1. With double- digit annual percentage increases in the cost of health insuranc
ID: 3378708 • Letter: 1
Question
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
Explanation / Answer
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
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