Recall from Unit 4 that when certain conditions are met the sampling distribution of the sample proportion is Normal Unit 7 Tests for Proportions np is at least 10 n 1 p is at least 10 This means that IF these conditions are met we can compute the z test statistic to determine the probability of a sample proportion as extreme as the one we got p hat is your sample proportion p is the population proportion under the null hypothesis 0 n is your sample size We use the population proportion under the null hypothesis np0 is at least 10 n 1 p0 is at least 10 There are two different ways to do null hypothesis significance testing when your null hypothesis is about a population proportion use the binomial distribution to get an exact p value use the Normal approximation to get a p value Remember that you can do the second one only when the conditions for the Normal approximation are met chi square test of goodness of fit The value of X2 depends on how far your sample data is from what is expected under the null hypothesis A large value of X2 corresponds to a small p value Here is the actual formula The first step in computing X2 is to compare the observed frequency of each outcome with the expected frequency of each outcome The expected frequency of each outcome is just the proportion of times that the outcome occurs according to the null hypothesis times the sample size The next step is just to compute the difference observed expected for each outcome Like with t we need to get the df This is equal to the number of categories 1 we can use a z test statistic to test a null hypothesis about a single population proportion we the chi square test of goodness of fit to test a null hypothesis about the entire distribution of a categorical variable In fact we can use chi square for something else too To test whether a categorical variable has the same distribution in two or more populations When we use chi square to answer this it is called the chi square test of independence The null hypothesis is always that the two variables are independent Now we use the same chi square formula as before summing over all the cells Finally an important warning In order for a chi square test to be valid in either the goodness of fit setting or the independence setting some assumptions have to be met All expected counts have to be 1 or more No more than 20 of cells can have expected counts of less than 5 This means that if you have a 2x2 contingency table no cell s expected count can be less than 5