Wikipedia:Reference desk/Archives/Mathematics/2006 December 16

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December 16

Homogeneity of Error Variance within groups

If I am doing regression of data that can be divided into two groups, and each group has a different variance (e.g., see http://members.aol.com/IMSAP/homogeneity.html), how can I still do regression? Any normalization/transformation etc that might help? 64.178.108.204 05:31, 16 December 2006 (UTC)

Perhaps look at weighted least squares. You weight the two groups differently. The observations in each group get a weight that is 1 / (variance of that group). --Spoon! 07:37, 16 December 2006 (UTC)

My variables are indicator variables, with values only 0 or 1. Can I still use weighting? 64.178.108.204 15:13, 16 December 2006 (UTC)

I think so. --Spoon! 05:36, 18 December 2006 (UTC)

the phase of a Fourier transformed sinusoid is a Dirac's delta

is it right? Btw, freq. channel is a syonin. of freq. bin ? tia --Ulisse0 15:54, 16 December 2006 (UTC)

did you look at our fourier transform#Table of important Fourier transforms? --Spoon! 21:48, 16 December 2006 (UTC)

yes, thanks, but $\arg \left({\sqrt {2\pi }}{\frac {\delta (\omega \!-\!a)\!-\!\delta (\omega \!+\!a)}{2i}}\right)\,={\frac {\pi }{2}}(\delta (\omega \!-\!a)\!-\!\delta (\omega \!+\!a))?$ --Ulisse0 22:20, 16 December 2006 (UTC)

Speed distance equation

Hello,

I am going to a party tonight that is about an hour away from my house. The longest part is a 26.9 mile stretch that I will be traveling at 60 mph on. I could watch the odometer (my car has no trip meter), but I would really rather have my passenger set an alarm so I know when to start looking for exits. This, of course, would only be a general time frame I would use and I wouldn't expect it to be extremely accurate (based on speed fluctuations).

So, assuming I will be traveling at 60 mph, how long will it take me to travel 26.9 miles?

Being a 'teach a man to fish' type guy, I would prefer if you just explain the equation to me (or even just link me somewhere!) so I can use it in the future. Happy holidays, and thanks in advance!

172.130.196.167 22:41, 16 December 2006 (UTC)

It's a standard velocity vs. time problem. The relevant equation is $s=vt$, with $s$ being your displacement (in this case, the distance traveled), $v$ being your velocity, and $t$ the time it takes to travel the distance.

Plugging in the numbers, you get $26.9/60$, which is 0.448 hours. Converting that to minutes, you get 26.9 minutes.

If you think about it, if you travel at 60 mph, you will travel 60 miles every 60 minutes, or a mile a minute. Titoxd^(?!?) 22:51, 16 December 2006 (UTC)

Dimensional analysis is the way to go here. 60 mpm is 60 miles per hour, or 60 (miles/hour). You want to end up with a measurement of time, i.e. in hours, so find a way to arrange "60 (miles/hour)" and "26.9 miles" in such a way that you get an answer in hours. The one way to do that is ${\frac {26.9\;\mathrm {miles} }{60\;(\mathrm {miles} /\mathrm {hour} )}}={\frac {26.9}{60}}\;\mathrm {hours}$. Then proceed as above. EdC 02:53, 17 December 2006 (UTC)

One hint, instead of doing math each time you look at the odometer, you can add 26.9 to the odometer reading when you start, write that number down, then watch for when that number comes up. That way, you only need to do the math once. StuRat 23:06, 16 December 2006 (UTC)

Thanks for the help! Cheers!

172.144.200.236 00:22, 17 December 2006 (UTC)