Question

1.

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 = ?

2.

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 Selectless than .005between .005 and .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 2
   
   What is your conclusion?
   SelectConclude health insurance coverage is not independent of the size of the companyCannot reject the assumption that health insurance coverage and size of the company are independentItem 3
   
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	%

Answer 1

1)

applying chi square goodness of fit test:

	relative	observed	Expected	residual	Chi square
category	frequency(p)	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
1-7	0.3000	93.00	90.00	0.32	0.100
8-14	0.2700	63.00	81.00	-2.00	4.000
15-21	0.2500	88.00	75.00	1.50	2.253
22-28	0.1800	56.00	54.00	0.27	0.074
total	1.000	300	300		6.4274
test statistic X2 =		6.43

2)

a)

Applying chi square test of independence:

Expected	Ei=row total*column total/grand total	Yes	No	Total
	small	42.00	8.00	50
	medium	63.00	12.00	75
	large	84.00	16.00	100
	total	189	36	225
chi square    χ2	=(Oi-Ei)2/Ei	Yes	No	Total
	small	2.3810	12.5000	14.8810
	medium	0.3968	2.0833	2.4802
	large	0.2976	1.5625	1.8601
	total	3.0754	16.1458	19.2212
test statistic X2 =		19.22

p value is less than .005

Conclude health insurance coverage is not independent of the size of the company

b)

small	36%
medium	9%
large	11%