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A mental health service agency is conducting a study of how clients pay for the counseling services they receive. The agency acc

fomenos

Answer:

There is a relationship between the age of the client and the payment method used.

Step-by-step explanation:

Hello!

You have two categorical variables measured in a sample of 400 clients and want to test if there is a relation between them.

X₁: "Payment method used by clients" (Categorized: In-Person, by Mail, by Credit Card, and by Third-Party Insurance)

X₂: "Age of the client" (Categorized: 20-30, 31-40, 41-50, Over 50)

The statistic test you can use to test the relation between these two variables is a Chi-Square test for independence.

The data for this test is usually in a contingency table, where one of the variables will be in the rows and the other in the columns. (See attachment)

This test is alway One-tailed (right). This means, a small statistic value means that the observed data and the expected data are similar, in other words, that there is a relation between the two variables. On the other hand, a large statistic value means that the two variables are too different and there is no relation between them.

To do so you need to compare the observed data against the expected frequencies for each category. (The expected frequencies are the values for each category if the null hypothesis holds)

The hypothesis for this test is:

H₀: The two categorical variables are independent.

H₁: The two categorical variables are not independent.

α:0.01

Statistic:

χ²=∑ $\frac{(o_{ij} - eij)^{2} }{e_{ij} }$

Where:

$o_{ij}$ is the observed value of the i-row and j-column

$e_{ij}$ is the expected value of the i-row and j-column

Critical region One-tailed (right):

Degrees of freedom:

For this test the degrees of freedom are given by the number of categories of each variable.

χ² $_{(i-1)(j-1);1-\alpha }$

χ² $_{(4-1)(4-1);1-\0.01 }$

χ² $_{9;0.99}$ = 21.67

If the calculated value is equal or grether than 21.67, you'll reject the null hypothesis

If the calculated value is less than 21.67 you will not reject the null hypothesis.

To calculate the statistic you need to first calculate the expected frecuencies for each category. The formula fot the expected frequencies is:

$e_{ij}= \frac{o_{i.}*o_{.j} }{N}$

Where:

N= is the sample total

$e_{ij}$ is an expected frequency

$o_{i.}$ is the marginal row frequency

$o_{.j}$ is the marginal column frequency

For example:

$e_{12}= \frac{o_{1.}*o_{.2} }{N}$

$e_{12}= \frac{86*218 }{400}$ = 46.87

(I've attached a file with all expected frequencies calculated)

With all expected frequencies calculated, all you need to do now is calculate the statistic. It's quite a long calculation.

χ²= $\frac{(o_{11}*e_{11})^2 }{e_{11} } + \frac{(o_{i2}*e_{12})^2 }{e_{12} } + \frac{(o_{13}*e_{13})^2 }{e_{13} } +\frac{(o_{14}*e_{14})^2 }{e_{14} }$ + $\frac{(o_{21}*e_{21})^2 }{e_{21} } + \frac{(o_{22}*e_{22})^2 }{e_{22} } + \frac{(o_{23}*e_{23})^2 }{e_{23} } +\frac{(o_{24}*e_{24})^2 }{e_{24} }$ + $\frac{(o_{31}*e_{31})^2 }{e_{31} } + \frac{(o_{32}*e_{32})^2 }{e_{32} } + \frac{(o_{33}*e_{33})^2 }{e_{33} } +\frac{(o_{34}*e_{34})^2 }{e_{34} }$ + $\frac{(o_{41}*e_{41})^2 }{e_{41} } + \frac{(o_{42}*e_{42})^2 }{e_{42} } + \frac{(o_{43}*e_{43})^2 }{e_{43} } +\frac{(o_{44}*e_{44})^2 }{e_{44} }$

(...)And the statistic value is:

χ²=58.01

Since the calculathed Chi-Square value (χ²=58.01) is greater than the critical value (χ² $_{9;0.99}$ = 21.67) you can reject the null hypothesis.

So there is enough evidence to say that the age of the clients and the method of payment they chose to use are not independent. In other words, there is a relationship between the age of the client and the payment method used.

I hope you have a SUPER day!

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5 0

3 years ago

Solve the system of linear equations by substitution.<br><br> 6x−9=y<br> y=−3x

Julli [10]

Answer:

x = 1

y = - 3

Step-by-step explanation:

5 0

3 years ago