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1. With double- digit annual percentage increases in the cost of health insuranc

ID: 3327550 • 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.

Please tell me exactly what to type in Excel to get these answers and if there are any formulas that can be used to get them. Thanks!

Health Insurance

Size of Company

Yes

No

Total

Small

50

25

75

Medium

80

20

100

Large

115

10

125

Total

245

55

300

Explanation / Answer

Sol:

perform chi sq test for independence

HO:employee health insurance coverage is INDEPENDENT of the size of the company

H1:employee health insurance coverage is DEPENDENTof the size of the company

alpha=0.05

its 2*3 cntingency table

calculate expected frequencies as

E=row total*colum total/grandtotal

Given frequencies are O

chi sq test stat=TOTAL(O-E)^2/E

Degrees of freedom=(r-1)(c-1)

r-rows

c-columns

=(3-1)(2-1)

=2*1

df =2

chi sq critical value=5.99 for 2 df and 0.05 level of significance

chi sq cal>chi sq cri

20.37>5.99

Decison:

Reject Null hypothesis

Accept alternative Hypothesis

Conclusion:

there is no sufficient evidence at 5% level of significance to conlcude thatemployee health insurance coverage is independent of the size of the company

Solutionb:

p=0.0038

p<0.05

Reject Null hypothesis

Accept alternative Hypothesis

Conclusion:

there is no sufficient evidence at 5% level of significance to conlcude thatemployee health insurance coverage is independent of the size of the company

p value=

HEALTHINSURANCE SIZEOFCOMPANY YES NO TOTAL SMALL 50 25 75 MEDIUM 80 20 100 LARGE 115 10 125 TOTAL 245 55 300 CHI SQ STAT CALCULATION O E O-E (O-E)^2 (O-E)^2/E 50 61.25 -11.25 126.5625 2.066327 25 13.75 11.25 126.5625 9.204545 80 81.66667 -1.66667 2.777778 0.034014 20 18.33333 1.666667 2.777778 0.151515 115 102.0833 12.91667 166.8403 1.634354 10 22.91667 -12.9167 166.8403 7.280303 CHISQSTAT 20.37106