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# Statistics for Psychology PSYC 233

FSU

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This 4 page Class Notes was uploaded by Marlene Abernathy DDS on Monday October 12, 2015. The Class Notes belongs to PSYC 233 at Fayetteville State University taught by David Wallace in Fall. Since its upload, it has received 26 views. For similar materials see /class/221596/psyc-233-fayetteville-state-university in Psychlogy at Fayetteville State University.

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Date Created: 10/12/15

Lesson 20 Chi Square Test of Independence Outline Measuring Independence observed frequencies expected frequencies chisquare Measuring Independence The Chisquare test of independence is similar to the test we just learned in the last lesson However instead of measuring frequencies along only one dimension we will measure frequencies for two variables at the same time Our test is designed to test whether or not these two variables are independent not related If we reject the null and say the variables are not independent of one another then we have established that the two variables are related The test of independence starts with frequencies or counts we observe in our sample or the observed freguencies For this example a sample of 50 people is used to record personality and color preference is measured Blue Red Yellow Extroverted 5 20 5 Introverted 10 5 5 Observed Frequencies Is there a relationship between personality type and color preference Our hypotheses will state exactly that with the null as usual stating that there is no effect or no relationship H1 There is a relationship between color preference and personality type variables are not independent H0 There is no relationship between color preference and personality type variables are independent Since we have two variables our degrees of freedom will change df R l C l 6 where R is the number of rows and C is the number of columns in our table There are two rows going across and three rows going down So degrees of freedom for this example are df2 13 112 2 We will find the critical value using the same table we used in the goodnessoffit test In our example Xinml 599 When we start to compute the statistic it will be similar to the goodnessoffit test as well However we will need a formula to compute the expected frequencies instead of just dividing our sample size equally between groups We will need to compute the expected value separately for each observed value in our sample So in our example we must compute six different expected frequencies Note that our expected frequencies with this test are the values we expect if the null is true Thus the expected frequencies are the values we expect if there is no relationship or the variables are independent z n frequencies for the c column and r row You multiply these values together They are the total frequency for the row and column of the individual expected value we are looking for with the computation So you must first add up the frequencies for each row and column 6 The value n is the total or 50 here In the numerator we have the Blue Red Yellow Extroverted 5 20 5 30 Introverted 10 5 5 20 15 25 10 Each value of our six observed values will have separate column frequency and row frequency For example the observed value in the first row and column 5 has a row frequency of 30 and a column frequency of 15 Blue Red Yellow Extroverted E 20 5 m Introverted 10 5 5 20 E 25 10 3015 9 50 50 So the expected frequency is f Continue nding each expected frequency in this same way for each of the observed values Notice that when we compute a different expected frequency our row and column frequencies will change So for the next value extroverted and red Blue Red Yellow Extroverted 5 I 5 Introverted 10 5 5 20 15 10 3025 750 f 15 50 50 Again you continue the process until you have found all the expected frequencies There is a shortcut for finding the rest of these values if you feel comfortable with the statistic Once we compute these two expected frequencies all the others are determined hence the two degrees of freedom All of our expected frequencies must have the same row and column sums as our observed frequencies So once we have computed these two expected frequencies all of the other values can be found by subtracting out the row or column total The remaining unknown expected value in the first column then must be the number that will make that first column add up to 15 Since the first expected value is 9 then the remaining number must be 15 7 9 6 Whether you compute each individual expected frequency or use the short cut you will get a complete table of expected frequencies Blue Red Yellow Extroverted 9 15 6 30 Introverted 6 10 4 20 15 25 10 Expected Frequencies The process for finding the final Chisquare value is the same as before We will find the difference between our expected and observed frequencies square the difference and then divide by the expected frequency for each value in our table 12 2 2 5 92 l20 152 l5 62 l10 62 l5 102 l5 42 I 9 I 4 15 6 6 10 I 42 52 12 42 52 12 9 15 6 6 10 4 2 16 25 1 16 25 1 I 9 15 6 6 10 4 12 17816701672672525 2 I 904 Since this value is greater than our critical value we will reject the null here and say there is a relationship Thus knowing someone s personality type gives you information about their likely color preference

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