Chapter 11.2, Problem 50E

a)

To determine

To determine which chi-square test is appropriate for given setting.

Answer to Problem 50E

We should use Chi-square test for homogeneity.

Explanation of Solution

Given:

First, we need to know which test should be apply for a specific scenario. There are three tests:

Chi-square goodness-of-fit test, chi-square test for homogeneity, chi-square test for independence.

We will know in details about these all tests.

When we are interested in the distribution of one variable, then we will use a chi-square goodness-of-fit test. When we are interested in the distribution of two variables in which there are multiple independent samples, then in this scenario we have to use a chi-square test for homogeneity. When we are interested in the distribution of two variables and if there is a single sample, then we would like to use a chi-square test for independence.

For given setting, we are given in two variables: Type of university and Distance from home. There are two random sample. Therefore, we should use Chi-square test for homogeneity.

b)

To determine

To state the null and alternative hypotheses.

Answer to Problem 50E

  $H_0$ : The distribution of distance from home is same for each university.

  $H_a$ : The distribution of distance from home is not same for each university.

Explanation of Solution

Given:

Calculation:

The null and alternative hypotheses:

  $H_0$ : The distribution of distance from home is same for each university.

  $H_a$ : The distribution of distance from home is not same for each university.

c)

To determine

To verify the conditions for inference.

Answer to Problem 50E

All conditions satisfied.

Explanation of Solution

Given:

Calculation:

The conditions of chi square test are: Random, Independence and Large counts.

There are two random sample selected, so condition of random sample is satisfied.

The sample size 43363 students at public universities is less than 10% of all students from public universities. Also, the sample size 257329 students at private universities is less than 10% of all students from private universities. So, the condition of independent is satisfied.

The expected count of each survey is at least 5 so, condition of large count is satisfied.

d)

To determine

To explain conclusion.

Answer to Problem 50E

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

Explanation of Solution

Given:

Calculation:

The p-value = 0.0000. Therefore, there is very less chance to obtain a similar or more extreme sample results or more extreme, when the distribution of distance from home is same for each university type.

Decision: P-value < 0.05, reject H0.

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