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

Answer:

1) H0: There is significant evidence of independence between marital status and diagnostic

H1: There is significant evidence of an association between marital status and diagnostic

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

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

3) $\chi^2 = \frac{(21-33.143)^2}{31.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

4) $p_v = P(\chi^2_{2} >19.72)=0.00005222$

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

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

Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated between the two variables analyzed.

Step-by-step 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:

                    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 significant evidence of independence between marital status and diagnostic

H1: There is significant evidence of an association between marital status and diagnostic

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

Part 2

The statistic to check the hypothesis is given by:

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

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{80*116}{280}=33.143$

$E_{2} =\frac{100*116}{280}=41.429$

$E_{3} =\frac{100*116}{280}=41.429$

$E_{4} =\frac{80*164}{280}=46.857$

$E_{5} =\frac{100*164}{280}=58.571$

$E_{6} =\frac{100*164}{280}=58.571$

And the expected values are given by:

                    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

Part 3

And now we can calculate the statistic:

$\chi^2 = \frac{(21-33.143)^2}{31.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

Part 4

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

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

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=0.00005222$

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

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

Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated between the two variables analyzed.