I have the following problem:

20 % of the exercises daily, with no difference between men or women. A random sample of 32 men and 32 women are asked whether they exercise.

What's the approximate probability that in this sample at least 10 % more men than women exercise?

I'm not sure how to approach this.

It seems that both men and women have a distribution of $\sim N(0.2, 0.005)$. And that I'm being asked what the probability of having $\frac{M}{F} \geq 1.1 $ is for this. But I'm not sure how to go about that, of if that's the correct approach to solve this problem.

  • 1
    $\begingroup$ Are you being "asked" as course work, or is the from actual research work? $\endgroup$ – Carl Apr 19 '17 at 17:51
  • $\begingroup$ @Carl sorry that phrasing was vague - it's neither coursework or research work, just a problem that I'm working on $\endgroup$ – baxx Apr 19 '17 at 17:52
  • $\begingroup$ OK, still a bit vague. Is the only information "exercise daily"? That seems a bit stilted. How would someone who exercises every day except Sunday be rated? What are the categories/classifications in the data? $\endgroup$ – Carl Apr 19 '17 at 17:57
  • $\begingroup$ @Carl yes it's pretty contrived - but for all sense and purpose that could be replaced with anything. 20 % like dogs more than cats, 20 % like fruit etc. There is no further information, I'm happy to alter the original question to something that seems less ambiguous $\endgroup$ – baxx Apr 19 '17 at 17:59

Using the normal approximation, $\hat{p} \sim N(p=.2, \frac{p(1-p)}{n}=.005)$ and assuming independence between $\hat{p}_{m}$ and $\hat{p}_{f}$, then $$\hat{p}_m - \hat{p}_r \sim N(0, .005 + .005=.10^2)$$ Using R as a calculator:

> pnorm(.10, 0, .10, lower.tail = F)
[1] 0.1586553
  • $\begingroup$ Thanks - I'm not sure what the actual calculations for this would be (I'm trying to do them by hand). So that I have $\hat{p} \sim N(0, 0.1^2)$ and need to find the probability that this is greater than $0.1$. So that I would use $P(x > 0.1)$ then standardise for $P(z > \frac{0.1 - 0}{\frac{\sigma}{\sqrt{n}}})$ where $\sqrt{n} = 1$. This gives a $z$ value of $1$, and from a table this reads off the probability of $0.1587$. If this is correct then may I add it to the answer please and it's done. Thanks. $\endgroup$ – baxx Apr 19 '17 at 18:18
  • $\begingroup$ I've edited how to calculate by hand and accepted the answer - if you could check the edit that would be appreciated. $\endgroup$ – baxx Apr 19 '17 at 18:30
  • 1
    $\begingroup$ OK, fine with me. $\endgroup$ – HStamper Apr 19 '17 at 20:40
  • $\begingroup$ I edited the post and the edit was refused because it 'deviated too much from the intent of the original post' :P Thanks though. I'll leave this here just for my own reference and perhaps that of others who find it though, as people often (unfortunately!) have to do these by hand... Calculating by hand, using tables, would be as follows; We have $\hat{p} \sim N(0, 0.1^2)$ And want $P(p > 0.1)$ Standardising and using the $z$ tables as $P(z > \frac{0.1 - 0}{\sigma})$, gives $z$ value of $1$ and from the table a probability of $0.1578$. $\endgroup$ – baxx Apr 19 '17 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.