Question

Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sample data below show the number of companies providing health insurance for small, medium, and large companies. For the purposes of this study, small companies are companies that have fewer than employees. Medium-sized companies have to employees, and large companies have or more employees. The questionnaire sent to employees asked whether or not the employee had health insurance and then asked the enployee to indicate the size of the company.

Health Insurance
Size of Company Yes No Total
Small 32 18 50
Medium 68 7 75
Large 89 11 100

Conduct a test of independence to determine whether health insurance coverage is independent of the size of the company. What is the p-value?

Answer 1

`\chi^2 = ((32-42)^2)/(42)+((18-8)^2)/(8)+((68-63)^2)/(63)+((7-12)^2)/(12)+((89-84)^2)/(84)+((11-16)^2)/(16)=19.221`

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) >19.221)=0.000067`

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

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example
`\alpha=0.05`, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

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".

Solution to the problem

Assume the following dataset:

Size Company/ Heal. Ins. Yes No Total

Small 32 18 50

Medium 68 7 75

Large 89 11 100

_____________________________________

Total 189 36 225

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

H0: independence between heath insurance coverage and size of the company

H1: NO independence between heath insurance coverage and size of the company

The statistic to check the hypothesis is given by:

`\sum_(i=1)^n ((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 = (total col * total row)/(grand total)`

And the calculations are given by:

`E_(1) =(50*189)/(225)=42`

`E_(2) =(50*36)/(225)=8`

`E_(3) =(75*189)/(225)=63`

`E_(4) =(75*36)/(225)=12`

`E_(5) =(100*189)/(225)=84`

`E_(6) =(100*36)/(225)=16`

And the expected values are given by:

Size Company/ Heal. Ins. Yes No Total

Small 42 8 50

Medium 63 12 75

Large 84 16 100

_____________________________________

Total 189 36 225

And now we can calculate the statistic:

`\chi^2 = ((32-42)^2)/(42)+((18-8)^2)/(8)+((68-63)^2)/(63)+((7-12)^2)/(12)+((89-84)^2)/(84)+((11-16)^2)/(16)=19.221`

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) >19.221)=0.000067`

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

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example
`\alpha=0.05`, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.