- Find Test Statistic - Find P-Value Car manufacturers are interested in whether
ID: 3363887 • Letter: #
Question
- Find Test Statistic
- Find P-Value
Car manufacturers are interested in whether there is a relationship between the size of car an individual drives and the number of people in the driver's family (that is, whether car size and family size are independent). To test this, suppose that 794 car owners were randomly surveyed with the following results. Conduct a test for independence at the 5% level. Family SizeSub & CompactMid-size Full-sizeVan & Truck 35 50 51 29 19 19 20 19 39 69 100 69 35 80 90 70 5+Explanation / Answer
Given table data is as below MATRIX col1 col2 col3 TOTALS TOTALS row 2 19 35 39 35 128 row 3 19 50 69 80 218 row 3 20 51 100 90 261 row 4 19 29 69 70 187 TOTALS 77 165 277 275 794 ------------------------------------------------------------------ calculation formula for E table matrix E-TABLE col1 col2 col3 col4 row 2 row1*col1/N row1*col2/N row1*col3/N row1*col4/N row 3 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 row 4 row4*col1/N row4*col2/N row4*col3/N row4*col4/N ------------------------------------------------------------------ expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 col3 col4 row 2 12.413 26.599 44.655 44.332 row 3 21.141 45.302 76.053 75.504 row 3 25.311 54.238 91.054 90.397 row 4 18.135 38.86 65.238 64.767 ------------------------------------------------------------------ calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 19 12.413 6.587 43.389 3.495 35 26.599 8.401 70.577 2.653 39 44.655 -5.655 31.979 0.716 35 44.332 -9.332 87.086 1.964 19 21.141 -2.141 4.584 0.217 50 45.302 4.698 22.071 0.487 69 76.053 -7.053 49.745 0.654 80 75.504 4.496 20.214 0.268 20 25.311 -5.311 28.207 1.114 51 54.238 -3.238 10.485 0.193 100 91.054 8.946 80.031 0.879 90 90.397 -0.397 0.158 0.002 19 18.135 0.865 0.748 0.041 29 38.86 -9.86 97.22 2.502 69 65.238 3.762 14.153 0.217 70 64.767 5.233 27.384 0.423 ^2 o = 15.825 ------------------------------------------------------------------ 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 =16.919 since our test is right tailed,reject Ho when ^2 o > 16.919 we use test statistic ^2 o = (Oi-Ei)^2/Ei from the table , ^2 o = 15.825 critical value the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 4 -1 ) * ( 4 - 1 ) = 3 * 3 = 9 is 16.919 we got | ^2| =15.825 & | ^2 | =16.919 make decision hence value of | ^2 o | < | ^2 | and here we do not reject Ho ^2 p_value =0.071 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: 15.825 critical value: 16.919 p-value:0.071 decision: do not reject HoRelated Questions
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