Scenario-B: In the following table, individuals are cross-classified by their ag

Question

Scenario-B: In the following table, individuals are cross-classified by their age group and income level.

USE THE SCENARIO ABOVE TO ANSWER ALL QUESTIONS.

1. To test that age group and income are independent, the null and alternative hypothesis are :

2. For the chi-square test of independence, the degrees of freedom are:

D. 9

3. For the chi-square test of independence, the p value is:

D. 0.028

4. Using the p-value approach and = 0.05, the decision and conclusion are:

D. do not reject the null hypothesis; age and income are independent

5. Using the p-value approach and = 0.01, the decision and conclusion are:

   Income Age Low Medium High 21–35 120 100 75 36–50 150 160 100 51–65 160 180 160

Explanation / Answer

Given table data is as below

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calculation formula for E table matrix

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expected frequecies calculated by applying E - table matrix formulae

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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

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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, = 0.01
from standard normal table, chi square value at right tailed, ^2 /2 =13.277
since our test is right tailed,reject Ho when ^2 o > 13.277
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 10.838
critical value
the value of |^2 | at los 0.01 with d.f (r-1)(c-1)= ( 3 -1 ) * ( 3 - 1 ) = 2 * 2 = 4 is 13.277
we got | ^2| =10.838 & | ^2 | =13.277
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.028
ANSWERS
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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: 10.838
critical value: 13.277
p-value:0.028
decision: do not reject Ho                                      

col1 col2 col3 row 1 120 100 75 295 row 2 150 160 100 410 row 3 160 180 160 500 TOTALS 430 440 335 N = 1205

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