Question

For each of the following situations give the degrees of freedom and an appropriate bound on the P-value (give the exact value if you have software available) for the χ2 statistic for testing the null hypothesis of no association between the row and column variables.

(a) A 2 by 2 table with χ2 = 0.89.

df =

P-value =

(b) A 4 by 4 table with χ2 = 18.96.

df =

P-value =

(c) A 2 by 8 table with χ2 = 23.39.

df =

P-value =

(d) A 5 by 3 table with χ2 = 12.72.

df =

P-value =

Answer 1

Answer:

a) The statistic calculated is $\chi^2 = 0.89$

For this case the degrees of freedom are given by:

$df =(2-1)*(2-1)= 1$

And the p value would be given by:

$p_v = P(\chi^2_{1} >0.89)=0.345$

And we can use the folloing excel code to find it: "=1-CHISQ.DIST(0.89,1,TRUE)"

b) The statistic calculated is $\chi^2 = 18.96$

For this case the degrees of freedom are given by:

$df =(4-1)*(4-1)= 9$

And the p value would be given by:

$p_v = P(\chi^2_{9} >18.96)=0.0255$

And we can use the folloing excel code to find it: "=1-CHISQ.DIST(18.96,9,TRUE)"

c)The statistic calculated is $\chi^2 = 23.39$

For this case the degrees of freedom are given by:

$df =(2-1)*(8-1)= 7$

And the p value would be given by:

$p_v = P(\chi^2_{7} >23.39)=0.0015$

And we can use the folloing excel code to find it: "=1-CHISQ.DIST(23.39,7,TRUE)"

d) The statistic calculated is $\chi^2 = 12.72$

For this case the degrees of freedom are given by:

$df =(5-1)*(3-1)= 8$

And the p value would be given by:

$p_v = P(\chi^2_{8} >12.72)=0.1219$

And we can use the folloing excel code to find it: "=1-CHISQ.DIST(12.72,8,TRUE)"

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no association (independence) in the between row and column variables

H1: There is association in the between row and column variables

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Where O means obsevd values and E expected values.

The degrees of freedom can be calculated with :

$df = (rows-1)*(cols-1)$

Part a

The statistic calculated is $\chi^2 = 0.89$

For this case the degrees of freedom are given by:

$df =(2-1)*(2-1)= 1$

And the p value would be given by:

$p_v = P(\chi^2_{1} >0.89)=0.345$

And we can use the folloing excel code to find it: "=1-CHISQ.DIST(0.89,1,TRUE)"

Part b

The statistic calculated is $\chi^2 = 18.96$

For this case the degrees of freedom are given by:

$df =(4-1)*(4-1)= 9$

And the p value would be given by:

$p_v = P(\chi^2_{9} >18.96)=0.0255$

And we can use the folloing excel code to find it: "=1-CHISQ.DIST(18.96,9,TRUE)"

Part c

The statistic calculated is $\chi^2 = 23.39$

For this case the degrees of freedom are given by:

$df =(2-1)*(8-1)= 7$

And the p value would be given by:

$p_v = P(\chi^2_{7} >23.39)=0.0015$

And we can use the folloing excel code to find it: "=1-CHISQ.DIST(23.39,7,TRUE)"

Part d

The statistic calculated is $\chi^2 = 12.72$

For this case the degrees of freedom are given by:

$df =(5-1)*(3-1)= 8$

And the p value would be given by:

$p_v = P(\chi^2_{8} >12.72)=0.1219$

And we can use the folloing excel code to find it: "=1-CHISQ.DIST(12.72,8,TRUE)"