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The df.for Difference of Means test when wve pool the variance iS A(n1 I)+(n2 - 1)(ni n2)(n1 n2)-1Reset Selection...

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The df.for Difference of Means test when wve pool the variance iS A(n1 I)+(n2 - 1)(ni n2)(n1 n2)-1Reset Selection

The df.for Difference of Means test when wve pool the variance iS A(n1 I)+(n2 - 1) (ni n2) (n1 n2)-1 Reset Selection



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The formula for the pooled variance, $s_{p}^{2},$ is given on page $440 .$ Show that, if the sample sizes, $n_{1}$ and $n_{2},$ are equal, then $s_{p}^{2}$ is the mean of $s_{1}^{2}$ and $s_{2}^{2}$

Okay. This question wants us to find the expected value and variance for this new. Why variable given that's related to this negative binomial variable X. And since we already know the MGF for X no, it makes sense to find the MGF for why so m sub y of tea would be well, in general, we know that M sub a X plus B of tea is equal to e to the bt times M setbacks of 80. So we just need to find out what A and B are for this specific distribution. And we see that the cowfish in front of X is one so a is one and are be coefficient would then be negative are so we get e to the negative rt times m setbacks of won t And since we already know what EMS of exes, it's just eat of the negative rt times p e to the t over one minus eat of the verse. Everyone minus q E to the t All raised to the are for messing with this e to the our part, we get P e to the t times e to the negative t all over one minus Q E to the T and we can raise that whole thing to the Arno. And now those two e's multiply to be one on top, giving us the final form of them supply, which would just be p over one minus Q e to the t all raised to the our and that's our MGF. For why and now to find the expected value in variance to avoid having to do a bunch of quotient rules will actually use the log version. So else of y of tea is just the natural log of em Civil I E. T. Or, in our case, Ln of p over one minus Q E to the t all raised to the r and you can probably see why I chose Did you this because we can use our log properties to bring the are in front So we get our Ln of p over one minus que e to the T. And from here we can also split this log based on multiplication or division rather, but since it's just constant on top, it's fine to keep it in this form. So if we want, we could just write it as our Ellen. He minus are Ln one minus. Q E t. And we did this because e of y is equal to a y prime of zero. So oh, prime y of tea. Taking a T derivative of this would just be negative are over one minus que eat of the tee times negative q E to the T or simplifying. We get q r e to the T over one minus que e to the t. So plugging in zero here gives us e of y. We're just be equal to you are over one minus que or q r overpay. Would we simplify? And now we want to find the variance which we get from the second derivative. So for the second derivative up, this is al double prime. We have to use a quotient rule. So we get que are outside the whole thing, then derivative of the top, keeping the bottom of the same minus derivative of the bottom, keeping the top the same. And then we divide the whole thing by the square of the denominator. So then the variance would just be dysfunction, evaluated at zero. So we get Q R times one times one minus que minus a negative sign. So plus que and then we're dividing this whole thing by one minus que all squared. And now we use our Q and P relations here. So we get still que are outside, Then on top, this becomes a P plus Q And then on bottom well, one minus que is just pee. So we get peace squared and we could just leave it like that. So the variants of why we can just leave like that or we could actually just write this as q r times P plus one minus p over P squared, which cancels the peas on top here, just giving us that. The variants of why would then just be equal to you are divided by P squared, and that's our answer.


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