1 With double digit annual percentage increases in the cost

1. With double- digit annual percentage increases in the cost of health insurance, more and more workers are likely to lack health insurance coverage. 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 is reported as follows:
Health Insurance  
Size of Company Yes No Total
Small 50 25 75 Medium 80 20 100 Large 115 10 125 Total 245 55 300
a. Conduct a test of independence using critical-value approach to determine whether employee health insurance coverage is independent of the size of the company. State the Hypotheses and the conclusion. Use = .005. b. What is the p-value? What is your conclusion based on p-value approach? c. The USA Today article indicated employees of small companies are more likely to lack health insurance coverage. Use percentages based on the preceding data to support this conclusion.  
2. FlightStats, Inc., collects data on the number of flights scheduled and the number of flights flown at major airports throughout the United States. FlightStats data showed 56% of flights scheduled at Newark, La Guardia, and Kennedy airports were flown during a three-day snowstorm. All airlines say they always operate within set safety parameters— if conditions are too poor, they don’t fly. The following data show a sample of 600 scheduled flights during the snowstorm. Use the chi- square test with a .10 level of significance to determine whether or not flying/ not flying in a snowstorm is independent of Airliner. State the Hypotheses. What is your conclusion based on Critical-Value test? Is it any different from conclusion based on a p-value approach? Sample data follows:
Flight American Continental Delta United Yes 70 105 95 45 No 80 55 85 65

Solution

Null hypothesis: employee health insurance coverage is independent of the size of the company

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

Chi-Square Test

Observed Frequencies

Column variable

Calculations

Row variable

C1

C2

Total

fo-fe

R1

50

25

75

-11.2500

11.2500

R2

80

20

100

-1.6667

1.6667

R3

115

10

125

12.9167

-12.9167

Total

245

55

300

Expected Frequencies

Column variable

Row variable

C1

C2

Total

(fo-fe)^2/fe

R1

61.25

13.75

75

2.0663

9.2045

R2

81.66667

18.33333

100

0.0340

0.1515

R3

102.0833

22.91667

125

1.6344

7.2803

Total

245

55

300

Data

Level of Significance

0.005

Number of Rows

3

Number of Columns

2

Degrees of Freedom

2

Results

Critical Value

10.597

Chi-Square Test Statistic

20.37106

p-Value

0.0000

Reject the null hypothesis

Calculated chi square =23.371 > 10.597 table value at 0.005 level

The null hypothesis is rejected.

We conclude that employee health insurance coverage is not independent of the size of the company.

b. What is the p-value? What is your conclusion based on p-value approach?

P=0.0000

Calculated P=0.000 < 0.005 level.

The null hypothesis is rejected.

We conclude that employee health insurance coverage is not independent of the size of the company.

c. The USA Today article indicated employees of small companies are more likely to lack health insurance coverage. Use percentages based on the preceding data to support this conclusion.  

we support the conclusion because only 66.6% covered in small industries compared to 80% and 92% in other industries.

2. FlightStats, Inc., collects data on the number of flights scheduled and the number of flights flown at major airports throughout the United States. FlightStats data showed 56% of flights scheduled at Newark, La Guardia, and Kennedy airports were flown during a three-day snowstorm. All airlines say they always operate within set safety parameters— if conditions are too poor, they don’t fly. The following data show a sample of 600 scheduled flights during the snowstorm. Use the chi- square test with a .10 level of significance to determine whether or not flying/ not flying in a snowstorm is independent of Airliner. State the Hypotheses. What is your conclusion based on Critical-Value test? Is it any different from conclusion based on a p-value approach? Sample data follows:
Flight American Continental Delta United Yes 70 105 95 45 No 80 55 85 65

Null hypothesis: flying/ not flying in a snowstorm is independent of Airliner

Alternate hypothesis: flying/ not flying in a snowstorm is not independent of Airliner

Chi-Square Test

Observed Frequencies

Column variable

Calculations

Row variable

C1

C2

Total

fo-fe

R1

70

80

150

-8.7500

8.7500

R2

105

55

160

21.0000

-21.0000

R3

95

85

180

0.5000

-0.5000

R4

45

65

110

-12.7500

12.7500

Total

315

285

600

Expected Frequencies

Column variable

Row variable

C1

C2

Total

(fo-fe)^2/fe

R1

78.75

71.25

150

0.9722

1.0746

R2

84

76

160

5.2500

5.8026

R3

94.5

85.5

180

0.0026

0.0029

R4

57.75

52.25

110

2.8149

3.1112

Total

315

285

600

Data

Level of Significance

0.01

Number of Rows

4

Number of Columns

2

Degrees of Freedom

3

Results

Critical Value

11.345

Chi-Square Test Statistic

19.031

p-Value

0.000269

Reject the null hypothesis

Calculated chi square =19.0.31 > 11.345 table value at 0.10 level

The null hypothesis is rejected.

We conclude that employee flying/ not flying in a snowstorm is not independent of Airliner

Calculated P=0.0003 which is < 0.10 level.

The null hypothesis is rejected.

There is no different from conclusion based on a p-value approach.

Chi-Square Test
Observed Frequencies
	Column variable			Calculations
Row variable	C1	C2	Total		fo-fe
R1	50	25	75		-11.2500	11.2500
R2	80	20	100		-1.6667	1.6667
R3	115	10	125		12.9167	-12.9167
Total	245	55	300
Expected Frequencies
	Column variable
Row variable	C1	C2	Total		(fo-fe)^2/fe
R1	61.25	13.75	75		2.0663	9.2045
R2	81.66667	18.33333	100		0.0340	0.1515
R3	102.0833	22.91667	125		1.6344	7.2803
Total	245	55	300
Data
Level of Significance	0.005
Number of Rows	3
Number of Columns	2
Degrees of Freedom	2
Results
Critical Value	10.597
Chi-Square Test Statistic	20.37106
p-Value	0.0000
Reject the null hypothesis