# Test the hypothesis

• Dec 15th 2009, 04:00 AM
zorro
Test the hypothesis
Question : A computer system has 6 I/O channels and the system personnel are reasonably certain that the load on channels is balanced. If X is random variable denoting the index of the channel to which a given I/O operations is directed , that its pmf is assumed to be uniformly distributed. Out of N=150 I/O operation onserved, the number of operations directed to various channels were:

$n_0 = 22 ; \ n_1 = 23 ; \ n_2 = 29 ; \ n_3 = 31 ; \ n_4 = 26 ; \ n_5 = 19$

Test the hypothesis that the load on channels is balanced at 5 percent level of significance.
• Dec 15th 2009, 09:22 AM
novice
Quote:

Originally Posted by zorro
Question : A computer system has 6 I/O channels and the system personnel are reasonably certain that the load on channels is balanced. If X is random variable denoting the index of the channel to which a given I/O operations is directed , that its pmf is assumed to be uniformly distributed. Out of N=150 I/O operation onserved, the number of operations directed to various channels were:

$n_0 = 22 ; \ n_1 = 23 ; \ n_2 = 29 ; \ n_3 = 31 ; \ n_4 = 26 ; \ n_5 = 19$

Test the hypothesis that the load on channels is balanced at 5 percent level of significance.

You are given a uniform pmf, which mean that $p_0 = p_1 = p_2=p_3=p_4=p_5 =\frac{1}{6}$

The expected means $\mu_0=\mu_1=\mu_2=\mu_3=\mu_4=\mu_5= Np$

$\sigma = \sqrt{Npq}$

Use a two-tailed test at the 0.05 significance level and adapt the following decision rule:

Accept $H_0$ if all z scores of the sample mean is inside the range $z_{0.05}$ to $z_{0.95}$, i.e, all channels are balanced.
Reject $H_0$ otherwise.
• Dec 16th 2009, 08:45 PM
zorro
which test are u using ?
Quote:

Originally Posted by novice
You are given a uniform pmf, which mean that $p_0 = p_1 = p_2=p_3=p_4=p_5 =\frac{1}{6}$

The expected means $\mu_0=\mu_1=\mu_2=\mu_3=\mu_4=\mu_5= Np$

$\sigma = \sqrt{Npq}$

Use a two-tailed test at the 0.05 significance level and adapt the following decision rule:

Accept $H_0$ if all z scores of the sample mean is inside the range $z_{0.05}$ to $z_{0.95}$, i.e, all channels are balanced.
Reject $H_0$ otherwise.

Which test are u using

And what is 'q' is it '1-p' or some thing else please advice
• Dec 17th 2009, 10:26 AM
novice
Quote:

Originally Posted by zorro
Which test are u using

And what is 'q' is it '1-p' or some thing else please advice

q = 1-p
• Dec 17th 2009, 02:59 PM
zorro
which test are u using
Quote:

Originally Posted by novice
q = 1-p

Could u please tell me which test are u using.......is there a link u could provide me ...........just if i encounter similar problems in the future
• Dec 18th 2009, 08:14 AM
novice
Quote:

Originally Posted by zorro
Could u please tell me which test are u using.......is there a link u could provide me ...........just if i encounter similar problems in the future

You have small sample, Student t-test would be good. If you prefer, you may do F-test, or Chi-square test too.

Take a look at an example here:

http://www.mathhelpforum.com/math-he...s-testing.html
• Dec 18th 2009, 12:10 PM
zorro
I still have questions
Quote:

Originally Posted by novice
You have small sample, Student t-test would be good. If you prefer, you may do F-test, or Chi-square test too.

Take a look at an example here:

http://www.mathhelpforum.com/math-he...s-testing.html

But the student t test is $\frac{ \bar X - \mu}{ \sigma / \sqrt{n}}$ so how are u using $\sigma = \sqrt{Npq}$
• Dec 19th 2009, 03:13 AM
zorro
I got it now
Quote:

Originally Posted by zorro
But the student t test is $\frac{ \bar X - \mu}{ \sigma / \sqrt{n}}$ so how are u using $\sigma = \sqrt{Npq}$

I got it when X is a binomial distribution and N is very large and probability of individual trails is close to 1/2 is can be seen that with increasing n these distribution approaches a normal distribution which can be expressed as

z = $\frac{X - np}{\sqrt{npq}}$ as $n \to \infty$

Thanks mite
cheers to every one for helping me with this problem
(Beer)