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Primary Heating Fuel. According to Current Housing Reports, published by the U.S. Census Bureau, the primary heating fuel for all occupied housing units is distributed as follows.

Suppose that you want to determine whether the distribution of primary heating fuel for occupied housing units built after 2010 differs from that of all occupied housing units. To decide, you take a random sample of housing units built after 2010 and obtain a frequency distribution of their primary heating fuel.

a. Identify the population and variable under consideration here.

b. For each of the following sample sizes, determine whether conducting a chi-square goodness-of-fit test is appropriate and explain

your answers: 200; 300; 400.

c. Strictly speaking, what is the smallest sample size for which conducting a chi-square goodness-of-fit test is appropriate?

Short Answer

Expert verified

conducting a chi-square goodness-of-fit test is not appropriate when n = 200 , n= 300 , n= 400 .

Required smallest sample size is 334 for which conducting a chi-square goodness-of-fit test is appropriate.

Step by step solution

01

Step 1. Given 

Primary heating fuel Percentage

Utility gas 50.8

Fuel oil, kerosene 7.9

Electricity 33.9

Bottled, tank, or LPG 5.2

Wood and other fuel 1.9

None 0.3

02

Step 2. Part ( a )

Identify the number of people

Population:

Population is a complete set of people considered in the study. Here, the population is "all the houses built after 2010"

Find variables in the study

A variable is an attribute or attribute that can be measured. The amount of flexibility may vary for each unit. That is, the variable is defined as the element recorded in each case

From the description given, the variable is the fuel temperature of each unit "because the result of each effort can be measured and this value can vary from effort to individual.

03

Step 2. Part ( b )

Decide whether to make a chi-square of goodness-of-fit test appropriate or not.

Guess:

• All at least 1 expected frequency.

At least 20% of expected frequencies are less than 5.

• The selected sample should be a simple random sample

Sample size n= 200

The expected frequency is tabulated below :

Primary heating fuelPercentageExpected frequency , np
Utility gas50.8101.6
fuel oil , kerosene7.915.8
Electricity33.967.8
Bottled tank or LPG5.210.4
Wood and other fuel1.93.8
None0.30.6

From the table, it is observed that the expected frequency for None is less than 1 and 33 %

=26×100of the expected frequencies are less than 5. Therefore, the assumptions are

not satisfied when n=200

Thus, conducting a chi-square goodness-of-fit test is not appropriate when n = 200

04

Step 4. When sample size , n= 300

Sample size n= 300

The expected frequency is tabulated below :

Primary heating fuelPercentageExpected frequency , np
Utility gas50.8101.6
fuel oil , kerosene7.915.8
Electricity33.967.8
Bottled tank or LPG5.210.4
Wood and other fuel1.93.8
None0.30.6

From the table, it is observed that the expected frequency for None is less than 1 and 17 %

=16×100of the expected frequencies are less than 5. Therefore, the assumptions are

not satisfied when n=300

Thus, conducting a chi-square goodness-of-fit test is not appropriate when n = 300

05

Step 5. When sample size , n= 400 .

Sample size n= 400

The expected frequency is tabulated below :

Primary heating fuelPercentageExpected frequency , np
Utility gas50.8101.6
fuel oil , kerosene7.915.8
Electricity33.967.8
Bottled tank or LPG5.210.4
Wood and other fuel1.93.8
None0.30.6

From the table, it is observed that the expected frequency for None is less than 1 and 17 %

=16×100of the expected frequencies are less than 5. Therefore, the assumptions are

not satisfied when n=400

Thus, conducting a chi-square goodness-of-fit test is not appropriate when n = 400

06

Step 6. Part ( c ) .

Find the smallest sample size for which conducting a chi-square goodness-of-fit test is appropriate.

The required expected frequency for "None" should be greater than or equal to 1. That is , n0.3100>1Therefore, the smallest sample size for which conducting a chi-square goodness of-fit test is appropriate is,

n(0.003) = 1

n=10.003=333.33334

Thus, the required smallest sample size is 334

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Most popular questions from this chapter

Presidential Election. According to Dave Leip's Atlas of U.S. Presidential Elections, in the 2012 presidential election, 51.01%of those voting voted for the Democratic candidate (Barack H. Obama), whereas 57.50%of those voting who lived in Illinois did so. For that presidential election, does an association exist between the variables "party of presidential candidate voted for" and "state of residence" for those who voted? Explain your answer.

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?

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.

12.73 two and two

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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