<p>1 variable, 1 sample = test of goodness of fit
2 variables, 1 sample = test of independence/association
1 variable, multiple samples from different populations = test of homogenity</p>
<p>That's how I decide which chi square test to use ... but some people find it confusing -.-</p>
<p>You increase the power of the test by either increasing alpha, increasing the sample size, or both. </p>
<p>To my knowledge a chi-squared test of independence can be used on any data that doesn't require a goodness to fit. My class went over 30+ free responses and those with chi-squared never mentioned a test of homogeneity. They may have been other possible options, but we only used independence.</p>
<p>The variance in a t distribution will be greater than that of a z redistribution. The standard deviation for the population is know for a z test while is isn't for the t test. Since the sample's standard deviation is used as an estimate for a t test, the variance for the distribution is larger than that of a z.</p>
<p>For homogeneity, it is just like independence test, except your null and alternative hypotheses are different.</p>
<p>Say you asked some males and females ... uhh ... if they like pie, and you get "yes" or "no" for each person. (male and female are 2 samples; yes and no are your variables)</p>
<p>The null for independence is: gender and preference for pie are independent. (you take into account both male and female, both "yes" and "no" responses)</p>
<p>The null for homogeneity is: proportion of people who like pie is the same for both male and female. (you take into account both male and female, but only acknowledge the "yes" responses)</p>
<p>carry out the test the exactly the same way you would for independence. =)</p>
<p>The test is tomorrow. Good luck everyone! ^^</p>
<p>ok...you guys are making me scared...considering no one i knew in my school even bothered to study stats o.O...hope this upcoming test isn;t going to be harder than the other yrs...=(</p>