Question

With double-digit annual percentage increases in the cost of health insurance, more and more workers are likely to lack health insurance coverage (USA Today, January 23, 2004). The following sample data provide a comparison of workers with and without health insurance coverage for small, medium, and large companies. For the purposes of this study, small companies are companies that have fewer than 100 employees. Medium companies have 100 to 999 employees, and large companies have 1000 or more employees. Sample data are reported for 50 employees of small companies, 75 employees of medium companies, and 100 employees of large companies.

		Health Insurance
	Size of Company	Yes	No	Total
	Small	31	19	50
	Medium	63	12	75
	Large	86	14	100

1. Conduct a test of independence to determine whether employee health insurance coverage is independent of the size of the company. Use = .05. Use Table 12.4.
   
   Compute the value of the ² test statistic (to 2 decimals).
   
   
   The p value is Selectless than .005between .005 and .01between .01 and .025between .025 and .05between .05 and .10greater than .10
   
   What is your conclusion?
   SelectConclude health insurance coverage is not independent of the size of the companyCannot reject the assumption that health insurance coverage and size of the company are independent
   
2. The USA Today article indicated employees of small companies are more likely to lack health insurance coverage. Calculate the percentages of employees without health insurance based on company size (to the nearest whole number).
   

   Small	%
   Medium	%
   Large	%

Answer 1

Part a

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H₀: Employee health insurance coverage is independent of the size of the company.

Alternative hypothesis: H_a: Employee health insurance coverage is not independent of the size of the company.

We are given level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 3

Number of columns = c = 2

Degrees of freedom = df = (r – 1)*(c – 1) = 2*1 = 2

α = 0.05

Critical value = 5.991465

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Health Insurance
Size of company	Yes	No	Total
Small	31	19	50
Medium	63	12	75
Large	86	14	100
Total	180	45	225

Expected Frequencies
	Health Insurance
Size of company	Yes	No	Total
Small	40	10	50
Medium	60	15	75
Large	80	20	100
Total	180	45	225

Calculations
(O - E)
-9	9
3	-3
6	-6

(O - E)^2/E
2.025	8.1
0.15	0.6
0.45	1.8

Chi square = ∑[(O – E)^2/E] = 13.125

Test statistic = Chi square = 13.13

P-value = 0.001412

(By using Chi square table or excel)

The p-value is less than .005.

P-value < α = 0.05

So, we reject the null hypothesis

Conclusion:

There is insufficient evidence to conclude that Employee health insurance coverage is independent of the size of the company.

Part b

Required percentages are given as below:

Size of company	No	Total	Prop.	Percentage
Small	19	50	19/50 = 0.38	38%
Medium	12	75	12/75 = 0.16	16%
Large	14	100	14/100 = 0.14	14%