Question

A researcher would like to determine if the proportion of households without health insurance coverage differs with household income. Suppose the following data were collected from 700 randomly selected households. Complete parts a through c.	Heath Insurance
	Household Income	Yes	No
	Less than​ $25,000	50	20
	​$25,000 to​ $49,999	141	44
	​$50,000 to​ $74,999	201	35
	​$75,000 or more	179	30

a. Using

alphaαequals=​0.01,

perform a​ chi-square test to determine if the proportion of households without health insurance differs by income bracket.

Choose the correct null and alternative hypotheses below.

A.

Upper H 0H0​:

Not all​ p's are equal

Upper H 1H1​:

p 1 equals p 2 equals p 3 equals p 4p1=p2=p3=p4

B.

Upper H 0H0​:

p 1 equals p 2 equals p 3 equals p 4p1=p2=p3=p4

Upper H 1H1​:

p 1 not equals p 2 not equals p 3 not equals p 4p1≠p2≠p3≠p4

C.

Upper H 0H0​:

p 1 equals p 2 equals p 3 equals p 4p1=p2=p3=p4

Upper H 1H1​:

Not all​ p's are equal

D.

Upper H 0H0​:

p 1 not equals p 2 not equals p 3 not equals p 4p1≠p2≠p3≠p4

Upper H 1H1​:

p 1 equals p 2 equals p 3 equals p 4

What is the test statistic? X squared = ? (round to two decimal places as needed)

What is the critical value? X squared 0.01 (round to two decimal places as needed)

What is the conclusion?

Determine the p-value for the chi-square test statistic and interpret its meaning.

How does income appear to impact the likelihood that a household has insurance coverage?

Answer 1

a) for hypothesis: Option C is correct:

b)

Applying chi square test of homogeneity

Expected	Ei=row total*column total/grand total	Yes	No	Total
	less than 25000	57.1	12.9	70
	25000-49999	150.9	34.1	185
	50000-74999	192.5	43.5	236
	>75000	170.5	38.5	209
	total	571	129	700
chi square    χ2	=(Oi-Ei)2/Ei	Yes	No	Total
	less than 25000	0.883	3.908	4.7906
	25000-49999	0.650	2.879	3.5294
	50000-74999	0.375	1.658	2.0324
	>75000	0.425	1.883	2.3082
	total	2.3332	10.3274	12.6606
test statistic X2=		12.66
p value =	0.0054	from excel: chidist(12.6606,3)

degree of freedom(df) =(rows-1)*(columns-1)=		3
for 3 df and 0.01 level,critical value χ2=		11.34

since test statistic falls in rejection region we reject null hypothesis

p value =0.0054

this is the probability of getting a test statistic or more extreme if null hypothesis is true

higher household income will have higher proprotion to get insurance coverage