Some people develop an allergy to the (fictional) antibiotic exocillin after exp
ID: 3324430 • Letter: S
Question
Some people develop an allergy to the (fictional) antibiotic exocillin after exposure to it. A researcher wonders whether susceptibility to this allergy depends on blood type. A random sample yields the following counts for blood types O, A, B, AB.
O A B AB
Allergic 10 12 9 2
non-allergic 10 8 7 5
Explain how you can test whether difference in blood type indeed influences susceptibility to allergy. Perform the test, report the results, and interpret them.
Explanation / Answer
Given table data is as below MATRIX col1 col2 col3 col4 TOTALS row 1 10 12 9 2 33 row 2 10 8 7 5 30 TOTALS 20 20 16 7 N = 63 ------------------------------------------------------------------ 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 ------------------------------------------------------------------ expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 col3 col4 row 1 10.476 10.476 8.381 3.667 row 2 9.524 9.524 7.619 3.333 ------------------------------------------------------------------ calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 10 10.476 -0.476 0.227 0.022 12 10.476 1.524 2.323 0.222 9 8.381 0.619 0.383 0.046 2 3.667 -1.667 2.779 0.758 10 9.524 0.476 0.227 0.024 8 9.524 -1.524 2.323 0.244 7 7.619 -0.619 0.383 0.05 5 3.333 1.667 2.779 0.834 ^2 o = 2.2 ------------------------------------------------------------------ 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, = 0.05 from standard normal table, chi square value at right tailed, ^2 /2 =7.815 since our test is right tailed,reject Ho when ^2 o > 7.815 we use test statistic ^2 o = (Oi-Ei)^2/Ei from the table , ^2 o = 2.2 critical value the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 4 - 1 ) = 1 * 3 = 3 is 7.815 we got | ^2| =2.2 & | ^2 | =7.815 make decision hence value of | ^2 o | < | ^2 | and here we do not reject Ho ^2 p_value =0.532 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: 2.2 critical value: 7.815 p-value:0.532 decision: do not reject HoRelated Questions
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