(a) find the test statistic, ROUNDING TO 3 DECIMAL PLACES. Also find the p-value
ID: 3320232 • Letter: #
Question
(a) find the test statistic, ROUNDING TO 3 DECIMAL PLACES.
Also find the p-value using chi-square distribution.
(b) construct a conditional distribution of health level of education with above data . ROUND TO 3 DECIMALS PLACES PLEASE.
The following data represent the level of health and the level of education for a random sample of 1513 residents. Complete parts (a) and (b) below. Education Excellent 58 73 51 60 Good 130 105 108 Fair 95 118 97 79 Poor 129 93 106 94 Not a H.S. graduate H.S. graduate Some college Bachelor Degree or higher (a) Does the sample evidence suggest that level of education and health are independent at the = 0.05 level of significance? Conduct a P-value hypothesis test. State the hypotheses. Choose the correct answer below H1: At least one mean is different from what is expected. H1: At least one of the proportions are not equal H1: Level of education and health are dependent O B. Ho: p1 = p2 =P3 c. Hg:Level of education and health are independent. Calculate the test statistic. (Round to three decimal places as needed.) Enter your answer in the answer box and then click Check Answer. parts Clear All Check Answer remainingExplanation / Answer
option :C
Given table data is as below MATRIX col1 col2 col3 TOTALS TOTALS row 2 58 130 95 129 412 row 3 73 105 118 93 389 row 3 51 108 97 106 362 row 4 60 117 79 94 350 TOTALS 242 460 389 422 1513 ------------------------------------------------------------------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 65.898 125.261 105.927 114.913 row 3 62.219 118.268 100.014 108.498 row 3 57.901 110.059 93.072 100.968 row 4 55.981 106.411 89.987 97.621 ------------------------------------------------------------------
calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 58 65.898 -7.898 62.378 0.947 130 125.261 4.739 22.458 0.179 95 105.927 -10.927 119.399 1.127 129 114.913 14.087 198.444 1.727 73 62.219 10.781 116.23 1.868 105 118.268 -13.268 176.04 1.488 118 100.014 17.986 323.496 3.235 93 108.498 -15.498 240.188 2.214 51 57.901 -6.901 47.624 0.823 108 110.059 -2.059 4.239 0.039 97 93.072 3.928 15.429 0.166 106 100.968 5.032 25.321 0.251 60 55.981 4.019 16.152 0.289 117 106.411 10.589 112.127 1.054 79 89.987 -10.987 120.714 1.341 94 97.621 -3.621 13.112 0.134 ^2 o = 16.882 ------------------------------------------------------------------
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 = 16.882
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| =16.882 & | ^2 | =16.919
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.051
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: 16.882
critical value: 16.919
p-value:0.051
decision: do not reject Ho
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