enter image description here

I'd be really grateful if someone can help me. I found this problem in the textbook Probability and Statistics by Wapole. This question look something similar to a Hypothesis test, but I have no clue how to solve it. I have tried re-reading parts of the book, but I still can't figure out how to do this.

For a) I figure that Z = (X-u)/(std/sqrt(n)) would give me the probability, so I know whether the amount is significant or low probability...But what do I then do for b)??


Let's do some back of the envelope calculations.

We have 30 samples, and the std of each sample is known. So the std of the mean is going to be $5 / \sqrt(30) \approx 1\% $ so, roughly applying the central limit theorem (be careful when applying the CLT ! it's full of traps ! learn about them), the distribution of the mean of the 30 samples is going to be a Gaussian, of std 1%, centered around $\mu_a$ Seeing a value that close to 65% for the realization of that variable, it's quite likely that the hypothesis that $\mu_a \geq 65$ is wrong

For the second part, we want to compute the probability of the difference of two Gaussians of std 1% to be bigger than 5.5. Very roughly, the difference of the two Gaussians will have std $\approx\sqrt(2)\approx1.5$ so that 5.5 is between 3 and 4 stds of deviation. That's a pretty good separation and is strong evidence for $\mu_a \neq \mu_b$

Now that I've done the back of the enveloppe calculations, try to redo all of this more rigorously

  • $\begingroup$ I really don't get the last bit, so can you please explain that? Where did you get the sqrt(2)? The last question really confuses me as to what we are really trying to find/do. $\endgroup$
    – Kelbe
    Jul 3 '15 at 9:35
  • $\begingroup$ If X and Y are two Gaussians of mean $\mu_x$ and $\mu_y$ and variance 1, then their difference is a Gaussian of mean the difference of the means $\mu_x - \mu_y$ and of variance the sum of the variance $1+1=2$, so the std is $\sqrt(2)$ $\endgroup$ Jul 3 '15 at 14:29

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.