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Chapter 12: Q. 12.93 (page 518)

Although the chi-square homogeneity test and the chi-square independence test use essentially the same basic procedure, the context of the two tests are quite different. Explain the difference.

Short Answer

Expert verified
Chi-square homogeneity test:
  • It provides the procedure to compare more than two population proportions.
  • The null hypothesis states that more than two populations within one variable is equal.
  • The proportions are same for all populations

Chi-square independence test:

  • It provides the procedure to compare the two variables.
  • The null hypothesis states that the two variables are not associated.
  • There is only one population within two variables.

Step by step solution

01

Step 1. Given information

Given tests are chi-square homogeneity test and chi-square independence test

02

Step 2. The difference between chi-square homogeneity test and chi-square independence test: 

Difference between chi-square homogeneity test and chi-square independence test:
Chi-square homogeneity test:
  • It provides the procedure to compare more than two population proportions.
  • The null hypothesis states that more than two populations within one variable is equal.
  • The proportions are same for all populations

Chi-square independence test:

  • It provides the procedure to compare the two variables.
  • The null hypothesis states that the two variables are not associated.
  • There is only one population within two variables.

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Most popular questions from this chapter

For a χ2-curve with df=4, determine

a. χ0.0052

b.χ0.012

What are the small boxes inside the heavy lines of a contingency table called ?

US. Hospitals. Refer to Exercise 12.50.


24 or fewer25-7475 or moreTotal
General260158635575403
Psychiatric24242471737
Chronic132226
Tuberculosis0224
Other25177208410
Total310201042606580

a. Determine the conditional distribution of number of beds within each facility type.

b. Does an association exist between facility type and number of beds for U.S. hospitals? Explain your answer.

c. Determine the marginal distribution of number of beds for U.S. hospitals.

d. Construct a segmented bar graph for the conditional distributions and marginal distribution of number of beds. Interpret the graph in light of your answer to part (b),

e. Without doing any further calculations, respond true or false to the following statement and explain your answer. "The conditional distributions of facility type within number-of-beds categories are identical.

f. Determine the marginal distribution of facility type and the conditional distributions of facility type within number-of-beds categories.

g. What percentage of hospitals are general facilities?

h. What percentage of hospitals that have at least 75 beds are general facilities?

i. What percentage of general facilities have at least 75 beds?

Identify three techniques that can be tried as a remedy when one or more of the expected-frequency assumptions for a chi-square independence test are violated.

In each of Exercises 12.11-12.16, we have given the relative frequencies for the null hypothesis of a chi-square goodness-of-fir text and the sample size. In each case, decide whether Assumptions 1 and 2 for using that text are satisfied.

Sample size : n= 100.

Relative frequencies: 0.44 , 0.25 , 0.30 , 0.01.
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