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In a population of working couples,the man's weekly income X and the woman's weekly income Y are statistically independent and normally distributed.X has a mean of 250 and variance of 50,while Y has a mean of 200 and a variance of 60.What is the probability husband's income exceed his wife's income,on average?

what I tried to do was take 

H0:u1−u2<=0
H1:u1−u2 >0

where u1:population mean income of men and u2:population mean income of women and under H0 compute,

enter image description hereAnd after that get get the p-value because p-value is the minimum value of significance level that would reject Ho.When H0 is rejected I get the value where man's income exceeds womens income.But how to get n1 and n2 values?And I don't know if I am going in the correct way.Any help to get this done please.

clarkson
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1 Answers1

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This is not a hypothesis testing problem, it is a pure probability problem. We want the probability that $X\gt Y$, so we want the probability that $W\gt 0$, where $W=X-Y$.

By our conditions, $W$ is normally distributed, with mean $250-200=50$, and variance $50+50=110$. We have $$\Pr(W\gt 0)=\Pr\left(Z\gt \frac{0-50}{\sqrt{110}}\right),$$ where $Z$ is standard normal.

By symmetry, this probability is the same as $\Pr \left(Z\lt \frac{50}{\sqrt{110} }\right)$. This probability will turn out to be close to $1$.

André Nicolas
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  • first I thought of calculating probability P(X>200) and standardize because it says on average husband's income exceed his wife's income,but then since this was under a inference question it looked so simple.So I tried doing it in that way. – clarkson Jun 22 '13 at 18:21
  • The phrase "on average" should not have been used by the problem setter, it is in this context meaningless. Of course on average the husband's income exceeds the wife's by $50$, but then the the word probability should not be used. – André Nicolas Jun 22 '13 at 18:26