Chapter 11: Q. 50 (page 756)

Distance from home A study of first-year college students asked separate random samples of students from private and public universities the following question: “How many miles is this university from your permanent home?” Students had to choose from the following options: $5$ or fewer, $6$ to $10, 11$ to $50, 51$ to $100, 101$ to $500$ , or more than $500$ . Here is the two-way table summarizing the responses:
a. Should we use a chi-square test for homogeneity or a chi-square test for independence in this setting? Justify your answer.
b. State appropriate hypotheses for performing the type of test you chose in part (a). Here is Minitab output from a chi-square test.
c. Check that the conditions for carrying out the test are met.
d. Interpret the P-value. What conclusion would you draw?

Short Answer

Expert verified

a. We should use Chi-square test for homogeneity.

b. $H_{0}$ : Each university has the same distance from home distribution.

H_{a}

: The distribution of distance from home varies by university.

c. The chi square test has three conditions: randomness, independence, and large counts. Because two random samples were chosen, the random sample requirement was met.

d. There is convincing evidence that the distribution of distance from home is not same for each university.

Step by step solution

Part (a) Step 1 : Given Information

We have to determine which chi-square test is appropriate for given setting.

Part (a) Step 2 : Simplification

First, we must determine which test should be used in a certain case.

Therearethreeteststocomplete:

Chi-square goodness-of-fit test, chi-square homogeneity test, and chi-square independence test All of these tests will be detailed for us.

A chi-square goodness-of-fit test is used when we are interested in the distribution of a single variable.

In this circumstance, we must employ a chi-square test for homogeneity when we are interested in the distribution of two variables with numerous independent samples.

We would like to do a chi-square test for independence when we are interested in the distribution of two variables and there is only one sample. We are provided two variables for a certain setting: university kind and distance from home. There are two samples at random.

As a result, we should perform the Chi-square test to determine homogeneity.

Part (b) Step 1 : Given Information

We have to state the null and alternative hypotheses.

Part (b) Step 2 : Simplification

The following are the null and alternate hypotheses:

H_{0}

: Each university has the same distance from home distribution.

H_{a}

: The distribution of distance from home varies by university.

Part (c) Step 1 : Given Information

We have to verify the conditions for inference.

Part (c) Step 2 : Simplification

The chi square test has three conditions: randomness, independence, and large counts. Because two random samples were chosen, the random sample requirement was met.

The sample size of

43363

students at public universities represents less than

10 %

of all public university students. Furthermore, the sample size of

257329

students from private universities represents less than

10 %

of all private university students.

As a result, the condition of independence is met. Because each survey's projected count is at least

5

, the criterion of a large count is met.

Part (d) Step 1 : Given Information

We have to explain conclusion.

Part (d) Step 2 : Simplification

The p-value =

0.0000

. When the distribution of distance from home is the same for each university type, there is a very little likelihood of getting similar or more extreme sample outcomes.