Question

Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of 280 adults is conducted, and each participant is classified as not alcoholic, diagnosed alcoholic, or undiagnosed alcoholic, and also by marital status.

Diagnosed Alcoholic Undiagnosed Alcoholic Not Alcoholic

Married 21 37 58
Not Married 59 63 42

Is there significant evidence of an association? Run the appropriate test at the 5% level of significance.

1. Write out the appropriate null hypothesis and the alternative hypothesis.
2. What is the appropriate test statistic that needs to be calculated?
3. Calculate and report the appropriate test statistic. To get full credit, you must show your work.
4. Decide to either fail to reject the null hypothesis or reject the null hypothesis, and describe the result and the statistical conclusion in an understandable way.

Answer 1

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

2) The statistic to check the hypothesis is given by:

`\sum_(i=1)^n ((O_i -E_i)^2)/(E_i)`

3)
`\chi^2 = ((21-33.143)^2)/(33.143)+((37-41.429)^2)/(41.429)+((58-41.429)^2)/(41.429)+((59-46.857)^2)/(46.857)+((63-58.571)^2)/(58.571)+((42-58.571)^2)/(58.571) =19.72`

4)
`df=(rows-1)(cols-1)=(3-1)(2-1)=2`

And we can calculate the p value given by:

`p_v = P(\chi^2_(2) >19.72)=5.22x10^(-5)`

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Explanation:

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".

Assume the following dataset:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 21 37 58 116

Not Married 59 63 42 164

Total 80 100 100 280

Part 1

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

H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

The level os significance assumed for this case is
`\alpha=0.05`

Part 2

The statistic to check the hypothesis is given by:

`\sum_(i=1)^n ((O_i -E_i)^2)/(E_i)`

Part 3

The table given represent the observed values, we just need to calculate the expected values with the following formula
`E_i = (total col * total row)/(grand total)`

And the calculations are given by:

`E_(1) =(80*116)/(280)=33.143`

`E_(2) =(100*116)/(280)=41.429`

`E_(3) =(100*116)/(280)=41.429`

`E_(4) =(80*164)/(280)=46.857`

`E_(5) =(100*164)/(280)=58.571`

`E_(6) =(100*164)/(280)=58.571`

And the expected values are given by:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 33.143 41.429 41.429 116

Not Married 46.857 58.571 58.571 164

Total 80 100 100 280

And now we can calculate the statistic:

`\chi^2 = ((21-33.143)^2)/(33.143)+((37-41.429)^2)/(41.429)+((58-41.429)^2)/(41.429)+((59-46.857)^2)/(46.857)+((63-58.571)^2)/(58.571)+((42-58.571)^2)/(58.571) =19.72`

Part 4

Now we can calculate the degrees of freedom for the statistic given by:

`df=(rows-1)(cols-1)=(3-1)(2-1)=2`

And we can calculate the p value given by:

`p_v = P(\chi^2_(2) >19.72)=5.22x10^(-5)`

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.