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.

	Health Insurance
Household Income	Yes	No
Less than $25,000	52	23
$25,000 to $49,999	143	41
$50,000 to $74,999	200	37
$75,000 or more	180	24

a. Using α=0.01, perform a​ chi-square test to determine if the proportion of households without health insurance differs by income bracket.

What is the test​ statistic?

X² =

​(Round to two decimal places as​ needed.)

What is the critical​ value?

X²α0.01=

​(Round to two decimal places as​ needed.)

b. Interpret the meaning of the​ p-value.

What is the​ p-value?

​p-value =

​(Round to three decimal places as​ needed.)

Answer 1

Solution:

(Part a)

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H₀: The proportion of households without health insurance do not differs by income bracket.

Alternative hypothesis: H_a: The proportion of households without health insurance differs by income bracket.

We are given level of significance = α = 0.01

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 4

Number of columns = c = 2

Degrees of freedom = df = (r – 1)*(c – 1) = 3*1 = 3

α = 0.01

Critical value = 11.34487

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	C1	C2	Total
R1	52	23	75
R2	143	41	184
R3	200	37	237
R4	180	24	204
Total	575	125	700

Expected Frequencies
	Column variable
Row variable	C1	C2	Total
R1	61.60714	13.39286	75
R2	151.1429	32.85714	184
R3	194.6786	42.32143	237
R4	167.5714	36.42857	204
Total	575	125	700

(O - E)
-9.60714	9.607143
-8.14286	8.142857
5.321429	-5.32143
12.42857	-12.4286

(O - E)^2/E
1.498157	6.891524
0.438698	2.018012
0.145458	0.669108
0.921812	4.240336

Chi square = ∑[(O – E)^2/E] = 16.82311

Chi square = 16.82

(Part b)

P-value = 0.000768

P-value = 0.001

(By using Chi square table or excel)

P-value < α = 0.01

So, we reject the null hypothesis

There is sufficient evidence to conclude that the proportion of households without health insurance differs by income bracket.