# Thread: Consider a random sample of 10 patients..

1. ## Consider a random sample of 10 patients..

Question : Consider a random sample of 10 patients using a drug A for headache remedy yields sample mean and variance as 12.3 minutes and 6.54 minutes, while a random sample of 10 other patients using drug B gives sample mean and variance as 10.1 minutes and 4.1 minutes. Find a 95% confidence interval for the true difference in the mean relief times.

--------------------------My work ---------------------------

$\displaystyle n_1 = 10$
$\displaystyle \bar x_1 = 12.3$

$\displaystyle \sigma^2 _1 = 6.54$

$\displaystyle n_2 = 10$
$\displaystyle \bar x_2 = 10.1$

$\displaystyle \sigma^2 _2 = 4.1$

$\displaystyle \alpha$ = 0.05 degree of freedom = 0.95

$\displaystyle t = \frac{x_1 - x_2}{\sqrt{\frac{\sigma^2 _ 1}{n_1} + \frac{\sigma^2 _2}{n_2}}}$

Is this correct what i am trying to do ???

BUT those seem to be sample and not population variances.
There are two cases.
Most people assume the POPULATION variances are equal.
In that case you POOL the twho sample variances.
That would give you a t distribution with 10+10-2 degrees of freedom.

That also looks more like a test statistic, which you can pivot on and slove for the pop means.

Here is the case when you do NOT assume equal pop variances...http://en.wikipedia.org/wiki/Welch's_t_test

$\displaystyle \bar X_1 -\bar X_2\pm t{\sqrt{\frac{S^2 _ 1}{n_1} + \frac{S^2 _2}{n_2}}}$

Here is the other case

$\displaystyle \bar X_1 -\bar X_2\pm tS_P{\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

3. So mite which one is the correct one to use for this problem

4. Originally Posted by zorro
So mite which one is the correct one to use for this problem
That is your decision to make. Perhaps after a discussion with your instructor.

5. Btw, what is a mite?
Mite - Wikipedia, the free encyclopedia
ewwwwww

6. u have atlast got it mite
cheers on a new discovery