With double-digit annual percentage increases in the cost of health insurance, more and more workers are likely to lack health i

Question

2 years ago

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With double-digit annual percentage increases in the cost of health insurance, more and more workers are likely to lack health i

nsurance 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 33 17 50
Medium 68 7 75
Large 88 12 100
Conduct a test of independence to determine whether employee health insurance coverage is independent of the size of the company. Use = .05.
1. Compute the value of the 2 test statistic (to 2 decimals).
2. The p value is :_____.
a. less than .005
b. between .005 and .01
c. between .01 and .025
d. between .025 and .05
e. between .05 and .10
f. greater than .10
3. What is your conclusion?
a. health insurance coverage is not independent of the size of the company
b. Cannot reject the assumption that health insurance coverage and size of the company are independent
4. 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 %

Mathematics

1 answer:

Dahasolnce [82] · Answer 1 · 2022-10-24T23:37:38+03:00

Answer:

1. χ² = 15.3902

2. The p value is :_____.a. less than .005

3. Conclusion

a. health insurance coverage is not independent of the size of the company

4. The percentages of employees

Small %= 33/50= 0.66

Medium %= 68/75 = 0.91

Large %= 88/100= 0.88

Step-by-step explanation:

1) We set up our null and alternative hypothesis as

H0: the employee health insurance coverage is independent of the size of the company

against the claim

Ha: the employee health insurance coverage is not independent of the size of the company

2) the significance level alpha is set at 0.05

3) the test statistic under H0 is

χ²= ∑ (O - E)²/ E where O is the observed and E is the expected frequency

which has an approximate chi square distribution with ( 3-1) (2-1)= 2 d.f

4) Computations:

Under H0 , the observed frequencies are :

Observed Expected E (O-E) (O-E)² (O-E)²/E

33 42 -9 81 1.928

68 63 5 25 0.3968

88 84 4 16 0.1904

17 8 9 81 10.125

7 12 -5 21 1.75

12 16 -4 16 1

15.3902

Expected Values are calculated using the formula :

Row Total * Column Total / sample size

E1= (33+17) (33+ 68+88)/ 50+75+100= 42

E4= (33+17) (17+ 7+ 12)/ 50+75+100=8

E2= (68+7) (33+ 68+88)/ 50+75+100= 63

E5= (68+7) (17+ 7+ 12)/ 50+75+100= 12

E3= (88+12) (33+ 68+88)/ 50+75+100= 84

E6= (88+12) (17+ 7+ 12)/ 50+75+100= 16

5) The critical region is χ² ≥ χ² (0.05)2 = 5.99

6) Conclusion:

The calculated χ² = 15.3902 falls in the critical region χ² ≥ 5.99 so we reject the null hypothesis that the employee health insurance coverage is NOT independent of the size of the company.

2. The p value is :_____.

a. less than .005

The p-value is .000385.