Question

In: Statistics and Probability

A researcher would like to determine if the proportion of households without health insurance coverage differs...

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?

Solutions

Expert Solution

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


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