Questions and Answers #3 Chi-squared test

Questions and Answers Sheet 3 Chi-squared test Question #1 You are conducting a test of homogeneity for the claim that two different populations have the same proportions of the following two characteristics. Here is the sample data. Category A ( B Population 1 Population 2 11 14 ) 56 105 What is the chi-square test-statistic for this data? 𝜒2 = Report all answers accurate to three decimal places. Answer: We have been given the observed frequencies ( 𝑂𝑖 ) as, Category A ( B Total Population 1 Population 2 Total 11 14 56 105 67 119 25 161) 186 We compute the expected frequency as, Expected frequency = (Column total×Row total) Grand total The expected frequencies ( 𝐸𝑖 ) are given by, Category A ( B Total Population 1 Population 2 Total 67  25 119  25 = 9.005376 = 15.99462 25 186 186 ) 67 161 119 161 = 57.99462 = 103.0054 161 186 186 67 119 186 The chi-square test statistic is computed as, 𝜒2 = ∑ 𝑖 = (𝑂𝑖 − 𝐸𝑖 )2 𝐸𝑖 (11 − 9.005376)2 (56 − 57.99462)2 (14 − 15.99462)2 (105 − 103.0054)2 + + + 9.005376 57.99462 15.99462 103.0054 =0.798 Thus, the chi-square test statistic is 0.798. Question #2 You are conducting a test of homogeneity for the claim that two different populations have the same proportions of the following two characteristics. Here is the sample data. Category A ( B Population 1 Population 2 36 125 ) 22 68 What is the chi-square test-statistic for this data? 𝜒2 = Report all answers accurate to three decimal places. Answer: The Chi-square test of independence is used to determine if there is a significant relationship between two nominal variables. The chi-square test of independence can be used to examine this relationship. The null hypothesis for this test is that there is no relationship between gender and empathy. Category A ( B Column total Population 1 Population 2 36 125 22 68 58 193 Row total 161 ) 90 N = 251 Chi-square cal = = N ((ad − bc) ) / (a + b)(b + d )(d + c )(c + a ) = 251((36*68) − (22*125)) /(36 + 125)(125 + 68)(68 + 22)(22 + 36) Chi square =0.1411 Question #3 You are conducting a test of independence for the claim that there is an association between the row variable and the column variable. ( A B X 35 53 Y 29 38 Z 6 ) 5 The expected observations for this table would be X ( Y Z A B ) What is the chi-square test-statistic for this data? (2 decimals) 2 = Report all answers accurate to two decimal places. Answer: From the given information, it is clear that, The observed data are, X 35 53 88 A ( B Total Y 29 38 67 Z 6 5 11 Total 70 96 ) 166 We have to find the chi-square test statistic , Formula for the chi-square test statistic is, 𝜒2 = ∑ (𝑂𝑖 −𝐸𝑖 )2 𝐸𝑖 Where O-i is observed value and 𝐸𝑖 is expected value First we have to calculate the expected values, Formula for the expected value is, 𝐸= 𝑅𝑜𝑤 𝑡𝑜𝑡𝑎𝑙×𝑐𝑜𝑙𝑢𝑚𝑛 𝑡𝑜𝑡𝑎𝑙 A ( B 𝑇𝑜𝑡𝑎𝑙 X Y Z 70(88) 70(67) 70(11) = 37.11 = 28.25 = 4.64 166 166 166 ) 96(88) 96(77) 96(11) = 50.89 = 38.75 = 6.36 166 166 166 Now, we have to calculate the chi-square test statistic, That is, X Y A ( B Z (35 − 37.11) (29 − 28.25) (6 − 4.64) 2 = 0.1199 = 0.0265 = 1.8496 37.11 28.25 4.64 ) (53 − 50.89) 2 (38 − 38.75) 2 (5 − 6.36) 2 = 0.0875 = 0.0145 = 0.2908 50.89 38.75 6.36 2 2 Question #4 You are conducting a test of independence for the daim that there is an association between morning beverage and age. ( less that 25 25 or older Juice 19 27 Coffee 18 30 Pop 5 2 Tea 32 ) 43 What is the chi-square test-statistic and the corresponding p-value for this data? 𝜒 2 =?, [three decimal accuracy] p-value=?, [three decimal accuracy] Answer: a) Null and alternative hypothesis: 𝐻0 : There is no association between the morning beverage and age. 𝐻𝛼 : There is association between the morning beverage and age. b) Observed frequency (O) table: less that 25 ( 25 or older Total Juice 19 27 46 Coffee 18 30 48 Pop 5 2 7 Tea 32 43 753 74 102) 176 The Expected frequency (E) table: This is a Chi-square test Expected frequencies and Test statistics can be calculated as follow: Expected frequency = (row total*column total)/Grand total less that 25 ( 25 or older Total Juice 19.341 26.659 46 Coffee 20.182 27.818 48 Chi-square test statistic 𝜒 2 = ∑ Pop Tea 2.943 31.534 74 4.057 43.466 102) 7 753 176 (𝑂−𝐸)2 𝐸 The Chi-square test statistic = 28.2658 4) df=n-1=3-1=2, 𝛼 = 0.05 . The P-value is 0.00001 5) P-value < 0.05 we reject the null hypothesis. c) Chi-square test statistic (𝜒)2 : Observed frequency (O) 19 18 5 32 ( 27 30 2 43 Expected frequency (E) 19.34 20.18 2.94 31.53 26.66 27.82 4.06 43.47 (O-E) 2 (O-E) 2 /E 0.1156 0.0134 4.7524 0.2355 4.2436 1.4434 0.2209 0.0070 0.1156 0.0043 ) 4.7524 0.1708 4.2436 1.0452 0.2209 0.0051 2 Chi-square test statistic (  ) = 2.9247 degree of freedom (df) = (𝑟𝑜𝑤 − 1) ∗ (𝑐𝑜𝑙𝑢𝑚𝑛 − 1) = (2 − 1) ∗ (4 − 1) =3 significance level 𝛼 = 0.05 d) P-value = 0.403 e) SInce P- value > significance level (𝛼),we fail to reject the null hypothesis. There is no association between the morning beverage and age. Question #5 You intend to conduct a test of homogeneity for a contingency table with 8 categories in the column variable and 2 categories in the row variable. You collect data from 352 subjects. What are the degrees of freedom for the 𝜒 2 distribution for this test? Answer: Provided information is, a test of homogeneity for a contingency table is conducted. The contingency table is having 8 categories in column and 2 categories in row. For test of homogeneity chi-square test is used, the chi-square test statistic follows chi-square distribution with degrees of freedom as (number of rows-1)(number of columns-1) = (8-1)(2-1) = 7 Thus, degrees of freedom for chi-square distribution is 7. Question #6 You obtain a sample chi-square value of 𝜒 2 = −5.49. On the basis of this value, you know that a. there is a negative association between your two variables. b. the observed frequencies are higher than the expected frequencies across at least half of your cells. c. you have made a calculation error; chi-square values cannot be negative. d. the observed frequencies are lower than the expected frequencies across all cells. Answer: Given: The observed/calculated value of 𝜒 2 = −5.49. The value of the chi-square test statistic is calculated as: 𝜒2 = ∑ (𝑂−𝐸)2 𝐸 Explanation to correct option: c. The observed value is calculated as the sum of the square of the difference between the observed and expected frequencies divided by the expected frequency. Therefore, the square will make the sum positive, and hence, the observed value can not be negative and it is clear that some errors in the calculations have been made and the value cannot be negative. Thus, option c. is correct. Explanation to incorrect options: a. The calculated value of the Chi-square test helps in finding the relationship and independence between the two variables. As the value of the test statistic is the sum of the square of the difference between observed and expected frequency. Therefore, the observed value can not be negative, and hence, there will not be a negative association between the variables. Thus, option a. is incorrect. b. The given observed value of the Chi-square test is -5.49. Therefore, It cannot be said that the observed frequencies are higher than the expected frequencies across at least half of your cells because the value will be calculated by squaring the difference between the frequencies. Hence, the higher value of observed frequencies or expected frequencies does not affect the sign of the calculated test statistics. Thus, option b. is incorrect.