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Probability on hair growth question

  1. May 9, 2004 #1
    There is a problem (the solution should use limit probability statements):

    One medical company invented some pills that should accelerate the growth of hair. The company wanted to try the pills on people - and so 2 groups of volunteers (each had 100 persons) were chosen to participate in the test. Persons in 1st group didn't take the pills and the experts noticed that the average monthly increment of the length of their hair was 3,5 cm, person in the 2nd group took the pills and the experts found out that the average monthly increment was 3,9 cm. It's known that the diffusibility (denoted D(X) for random variable X) of monthly increment of the length of hair is 0,5 cm^2. It was only a luck or the pills are really effective? What is the probability that by using the pills the hair will grow faster?

    I would like to know, how to use the limit statements from the probability theory (as Central Limit Statement, the Rule of Great Numbers...) on this problem. I know that it isn't so difficult, but this is new for me and I'm not able to applicate it yet.

    Thank you very much for help.
  2. jcsd
  3. May 11, 2004 #2
    Assuming a normal distribution (n is also enough high) it is possible to use a z significance test,by setting alpha=0.05 (an usual value for such tests though in many situations alpha is set to 0.01);the distribution in the group which received a placebo characterizes the null hypothesis H0.We must test whether there are sufficient reasons to believe that the pills really work (hypothesis H1 is accepted) from the results of the experiment with the second group.The value of alpha represent the probability to reject H0 when it is in fact true,this is why it is chosen very low.



    We have then [by standardising the normal distribution for the first group by taking z=(x-m1)/s1]:

    P(z>a)=0.05=(1/2)-P(0≤z≤a)=(1/2)-∫0a (1/√ (2π))*exp(-z2/2) dz

    Therefore Φ(a)=∫0a exp(-z2/2) dz=√ (2π)*(0.5-0.05)=1.1277


    0a exp(-z2/2) dz=[a-(a3/6)+(a5/40)-(a7/336)+...]

    Results (using the trial and error method for example) that a=1.645 (the same result can be obtained from the tables with the function Φ(z)).

    Let now Z=(m2-m1)/s1

    If Z>a then the null hypothesis is rejected (H1 is accepted).

    If Z≤a then H1 is rejected (H0 is accepted).

    In our case Z=(3.9-3.5)/0.707=0.56 < 1.645 therefore the null hypothesis is accepted (no significative effects for the pills used).
    Last edited: May 11, 2004
  4. May 11, 2004 #3
    Thank you...
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