Statistics

you want to compare the mean percentage of the two samples if yes then depending on the sample size you can go for one sample z or t test again depending if your data is normal.

WRONG!!!!!  You use a 2 proportion test.  You use a one sample z to compare one sample to a standard…continuous data and a t test for smaller samples.  A two sample t test for comparing the continuous mean of two samples.  The 2 proportion test is for two proportions which is the same as two percentages which is what the poster asked about.

have a query might be trivial
percentage is a discrete data then why use T-test
can’t we use chi sqaure to know if there is commonality between the standard and the sample in question ????
rgds,

Hi Indresh you need to ask yourself whether percentage is a discrete data.

Wrong

Joseph,
Go learn something before you give advice

Why should a proportion test be a Z-test? You use the binomial distribution to compute the confidence intervals
 
 

Staney boy-
Please read from MINITAB help, Formulas and methods. I’m not apologizing for squat. It is simply an application of the Z test. It uses the binomial mean and sd, but the test statistic is Z.
 
 
 
 
 
Hypotheses

 
 
 
 
The null hypothesis for a 2 proportion test is: H0: p1 – p2 = d0
where:
·    p1 = the proportion for the first population
·    p2 = the proportion for the second population
·    d0 = the hypothesized difference between the population proportions
You can choose any one of three alternative hypotheses:

H1: p1 – p2 > d0

One-tailed test

H1: p1 – p2 < d0

One-tailed test

H1: p1 – p2    ddd0

Two-tailed test

Test statistics

The calculation of the test statistic, Z, depends on the method used to estimate p.
By default, Minitab uses separate estimates of p for each population and calculates Z by:

Also note that you get the following warning from Minitab when doing the sample I suggested –
* NOTE * The normal approximation may be inaccurate for small samples.
This is because we are using the normal (z) as an approximation of the binomial and there are certain assumpltions that must be met. The assumprions are simple np and n(1-p) must both be greater than 5. There is a reason I gave you the example I did – I knew it did not meet the assumptions.
The correct way to do this is to get the confidence intervals of the two proportions using binomial and see if proportion 1 is inside of the confidence interval for proportion 2 and vice versa.

Hey Confucius,
Strange to see you show up on a site dedicated to rampant profitering. Last time I checked, you were against such things.
Be that as it may, let me give you my new favorite Confucius type sayings –
1) Be careful when trying to sound like Confucius, people will not know if you are an idiot or a wise man who just sounds like an idiot.
2) Average men, with a little learning that they think is great learning, working in earnest together, are just a group of average men with a little learning.