Question

0 The following data represent the level Education of health and the level of education for Not a H.S. graduate a random sample of 1687 residents H.S. graduate Complete parts (a) and (b) below. Some college Bachelor Degree or higher Excellent 74 92 63 75 Good 134 121 148 137 Fair 109 95 80 105 Poor 118 93 126 117 (a) Does the sample evidence suggest that level of education and health are independent at the a=0.05 level of significance? Conduct a P-value hypothesis test. State the hypotheses. Choose the correct answer below. O A. Ho: H4 = E, and H2 = E, and H3 = Ez and H=E4 Hy At least one mean is different from what is expected. OB. Ho Level of education and health are independent. Hy Level of education and health are dependent OC. Ho P4 = P2 = P3 Hy At least one of the proportions are not equal. Calculate the test statistic. y? 1 Click to select your answer(s). Omul

Answer 1

Given table data is as below MATRIX col1 col2 col3 TOTALS TOTALS row 2 74 134 109 118 435 row 3 92 121 95 93 401 row 3 63 148 80 126 417 row 4 75 137 105 117 434 TOTALS 304 540 389 454 1687 ------------------------------------------------------------------ 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 frequencies calculated by applying E - table matrix formulae E-TABLE col1 col2 col3 col4 row 2 78.388 139.241 100.305 117.066 row 3 72.261 128.358 92.465 107.916 row 3 75.144 133.48 96.155 112.222 row 4 78.207 138.921 100.075 116.797 ------------------------------------------------------------------ calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 74 78.388 -4.388 19.255 0.246 134 139.241 -5.241 27.468 0.197 109 100.305 8.695 75.603 0.754 118 117.066 0.934 0.872 0.007 92 72.261 19.739 389.628 5.392 121 128.358 -7.358 54.14 0.422 95 92.465 2.535 6.426 0.069 93 107.916 -14.916 222.487 2.062 63 75.144 -12.144 147.477 1.963 148 133.48 14.52 210.83 1.579 80 96.155 -16.155 260.984 2.714 126 112.222 13.778 189.833 1.692 75 78.207 -3.207 10.285 0.132 137 138.921 -1.921 3.69 0.027 105 100.075 4.925 24.256 0.242 117 116.797 0.203 0.041 0 ᴪ^2 o = 17.498 ------------------------------------------------------------------ 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, α = 0.05 from standard normal table, chi square value at right tailed, ᴪ^2 α/2 =16.919 since our test is right tailed,reject Ho when ᴪ^2 o > 16.919 we use test statistic ᴪ^2 o = Σ(Oi-Ei)^2/Ei from the table , ᴪ^2 o = 17.498 critical value the value of |ᴪ^2 α| at los 0.05 with d.f (r-1)(c-1)= ( 4 -1 ) * ( 4 - 1 ) = 3 * 3 = 9 is 16.919 we got | ᴪ^2| =17.498 & | ᴪ^2 α | =16.919 make decision hence value of | ᴪ^2 o | > | ᴪ^2 α| and here we reject Ho ᴪ^2 p_value =0.041 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: 17.498 critical value: 16.919 p-value:0.041 decision: reject Ho

we have enough evidence to support the claim that level of education and health are associated.

b.

option:A is suitable Bar diagram as per the data given.