**Theorem 1**: Let *x̄* and ȳ be the sample means of two sets of data of size *n _{x}* and

*n*

_{y}respectively. If

*x*and y are normal, or

*n*and

_{x}*n*

_{y}are sufficiently large for the Central Limit Theorem to hold, and

*x*and y have the same variance, then the random variable

has distribution *T*(*n _{x} + n*

_{y}– 2) where

Proof: Let *σ* be the common standard deviation of *x* and y. Then *x̄* – ȳ has a normal distribution with mean *µ _{x} – µ*

_{y}and standard deviation

Defining *z* as follows, we know that *z* has distribution *N*(0, 1).

We also know that has distribution χ^{2}(*n _{x }*– 1) and has distribution χ

^{2}(

*n*

_{y }– 1), and so

has distribution χ^{2}(*n _{x }+ n*

_{y }– 2).

Defining *t* = *z*/*u*, where *m* = *n _{x }+ n*

_{y }– 2, it follows by Property A of Basic Concepts of t Distribution that

*t*has distribution

*T*(

*m*).

where *s* is defined as in the statement of the theorem.

Hi Charls,

Can u guide me : On interpreting the p value score and result for both hypothesis and that too for same case

Way1 : H0 = there is no change in awareness level of TG

Ha = there is significant change in awareness level of TG

Way2 : H0 = Awareness level has increased among TG post campaign

Ha = There is no change in awareness level How 2 Null hypothesis for same case

Can u plz suggest how two different hypothesis for same case can make change in result and interpretation. Please suggest

Hello,

It sounds like Way1 will require a two-tailed test (in Ha, awareness can increase or decrease), while Way2 requires a one-tailed test (you seem to be ruling out the case where aware decreases).

Charles

thank you very much sir for this informations.

you missed a square at this u

http://i0.wp.com/www.real-statistics.com/wp-content/uploads/2013/03/image3398.png

Thank you for catching this error. I have now corrected the mistake.

I appreciate your helping to make the website more accurate and easy to follow.

Charles