Question

А 10 40 B 20 30 у b The contingency table shown to the right gives a cross-classification of a random sample of values for two variables, x and y, of a population a. Find the expected frequencies. Note: You will first need to compute the row totals, column totals, and grand total. b. Determine the value of the chi-square statistic. c. Decide at the 5% significance level whether the data provide sufficient evidence to conclude that the two variables are associated a. Compute the expected frequencies and add them into the table given below. X A 10 B 20 40 30 (Round to two decimal places as needed.)

Answer 1

Answer Date: 31/07/2020 Part 1 a) The two-way contingency table is, yx a A 10 40 50 B 20 30 50 Row total 30 70 100 b Column total The expected frequency for ya and xA is, Row total > Column total Expected frequency = N 30 x 50 100 = 15

The expected frequency for ya and xB is, Row total > Column total Expected frequency = N 30 x 50 100 = 15 The expected frequency for yb and xA is, Expected frequency Row total x Column total N 70 x 50 100 = 35 =

The expected frequency for yb and xB is, Expected frequency = Row total > Column total N 70 x 50 100 = 35 = The expected frequencies table is, y\x a A 15 35 50 B 15 35 50 Row total 30 70 100 b Column total b)

The chi-square test statistic is, x? = 2 (f. - $.)? fe (10-15) (20 – 15)2 (40 – 35)2 + 15 15 35 = 4.762 (20 - 35) + + 35 The chi-square test statistic is 4.762). The null and alternative hypothesis is, The null hypothesis is that the two variables are not associated. The alternative hypothesis is that the two variables are associated. The chi-square critical value is,

First, compute the degree of freedom then find chi-square critical value. The information in given matrix is two rows and two columns for compute degrees of freedom. d.f. = (r - 1)(c-1) = (2-1) (2-1) = 1 From the critical value table, the degree of freedom which is 1 and the significance level 0.05, so chi-square critical values is 3.841. The chi-square critical value is 3.841 Decision The conclusion is that chi-square test statistic 4.762 greater than chi-square critical value 3.841, so the null hypothesis is rejected at 5% level of significance. The does provide sufficient evidence to indicate that the two variables are associated. The result is statistically significant.

Part 2 a) The two-way contingency table is, ylx a A 40 20 60 B 50 10 60 Row total 90 30 120 b Column total The expected frequency for ya and xA is, Expected frequency Row total > Column total N 90 x 60 120 = 45

The expected frequency for ya and xB is, Expected frequency = Row total > Column total N 90 x 60 120 = 45 = The expected frequency for yb and xA is, Expected frequency Row total > Column total N 30 x 60 120 = 15 =

The expected frequency for yb and xB is, Row total > Column total Expected frequency = N 30 x 60 120 = 15 The expected frequencies table is, уіх a 45 A B 45 15 15 60 60 Row total 90 30 b Column total 120 b)

The chi-square test statisticis, χ =Σ (fo- f.)? fe (40 - 45)2 (50 – 45)(20-15)2 (10–15)? 15 15 = 4.444 + + + 45 45 The chi-square test statistic is 4.444 c) The null and alternative hypothesis is, The null hypothesis is that the two variables are not associated. The alternative hypothesis is that the two variables are associated. The chi-square critical value is,

First, compute the degree of freedom then find chi-square critical value. The information in given matrix is two rows and two columns for compute degrees of freedom. d.f. = (r - 1)*(c-1) = (2-1) (2-1) = 1 From the critical value table, the degree of freedom which is 1 and the significance level 0.05, so chi-square critical values is 3.841. The chi-square critical value is 3.841 Decision The conclusion is that chi-square test statistic 4.444 greater than chi-square critical value 3.841, so the null hypothesis is rejected at 5% level of significance. The does provide sufficient evidence to indicate that the two variables are associated. The result is statistically significant.