed $mathcal {N} (62$ inches, $3.2$ inches$)$, what is the probability that a girl chosen
1st 2nd boy girl boy seen boy boy boy seen girl boy The net effect is that even if I don''t know which one is definitely a boy, the other child can only be a girl or a boy and that is always and
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