State we want to perform a test of h 0 there is no

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State:We want to perform a test ofH0: There is no association between anger level and heart disease in thepopulation of people with normal blood pressure.Ha: There is an association between anger level and heart disease in thepopulation of people with normal blood pressure.We will useα= 0.05.Here is the complete table of observed and expected counts for the CHDand anger study side by side. Do the data provide convincing evidence ofan association between anger level and heart disease in the population ofinterest?
+Example: Angry People and Heart DiseaseInference for RelationshipsPlan:If the conditions are met, we should conduct a chi-square test forassociation/independence.RandomThe data came from a random sample of 8474 people withnormal blood pressure.Large Sample SizeAll the expected counts are at least 5, so thiscondition is met.IndependentKnowing the values of both variables for one person inthe study gives us no meaningful information about the values of thevariables for another person. So individual observations areindependent. Because we are sampling without replacement, we needto check that the total number of people in the population with normalblood pressure is at least 10(8474) = 84,740. This seems reasonableto assume.
+Example: Cocaine Addiction is Hard to BreakInference for RelationshipsDo:Since the conditions are satisfied, we can perform a chi-test forassociation/independence. We begin by calculating the test statistic.Test statistic:χ2=(Observed -Expected)2Expected=(53-69.73)269.73+(110-106.08)2106.08+...+(606-618.81)2618.81=4.014+0.145+...+0.265=16.077Conclude:Because theP-valueis clearly less thanα= 0.05, we rejectH0andconclude that anger level and heart disease are associated in the population ofpeople with normal blood pressure.P-Value:The two-way table of anger level versus heart disease has 2 rows and 3 columns. Wewill use the chi-square distribution with df = (2 - 1)(3 - 1) = 2 to find theP-value.Table: Look at the df = 2 line in Table C. The observed statisticχ2= 16.077 is largerthan the critical value 15.20 forα= 0.0005. So theP-value is less than 0.0005.Technology: The commandχ2cdf(16.077,1000,2) gives 0.00032.
+Using Chi-Square Tests WiselyInference for RelationshipsBoth the chi-square test for homogeneity and the chi-square test forassociation/independence start with a two-way table of observed counts.They even calculate the test statistic, degrees of freedom, andP-value inthe same way.The questions that these two tests answer are different,however.A chi-square test for homogeneity tests whether the distribution of acategorical variable is the same for each of several populations ortreatments.Instead of focusing on the question asked, it’s much easier to look at how thedata were produced.If the data come from two or more independent random samples or treatmentgroups in a randomized experiment, then do a chi-square test for homogeneity.If the data come from a single random sample, with the individuals classifiedaccording to two categorical variables, use a chi-square test forassociation/independence.

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Term
Fall
Professor
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Tags
Statistics, Chi Square Test, AP Statistics, Statistical hypothesis testing, Statistical tests, Chi square distribution, Pearson s chi square test

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