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Babies die after mums given Viagra in Dutch trial
A Viagra in pregnancy trial has been urgently stopped after 11 newborn babies died. Women taking part in the Dutch study had been given the anti-impotence tablets to improve growth of their unborn children because they had poorly developed placentas.
It appears the drug, which promotes blood flow, may have caused lethal damage to the babies' lungs. Experts say a full investigation is needed to understand what happened. There is no suggestion that there was any wrong-doing.
Earlier trials in the UK and Australia and New Zealand did not find any evidence of potential harm from the intervention. But they also found no benefit.
[...] Foetal growth restriction caused by an underdeveloped placenta is a serious condition that currently has no treatment. It can mean babies are born prematurely, with a very low birth weight and poor chances of survival. A medication that could improve weight or prolong the time to delivery could have significant advantages for these very sick babies.
(Score: 4, Interesting) by exaeta on Thursday July 26 2018, @05:45PM
A 5 fold increase at that level IS statistically significant. Maybe not to six sigma or whatevet level of certainty you want.
Let's apply the Maximum Likelyhood Method for analysis:
93 test subjects in group A.
17 true results.
76 false results.
90 test subjects for group B.
3 true results.
87 false results.
so therefore for group A we have the probability equation:
(p)^17(1-p)^76 / (93 choose 17)
We can omit the last bit for finding the most likely value of p given that it is constant (but will need it if we want exact values).
Therefore for group A the p-value most likely is 0.18280.
For group B the value most likely is given by maximizing:
(p)^3(1-p)^87 / (90 choose 3)
p is most likely around 0.03333 ( by maximization).
Now we can perform cross substituition:
((0.18280)^17(1-0.18280)^76 / (93 choose 17))/((0.03333)^17(1-0.03333)^76 / (93 choose 17))
Result is 1.05...
Therefore there was only a 5% increase in likelihood...
So not statistically significant? Maybe! but we need to substituite the value in the control group too.
Control distribution is:
(p)^3(1-p)^87
so:
((0.03333)^3(1-0.03333)^87)/((0.18280)^3(1-0.18280)^87)
≈ 13,500
13,500 * 1.05 ≈ 14,100
Therefore there is about a 1-in-14,100 chance the result was caused by random variation. Good idea to stop the study, I think.
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