Wikipedia:Reference desk/Archives/Mathematics/2018 August 21

= August 21 =

Figuring out the proportion of Chinese mothers that abort due to female fetuses
Null Hypothesis H0 is mothers do not abort due to sex of fetus which results in a birth ratio of 105 boys to 100 girls

Alt Hypothesis H1 is some mothers do abort due to the sex of the fetus being a girl which results in a birth ratio of 111 boys to 100 girls.

There are 8654 births given by mothers who are born in China (it’s the mum who is born in China, not the baby). Of these, 4553 babies are boys and 4101 babies are girls.

Question 1: What is the probability that these results came from H0 being true?

Question 2: What is the probability that H0 is true given we have these results? Assume that either H0 is true or H1 is true.

Question 3: Are the numerical values the same for Q1 and Q2?

Question 4: How many mothers (what proportion of mothers) choose to abort the fetus if it is a girl? Note that mothers that abort will not result in a birth and thus not captured in the results here.

It's question 4 that is driving me insane. How do I figure out the proportion that aborts? Ohanian (talk) 11:12, 21 August 2018 (UTC)
 * Question 4 is actually just simple algebra, the estimated number is found by solving $$\frac{4553}{4553+4101+x} = \frac{105}{105+100}$$ for x, and then the proportion of mothers who choose to abort if the fetus is a girl is $$\frac{x}{4101+x}$$. For your other 3 questions, I would rewrite H1 to "some mothers do abort due to the sex of the fetus being a girl which results in a birth ratio that is greater than 105 boys to 100 girls.", and then as the sample is large, you can assume a normal distribution (by the De Moivre–Laplace theorem) for your statistical tests. Iffy★Chat -- 11:37, 21 August 2018 (UTC)

I found the answer


 * 105/(100 + 105)*n == 4553
 * n is 186673/21


 * 100/(100 + 105)*n*(1 - p) == 4101
 * (91060 (1 - p))/21 == 4101
 * p is 0.054239

Ohanian (talk) 12:01, 21 August 2018 (UTC)

I just realised that real life can be more complicated than I imagine because it is also theoretically possible that Chinese mothers abort both males and females fetus, just that they abort female fetuses more often than males fetuses. This makes the mathematics even more complicated. Ohanian (talk) 12:43, 21 August 2018 (UTC)


 * Q1 and Q2 ask for the probability that H0 is true (for Q2, given that either H0 or H1 is true). However, it is impossible to ask this question in a frequentist framework; you would have to use the Bayesian approach of starting with a prior (typically subjective) probability that H0 is true. In the more commonly used frequentist approach, you have to ask “If H0 were true, what would be the probability of observing something at least this far away from what was expected?” Loraof (talk) 22:10, 21 August 2018 (UTC)

Bayesian approach
Consider a huge population of N newborn babies, pN boys and N–pN girls. A random sample of 8654 babies consists of 4553 boys and 4101 girls. The task is to estimate p. The Bayesian distribution of p is a beta distribution having mean value = $$ \frac{4553+1}{8654+2} = 0.526109$$ and index of dispersion = $$ \frac{4101+1}{(8654+2)(8654+3)}=   0.0000547408$$. The standard deviation is $$ \sqrt{0.526109\times 0.0000547408}= 0.00536656$$.

Hypothesis H0 is that $$p=\frac{105}{105+100}=0.512195$$.

The z-value is $$z=\frac{0.526109-0.512195}{0.00536656} = 2.59272$$

So the hypothesis H0 is rejected.

Hypothesis H1 is that $$p=\frac{111}{111+100}=0.526066$$.

The z-value is $$z=\frac{0.526109-0.526066}{0.00536656} = 0.00801258$$

So the hypothesis H1 is not rejected.

Bo Jacoby (talk) 17:47, 24 August 2018 (UTC).

GCDs and LCMs of powers in GCD domains
Let a and b be 2 nonzero elements of a GCD domain R and n be a positive integer. Is it always true that if c is a GCD (resp. LCM) of a and b (unique up to multiplication by units in R), then cn is a GCD (resp. LCM) of an and bn? GeoffreyT2000 (talk) 22:42, 21 August 2018 (UTC)