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Chapter 12: Q. 12.74 (page 510)

In each of the given Exercises, we have given the number of possible values for two variables of a population. For each exercise, determine the maximum number of expected frequencies that can be less than 5 in order that Assumption 2 of Procedure 12.2 on page 506 to be satisfied. Note: The number of cells for a contingency table with m rows and n columns is m⋅n.

12.74 six and seven

Short Answer

Expert verified

Ans: The maximum number of expected frequencies that can be less than 5 by using Assumption 2 is 8.

Step by step solution

01

Step 1. Given information.

given,

six and seven

02

Step 2. First, determine the maximum number of expected frequencies that can be less than 5 by using Assumption 2.  

Assumption:

  1. All expected frequencies are at least 1.
  2. At most 20% of the expected frequencies are less than 5.
  3. All the selected samples should be a simple random samples.
03

Step 3. Now,  

According to the given information, the number of possible values for the first variable is 6 and the number of possible values for the second variable is 7. Therefore, there are 42(=6×7)expected frequencies.

The maximum number of expected frequencies that can be less than 5, by using Assumption 2 is,

20%of42=20100×42=8.4≈8

Hence, the maximum number of expected frequencies that can be less than 5 by using Assumption 2 is 8.

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Most popular questions from this chapter

If a variable of two populations has only two possible values, the chi-square homogeneity test is equivalent to a two-tailed test that we discussed in an earlier chapter. What test is that?

In each of the given Exercises, we have given the number of possible values for two variables of a population. For each exercise, determine the maximum number of expected frequencies that can be less than 5 in order that Assumption 2 of Procedure 12.2 on page 506 to be satisfied. Note: The number of cells for a contingency table with m rows and n columns is m⋅n.

12.71 two and three

For what purpose is a chi-square homogeneity test used?

The U.S. Census Bureau publishes census data on the resident population of the United States in Current Population Reports. According to that document, 7.3% of male residents are in the age group20−24 years.
a. If there is no association exists between age group and gender, what percentage of the resident population would be in the age group20−24 years? Explain your answer.
b. If there is no association exists between age group and gender, what percentage of female residents would be in the age group20−24 years? Explain your answer.
c. There are about 158million female residents of the United States. If no association exists between age group and gender, how many female residents would there be in the age group20−24 years?
d. In fact, there are some 10.2 million female residents in the age group20−24 years. Given this number and your answer to part (c), what do you conclude?

Fill in the blank: If a variable has only two possible values, the chi-square homogeneity test provides a procedure for comparing several population......................................
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