LearningActivities ColorBlindness

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* 100 men are selected at random. What is the distribution of <math>X_m</math> = the number of these men that are colorblind?
* 100 men are selected at random. What is the distribution of <math>X_m</math> = the number of these men that are colorblind?
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:: <math>X_m</math>~<math>Binomial(100,p)</math>.
 
{{hidden| See the Answer| <math>X_m</math>~<math>Binomial(100,p)</math> }}
{{hidden| See the Answer| <math>X_m</math>~<math>Binomial(100,p)</math> }}
* 100 women are selected at random. What is the distribution of <math>X_f</math> = the number of these women that are colorblind?  
* 100 women are selected at random. What is the distribution of <math>X_f</math> = the number of these women that are colorblind?  
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::'''Hint''': the chance that an individual woman is colorblind is <math>p^2</math>, why?
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{{hidden| See a Hint| the chance that an individual woman is colorblind is <math>p^2</math>, why? }}
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::'''Solution''': <math>X_f</math>~<math>Binomial(100,p^2)</math>
+
 
 +
{{hidden| See the Answer| <math>X_f</math>~<math>Binomial(100,p^2) }}
* To estimate the probability that a randomly selected woman is colorblind, you might use the proportion of colorblind women in a sample of n women. What is the variance of this estimator?  
* To estimate the probability that a randomly selected woman is colorblind, you might use the proportion of colorblind women in a sample of n women. What is the variance of this estimator?  
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:: <math>X_f</math>~<math>Binomial(n,p^2)</math>. Thus <math>Var(\frac{X_f}{n})=\frac{p^2(1-p^2)}{n}</math>.
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 +
{{hidden| See the Answer| <math>X_f</math>~<math>Binomial(n,p^2)</math>. Thus <math>Var(\frac{X_f}{n})=\frac{p^2(1-p^2)}{n}</math> }}
* Alternatively, to estimate the probability that a randomly selected woman is colorblind, you might use the square of the proportion of colorblind men in a sample of n men. Explain why this estimate makes sense. What is the variance of this estimator?  
* Alternatively, to estimate the probability that a randomly selected woman is colorblind, you might use the square of the proportion of colorblind men in a sample of n men. Explain why this estimate makes sense. What is the variance of this estimator?  
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:: '''Hint''': The moment generating function can be used to find the fourth moment about the origin.
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{{hidden| See a Hint| The moment generating function can be used to find the fourth moment about the origin. }}
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:: '''Hint''': We want to estimate <math>p^2</math> and <math>\frac{X_m}{n}</math> estimates <math>p</math> so it makes sense to use <math>(\frac{X_m}{n})^2</math> as the estimator (in fact it will be the [http://wiki.stat.ucla.edu/socr/index.php/AP_Statistics_Curriculum_2007_Estim_MOM_MLE#Maximum_Likelihood_Estimation_.28MLE.29 maximum likelihood estimate]). We have <math>Var[( \frac{X_m}{n} )^2 ] = n^{-4}[E(X_m^4 ) - (E(X_m^2 ))^2 ]</math>. Take <math>q=1-p</math>. Then the fourth moment about the origin of a binomial is <math>E(X^4)= np(q-6pq^2+7npq-11np^2q+6n^2p^2q+n^3p^3)</math> and the second moment is <math>E(X^2)=np(q+np)</math>. Thus <math>Var[( \frac{X_m}{n} )^2 ] = n^{-3}(pq + 6(n-1)p^2q^2 + 4n(n-1)p^3q)</math>.  
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 +
{{hidden| See the Answer| We want to estimate <math>p^2</math> and <math>\frac{X_m}{n}</math> estimates <math>p</math> so it makes sense to use <math>(\frac{X_m}{n})^2</math> as the estimator (in fact it will be the [http://wiki.stat.ucla.edu/socr/index.php/AP_Statistics_Curriculum_2007_Estim_MOM_MLE#Maximum_Likelihood_Estimation_.28MLE.29 maximum likelihood estimate]). We have <math>Var[( \frac{X_m}{n} )^2 ] = n^{-4}[E(X_m^4 ) - (E(X_m^2 ))^2 ]</math>. Take <math>q=1-p</math>. Then the fourth moment about the origin of a binomial is <math>E(X^4)= np(q-6pq^2+7npq-11np^2q+6n^2p^2q+n^3p^3)</math> and the second moment is <math>E(X^2)=np(q+np)</math>. Thus <math>Var[( \frac{X_m}{n} )^2 ] = n^{-3}(pq + 6(n-1)p^2q^2 + 4n(n-1)p^3q)</math>. }}
# For large samples, is it better to use a sample of men or a sample of women to estimate the probability that a randomly selected women is colorblind? Explain.  
# For large samples, is it better to use a sample of men or a sample of women to estimate the probability that a randomly selected women is colorblind? Explain.  
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:: '''Hint''': Show that a normal approximation is valid for both and then compare the variances.
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{{hidden| See a Hint| Show that a normal approximation is valid for both and then compare the variances.}
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::'''Solution''': For large n the ratio of the variances for the estimate in part c to the estimate in part d is <math>\frac{Var(\frac{X_f}{n})}{Var((\frac{X_m}{n})^2 )} \sim \frac{p^2(1-p^2)}{4p^3q} = \frac{1+ p}{4p}</math>. When this ratio is greater than 1, the estimator based on the sample of men will be better. Since this happens for any <math>p < \frac{1}{3}</math>, which is clearly the case for colorblindness, it is better to use a sample of men to estimate the probability that a random woman is colorblind.
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{{hidden| See the Answer| For large n the ratio of the variances for the estimate in part c to the estimate in part d is <math>\frac{Var(\frac{X_f}{n})}{Var((\frac{X_m}{n})^2 )} \sim \frac{p^2(1-p^2)}{4p^3q} = \frac{1+ p}{4p}</math>. When this ratio is greater than 1, the estimator based on the sample of men will be better. Since this happens for any <math>p < \frac{1}{3}</math>, which is clearly the case for colorblindness, it is better to use a sample of men to estimate the probability that a random woman is colorblind.}}
===Conclusions===
===Conclusions===

Revision as of 20:06, 25 October 2011

Contents

Distributome Learning Activities - Distributome Colorblindness Activity

Overview

Colorblindness - Can you see the number in this image?

This Distributome Activity illustrates an application of probability theory to study Colorblindness.

Colorblindness results from an abnormality on the X chromosome. The condition is thus rarer in women since a woman would need to have the abnormality on both of her X chromosomes in order to be colorblind (whether a woman has the abnormality on one X chromosome is essentially independent of having it on the other).

Goals

The goal of this activity is to demonstrate an efficient protocol of estimating the probability that a randomly chosen individual may be colorblind.

Hands-on Activity

Suppose that p is the probability that a randomly selected man is colorblind.

  • 100 men are selected at random. What is the distribution of Xm = the number of these men that are colorblind?
See the Answer


  • 100 women are selected at random. What is the distribution of Xf = the number of these women that are colorblind?
See a Hint


See the Answer


  • Alternatively, to estimate the probability that a randomly selected woman is colorblind, you might use the square of the proportion of colorblind men in a sample of n men. Explain why this estimate makes sense. What is the variance of this estimator?
See a Hint


See the Answer


  1. For large samples, is it better to use a sample of men or a sample of women to estimate the probability that a randomly selected women is colorblind? Explain.

{{hidden| See a Hint| Show that a normal approximation is valid for both and then compare the variances.}

See the Answer


Conclusions

You can also use the delta method to find the approximate variance for the estimator above.


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