Question

Find the expected count and the contribution to the chi-square statistic for the left-parenthesis C comma F right-parenthesis cell in the two-way table below. Upper D Upper E Upper F Upper G Total Upper A 39 30 36 36 141 Upper B 77 88 70 55 290 Upper C 21 37 27 28 113 Total 137 155 133 119 544 Round your answer for the excepted count to one decimal place, and your answer for the contribution to the chi-square statistic to three decimal places. EXPECTED COUNT: CONTRIBUTION TO THE CHI-SQUARE STATISTICS:

Answer 1

Given table data is as below MATRIX col1 col2 col3 col4 TOTALS row 1 39 30 36 36 A1 row 2 77 88 70 55 A2 row 3 21 37 27 28 A3 TOTALS 137 155 133 119 N = 544 ------------------------------------------------------------------ calculation formula for E table matrix E-TABLE col1 col2 col3 col4 row 1 row1*col1/N row1*col2/N row1*col3/N row1*col4/N row 2 row2*col1/N row2*col2/N row2*col3/N row2*col4/N row 3 row3*col1/N row3*col2/N row3*col3/N row3*col4/N ------------------------------------------------------------------ expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 col3 col4 row 1 35.509 40.175 34.472 30.844 row 2 73.033 82.629 70.901 63.438 row 3 28.458 32.197 27.627 24.719 ------------------------------------------------------------------ calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 39 35.509 3.491 12.187 0.343 30 40.175 -10.175 103.531 2.577 36 34.472 1.528 2.335 0.068 36 30.844 5.156 26.584 0.862 77 73.033 3.967 15.737 0.215 88 82.629 5.371 28.848 0.349 70 70.901 -0.901 0.812 0.011 55 63.438 -8.438 71.2 1.122 21 28.458 -7.458 55.622 1.955 37 32.197 4.803 23.069 0.716 27 27.627 -0.627 0.393 0.014 28 24.719 3.281 10.765 0.435 chisqr^2 o = 8.667 ------------------------------------------------------------------ set up null vs alternative as null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent level of significance, alpha = 0.05 from standard normal table, chi square value at right tailed, chisqr^2 alpha/2 =12.592 since our test is right tailed,reject Ho when chisqr^2 o > 12.592 we use test statistic chisqr^2 o = Σ(Oi-Ei)^2/Ei from the table , chisqr^2 o = 8.667 critical value the value of |chisqr^2 alpha| at los 0.05 with d.f (r-1)(c-1)= ( 3 -1 ) * ( 4 - 1 ) = 2 * 3 = 6 is 12.592 we got | chisqr^2| =8.667 & | chisqr^2 alpha | =12.592 make decision hence value of | chisqr^2 o | < | chisqr^2 alpha | and here we do not reject Ho chisqr^2 p_value =0.193 ANSWERS --------------- null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent test statistic: 8.667 critical value: 12.592 =12.6 p-value:0.193 decision: do not reject Ho