Calculating the 1-way χ2 and 2-way χ2 Statistics Using statistics in small-scale language education research Jean Turner © Taylor & Francis 2014

1-way χ2 The fabricated Halloween data are used in this discussion—questionnaire data from 300 students regarding whether they plan to attend their school Halloween party. The first step in statistical logic is… © Taylor & Francis 2014

State the formal research hypotheses. Null: There is no statistically significant pattern of response to the questionnaire. (There is no prevailing tendency.) Alternative: There is a statistically significant pattern of response to the questionnaire. (There is a prevailing tendency.) © Taylor & Francis 2014

Set alpha. Alpha is set at .01 because people may make decisions about allocation of resources based on the findings. © Taylor & Francis 2014

Propose the statistic to use in the analysis. Hmmm…. There’s a single nominal variable. The data are frequency counts. The researcher wants to know if there’s a statistically significant pattern of response to the question. Sounds like the 1-way χ2 is appropriate. © Taylor & Francis 2014

Collect the data. Here’s a summary table of the data. Number of respondents who indicated “yes” Number of respondents who indicated “no” Number of participants who indicated “I haven’t decided yet” Number of participants who chose not to respond to the question   61 65 84 90 © Taylor & Francis 2014

Check the assumptions. The independent variable is nominal. OK The data are frequency counts. OK Each piece of information is independent from every other (no influenced votes!) This seems to be OK. Each individual is included in only one level of a variable. OK df > 1, so all expected frequencies must be greater than or equal to 5. (Three hundred people participated; expected frequency is 300/4 = 75. We’re OK with this assumption too.)   © Taylor & Francis 2014

Calculate the observed value of the statistic. Cell fo fe fo ‒ fe (fo ‒ fe)2 (fo ‒ fe)2 / fe A 61 75 B 65 C 84 D 90   © Taylor & Francis 2014

Cell fo fe fo ‒ fe (fo ‒ fe)2 (fo ‒ fe)2 / fe A 61 75 ‒14 B 65 ‒10 C 84 +9 D 90 +15   © Taylor & Francis 2014

Cell fo fe fo ‒ fe (fo ‒ fe)2 (fo ‒ fe)2 / fe A 61 75 ‒14 196 B 65 ‒10 100 C 84 +9 81 D 90 +15 225   © Taylor & Francis 2014

Cell fo fe fo ‒ fe (fo ‒ fe)2 (fo ‒ fe)2 / fe A 61 75 ‒14 196 196/75 = 2.613 B 65 ‒10 100 100/75 = 1.33 C 84 +9 81 81/75 = 1.08 D 90 +15 225 225/75 = 3   © Taylor & Francis 2014

The sum of the last column is χ2observed. © Taylor & Francis 2014

Calculating 2-way χ2 Research question: Is there a statistically significant relationship between teachers’ degree of scoring agreement depending on how well they know their students? © Taylor & Francis 2014

State the formal research hypotheses. Null hypothesis: There is no statistically significant relationship between the variables. Alternative hypothesis: There is a statistically significant relationship between the variables. © Taylor & Francis 2014

Set alpha. Alpha is set at .01 because the researcher believes important decisions may be made on the basis of the findings. © Taylor & Francis 2014

Propose the statistic to use in the analysis. There are two nominal variables. The data are frequency counts. The researcher wants to know if there’s a statistically significant relationship between the two variables. Sounds like the 2-way χ2 statistic is appropriate. © Taylor & Francis 2014

Check the assumptions. The independent variable and the moderator variable are nominal. OK The data are frequency counts. OK Each piece of information is independent of every other. (Depends on how the data were collected.) I think this is OK. Each individual is included in only one level of a variable. OK df > 1, so all expected frequencies must be greater than or equal to 5. Must be calculated and checked. © Taylor & Francis 2014

Calculate the observed value. Two raters agree T who knows Ss rates higher T who knows Ss rates lower 1st semester Ss 13 (cell A) 6 (cell B) 5 (cell C) 2nd semester Ss 11 (cell D) 17 (cell E) (cell F) 3rd semester Ss (cell G) 10 (cell H) 19 (cell I) © Taylor & Francis 2014

T who knows Ss rates higher T who knows Ss rates lower Two raters agree T who knows Ss rates higher T who knows Ss rates lower Marginals for the rows 1st semester Ss 13 (cell A) 6 (cell B) 5 (cell C) 24 2nd semester Ss 11 (cell D) 17 (cell E) (cell F) 41 3rd semester Ss (cell G) 10 (cell H) 19 (cell I) 35 Marginals for the columns 30 33 37 N = 100 © Taylor & Francis 2014

Cell fo fe fo ‒ fe (fo ‒ fe)2 (fo ‒ fe)2/ fe A 13 B 6 C 5 D 11 E 17 F G H 10 I 19 © Taylor & Francis 2014

Cell fo fe fo ‒ fe (fo ‒ fe)2 (fo ‒ fe)2/ fe A 13 (24)(30) / 100 = B 6 (24)(33) / 100 = C 5 (24)(37) / 100 = D 11 (41)(30) / 100 = E 17 (41)(33) / 100 = F (41)(37) / 100 = G (35)(30) / 100 = H 10 (35)(33) / 100 = I 19 (35)(37) / 100 = © Taylor & Francis 2014

Cell fo fe fo ‒ fe (fo ‒ fe)2 (fo ‒ fe)2/ fe A 13 (24)(30) / 100 = 7.20 B 6 (24)(33) / 100 = 7.92 C 5 (24)(37) / 100 = 8.88 D 11 (41)(30) / 100 = 12.30 E 17 (41)(33) / 100 = 13.53 F (41)(37) / 100 = 15.17 G (35)(30) / 100 = 10.50 H 10 (35)(33) / 100 = 11.55 I 19 (35)(37) / 100 = 12.95 © Taylor & Francis 2014

Cell fo fe fo ‒ fe (fo ‒ fe)2 (fo ‒ fe)2/ fe A 13 (24)(30) / 100 = 7.20 13 – 7.20 = 5.80 B 6 (24)(33) / 100 = 7.92 6 – 7.92 = ‒1.92 C 5 (24)(37) / 100 = 8.88 5 – 8.88 = ‒3.88 D 11 (41)(30) / 100 = 12.30 11 – 12.30 = ‒1.30 E 17 (41)(33) / 100 = 13.53 17 – 13.53 = 3.47 F (41)(37) / 100 = 15.17 13 – 15.17 = ‒2.17 G (35)(30) / 100 = 10.50 6 – 10.50 = ‒4.50 H 10 (35)(33) / 100 = 11.55 10 – 11.55 = ‒1.55 I 19 (35)(37) / 100 = 12.95 19 – 12.95 = 6.05 © Taylor & Francis 2014

Cell fo fe fo ‒ fe (fo ‒ fe)2 (fo ‒ fe)2/ fe A 13 (24)(30) / 100 =7.20 13 – 7.20 = 5.80 33.64 B 6 (24)(33) / 100 = 7.92 6 – 7.92 = ‒1.92 3.69 C 5 (24)(37) / 100 = 8.88 5 – 8.88 = ‒3.88 15.05 D 11 (41)(30) / 100 = 12.30 11 – 12.30 = ‒1.30 1.69 E 17 (41)(33) / 100 = 13.53 17 – 13.53 = 3.47 12.04 F (41)(37) / 100 = 15.17 13 – 15.17 = ‒2.17 4.71 G (35)(30) / 100 = 10.50 6 – 10.50 = ‒4.50 20.25 H 10 (35)(33) / 100 = 11.55 10 – 11.55 = ‒1.55 2.40 I 19 (35)(37) / 100 = 12.95 19 – 12.95 = 6.05 36.60 © Taylor & Francis 2014

Cell fo fe fo ‒ fe (fo ‒ fe)2 (fo ‒ fe)2/ fe A 13 (24)(30) / 100 =7.20 13 – 7.20 = 5.80 33.64 33.64 / 7.20 = 4.67 B 6 (24)(33) / 100 = 7.92 6 – 7.92 = ‒1.92 3.69 3.69 / 7.92 = .47 C 5 (24)(37) / 100 = 8.88 5 – 8.88 = ‒3.88 15.05 15.05 / 8.88 = 1.69 D 11 (41)(30) / 100 = 12.30 11 – 12.30 = ‒1.30 1.69 1.69 / 12.30 = .14 E 17 (41)(33) / 100 = 13.53 17 – 13.53 = 3.47 12.04 12.04 / 13.53 = .89 F (41)(37) / 100 = 15.17 13 – 15.17 = ‒2.17 4.71 4.71 / 15.17 = .31 G (35)(30) / 100 = 10.50 6 – 10.50 = -4.50 20.25 20.25 / 10.50 = 1.93 H 10 (35)(33) / 100 = 11.55 10 – 11.55 = ‒1.55 2.40 2.40 / 11.55 = .21 I 19 (35)(37) / 100 = 12.95 19 – 12.95 = 6.05 36.60 36.60 / 12.95 = 2.83 © Taylor & Francis 2014

Cell fo fe fo ‒ fe (fo ‒ fe)2 (fo ‒ fe)2/ fe A 13 (24)(30) / 100 =7.20 13 – 7.20 = 5.80 33.64 33.64 / 7.20 = 4.67 B 6 (24)(33) / 100 = 7.92 6 – 7.92 = ‒1.92 3.69 3.69 / 7.92 = .47 C 5 (24)(37) / 100 = 8.88 5 – 8.88 = -3.88 15.05 15.05 / 8.88 = 1.69 D 11 (41)(30) / 100 = 12.30 11 – 12.30 = ‒1.30 1.69 1.69 / 12.30 = .14 E 17 (41)(33) / 100 = 13.53 17 – 13.53 = 3.47 12.04 12.04 / 13.53 = .89 F (41)(37) / 100 = 15.17 13 – 15.17 = ‒2.17 4.71 4.71 / 15.17 = .31 G (35)(30) / 100 = 10.50 6 – 10.50 = ‒4.50 20.25 20.25 / 10.50 = 1.93 H 10 (35)(33) / 100 = 11.55 10 – 11.55 = ‒1.55 2.40 2.40 / 11.55 = .21 I 19 (35)(37) / 100 = 12.95 19 – 12.95 = 6.05 36.60 36.60 / 12.95 = 2.83 χ2 = ∑(fo ‒ fe)2/ fe = 13.14 © Taylor & Francis 2014

Using R to calculate 2-way χ2 Enter the values for the first variable. x = c(1,1,1,1,1,3,2,2,1,1,1,1,2,1,2,3,2,2,3,1,3,2,2,2,1,3,2,1,3, 2,1,2,2,3,1,1,3,2,3,3,2,2,2,2,3,2,3,2,3,2,3,2,2,1,3,3,2,1,1,2,2,2, 2,3,3,3,3,3,3,2,3,3,3,3,2,2,1,2,2,1,2,2,1,3,3,2,2,2,2,3,3,3,3,2,3, 2,3,1,1,3) Enter the values for the second variable. y = c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1, 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2, 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3, 3,3,3,3)   chisq.test (x,y) χ2 = 13.1338, df = 4, p-value = 0.01064 © Taylor & Francis 2014