Question

CH12

1. 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	32	18	50
	Medium	68	7	75
	Large	89	11	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 ?
   
   
   
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	%

   SMALL ? , MEDIUM? LARGE ?

2.

During the first 13 weeks of the television season, the Saturday evening 8:00 P.M. to 9:00 P.M. audience proportions were recorded as ABC 30%, CBS 27%, NBC 25%, and independents 18%. A sample of 300 homes two weeks after a Saturday night schedule revision yielded the following viewing audience data: ABC 93 homes, CBS 63 homes, NBC 88 homes, and independents 56 homes. Test with  = .05 to determine whether the viewing audience proportions changed. Use Table 12.4.

Round your answers to two decimal places.

χ ² = ??

Answer 1

Answer:

Given that,

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.

Given the table,

	Health Insurance
Size of company	Yes	No	Total
Small	32	18	50
Medium	68	7	75
Large	89	11	100

(a).

Conduct a test of independence to determine whether employee health insurance coverage is independent of the size of the company. Use  =0.05:

The null and alternative hypotheses to be tested are:

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

H1: Employee health insurance coverage is not independent of the size of the company.

The observed frequencies for a sample of 225 are:

	Health Insurance
Size of company	Yes	No	Total
Small	32	18	50
Medium	68	7	75
Large	89	11	100
Totals	189	36	225

=Observed frequency for contingency table category in row i and column j.

=Expected frequency for contingency table category in row i and column j based on the assumption of independence.

=42

Similarly, we can deduce the remaining expected frequencies.

Expected frequencies if the column variable is independent of the row variable:

	Health Insurance
Size of company	Yes	No
Small	42	8
Medium	63	12
Large	84	16

Computation of the Chi-square test statistic for determining whether the column variable is independent of the row variable:

Row variable	Column variable	Observed frequency	Expected Frequency	Difference
Small	Yes	38	42	-4	16	0.381
Small	No	18	8	10	100	12.5
Medium	Yes	68	63	5	25	0.397
Medium	No	7	12	-5	25	2.083
Large	Yes	89	84	5	25	0.298
Large	No	11	16	-5	25	1.563
Total		225				=4.341

With n rows and m columns in the contingency table, the test statistic has a chi-square distribution with (n-1)(m-1) degrees freedom provided that the expected frequencies are five or more for all categories.

Test statistic for independence is given by,

=4.341

The number of degrees of freedom is (3-1)(2-1)=2

The test for independence rejects H0 if the difference between observed and expected frequencies provide a large value for the test statistic.

Thus the test for independence is an upper-tail test.

Using the chi-square table,

The p-value:

The p-value at =4.341 and df=2 is 0.114.

Here p-value > 0.05.

We fail to reject the null hypothesis.

Conclusion:

Employee health insurance coverage is independent of the size of the company.

(b).

From part(a) we see that the employee health insurance coverage is independent of the size of the company.

So, based on the size of the company, the percentage of employees with health insurance coverage for small, medium, and large companies are as follows:

The percentage of employees with health insurance coverage for small companies is=(32/50)100

=64%

The percentage of employees with health insurance coverage for medium companies is=(68/75)100

=91%

The percentage of employees with health insurance coverage for large companies is=(89/100)100

=89%

Note:

** As per Chegg guidelines we should solve only the first question. So I have done it.
For the other question please post differently mentioning your requirement.