Business Mathematics 42 – Management Science I
Lesson 1
Chi-Square Applications
Learning Objectives
When you have completed this chapter, you will be able to:
1. List and understand the characteristics of chi-square distribution.
2. Conduct a test of hypothesis comparing an observed set of frequencies to an expected distribution.
3. Conduct a test of hypothesis to determine whether two classification criteria are related.
LO1. Characteristics of the Chi-Square Distribution
1. Chi-square values are never negative.
2. The chi-square distribution is positively skewed.
3. There is a family of chi-square distributions.
•
Each time the degrees of freedom change, a new distribution is formed.
•
As the degrees of freedom increases, the distribution approaches a normal distribution.
LO2. GOODNESS-OF-FIT-TEST
A goodness-of-fit test will show whether an observed set of frequencies could have come from a hypothesized
population distribution
A. The degrees of freedom are k-1, where k is the number of categories.
B. The formula for computing the value of chi-square is
𝑥2 = ∑
(𝑓𝑜 − 𝑓𝑒 )2
𝑓𝑒
LO2. GOODNESS-OF-FIT TEST (Equal Expected Frequencies)
Illustration:
The human resources director at Georgetown Paper, Inc., is concerned about absenteeism among hourly workers.
She decides to sample the company records to determine whether absenteeism is distributed evenly throughout the
six-day workweek. The hypotheses are:
𝐻0 : 𝐴𝑏𝑠𝑒𝑛𝑡𝑒𝑒𝑖𝑠𝑚 𝑖𝑠 𝑒𝑣𝑒𝑛𝑙𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑡ℎ𝑟𝑜𝑢𝑔ℎ𝑜𝑢𝑡 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘𝑤𝑒𝑒𝑘.
𝐻1 : 𝐴𝑏𝑠𝑒𝑛𝑡𝑒𝑒𝑖𝑠𝑚 𝑖𝑠 𝑛𝑜𝑡 𝑒𝑣𝑒𝑛𝑙𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑡ℎ𝑟𝑜𝑢𝑔ℎ𝑜𝑢𝑡 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘𝑤𝑒𝑒𝑘.
The sample results are: (use 1 percent significant level)
Step 1. State the null and alternate hypothesis
H0: Absenteeism is evenly distributed throughout the workweek.
H1: Absenteeism is not evenly distributed throughout the workweek.
Step 2. Select the level of significance
We selected the 0.01 significance level. The probability is 0.01 that a true null hypothesis will be rejected.
Step 3. Select the test statistics
The test statistic follows the chi-square distribution, designated as
𝑥2 = ∑
(𝑓𝑂 − 𝑓𝑒 )2
𝑓𝑒
Step 4. Formulate the decision rule
𝑑𝑓 = 𝑘 − 1 = 6 − 1 = 5
Critical value = 15.086
Decision Rule: Reject Null Hypothesis if the computed value of chi-square is greater than 15.086. Otherwise, do not
reject.
Step 5. Compute the value of chi-square and make a decision.
(𝑓𝑂 − 𝑓𝑒 )2
𝑥2 = ∑ [
]
𝑓𝑒
𝑥 2 = 0.80
Do not reject the null hypothesis. Absenteeism is distributed evenly throughout the week. The observed differences
are due to sampling variation.
Exercise:
Classic Golf, Inc., manages five courses in the Jacksonville, Florida, area. The director of golf wishes to study the
number of rounds of gold played per weekday at the five courses. He gathered the following sample information
shown below. At the 0.05 significance level, test the null hypothesis that there is no significant difference in the
number of rounds played by day of the week.
Step 1. State the null and alternate hypothesis
H0: There is no significant difference in the number of rounds played by day of the week. H1: There is significant
difference in the number of rounds played by day of the week.
Step 2. We selected the 0.05 significance level.
The probability is 0.05 that a true null hypothesis will be rejected.
Step 3. Select the test statistics
The test statistic follows the chi-square distribution, designated as
(𝑓𝑂 − 𝑓𝑒 )2
𝑥 =∑
𝑓𝑒
2
Step 4. Formulate the decision rule
𝑑𝑓 = 𝑘 − 1 = 5 – 1 = 4
Critical value = 9.488
Decision Rule: Reject Null Hypothesis if the computed value of chi-square is greater than 9.488. Otherwise, do not
reject.
Step 5. Compute the value of chi-square and make a decision.
(𝑓𝑂 − 𝑓𝑒 )2
𝑥2 = ∑ [
]
𝑓𝑒
𝑥 2 = 15.31
Reject the null hypothesis. There is significant difference in the number of rounds played by day of the week.
LO2. GOODNESS-OF-FIT TEST: (UNEQUAL EXPECTED FREQUENCIES)
The owner of a mail-order catalog would like to compare her sales with the geographic distribution of the
population. According to the United States Bureau of the Census, 21 percent of the population lives in the Northeast,
24 percent in the Midwest, 35 percent in the South, and 20 percent in the West. Listed below is the breakdown of a
sample of 400 orders randomly selected from those shipped last month. At the 0.01 significance level, does the
distribution of the orders reflect the population?
Step 1. State the null and alternate hypothesis
H0: 𝝅𝒏 = 𝟎. 𝟐𝟏,𝝅𝒎 = 𝟎. 𝟐𝟒,𝝅𝒏𝒔 = 𝟎. 𝟑𝟓,𝝅𝒘 = 𝟎. 𝟐𝟎
H1: The distribution is not as given
Step 2. Select the level of significance
We selected the 0.01 significance level. The probability is 0.01 that a true null hypothesis will be rejected.
Step 3. Select the test statistics
The test statistic follows the chi-square distribution, designated as
(𝑓𝑂 − 𝑓𝑒 )2
𝑥2 = ∑ [
]
𝑓𝑒
Step 4. Formulate the decision rule
𝑑𝑓 = 𝑘 − 1 = 4 – 1 = 3
Critical value = 11.345
Decision Rule: Reject Null Hypothesis if the computed value of chi-square is greater than 11.345. Otherwise, do not
reject.
Step 5. Compute the value of chi-square and make a decision.
(𝑓𝑂 − 𝑓𝑒 )2
𝑥 = ∑[
]
𝑓𝑒
2
𝑥 2 = 5.9345
Do not reject the null hypothesis. The distribution of order destinations reflects the population.
LO3. CONTINGENCY TABLE ANALYSIS
A contingency table is used to test whether two traits or characteristics are related.
A. Each observation is classified according to two traits.
B. The expected frequency is determined as follows
𝑓𝑒 =
(𝑟𝑜𝑤 𝑡𝑜𝑡𝑎𝑙)(𝑐𝑜𝑙𝑢𝑚𝑛 𝑡𝑜𝑡𝑎𝑙)
𝐺𝑟𝑎𝑛𝑑 𝑇𝑜𝑡𝑎𝑙
C. The degrees of freedom are found by:
𝑑𝑓 = (𝑅𝑜𝑤𝑠 − 1)(𝐶𝑜𝑙𝑢𝑚𝑛 − 1)
D. The usual hypothesis testing procedure is used.
ILLUSTRATION
A social scientist sampled 140 people and classified them according to income level and whether or not they played
a state lottery in the last month. The sample information is reported below. Is it reasonable to conclude that playing
the lottery is related to income level? Use the 0.05 significance level.
Step 1. State the null and alternate hypothesis
H0: There is no relationship between income and whether the person played the lottery.
H1: There is a relationship between income and whether the person played the lottery.
Step 2. Select the level of significance
We selected the 0.05 significance level. The probability is 0.05 that a true null hypothesis will be rejected.
Step 3. Select the test statistics
The test statistic follows the chi-square distribution, designated as
(𝑓𝑂 − 𝑓𝑒 )2
𝑥2 = ∑ [
]
𝑓𝑒
Step 4. Formulate the decision rule
𝑑𝑓 = (𝑅𝑜𝑤𝑠 – 1)(𝐶𝑜𝑙𝑢𝑚𝑛𝑠 – 1) = (2 − 1)(3 − 1) = 2
Critical value = 5.991
Decision Rule: Reject Null Hypothesis if the computed value of chi-square is greater than 5.991. Otherwise, do not
reject.
EXERCISE
A study regarding the relationship between age and the amount of pressure sales personnel feel in relation to their
jobs revealed the following sample information. At the .01 significance level, is there a relationship between job
pressure and age?
Degrees of Job Pressure
Source: Introduction to Business Statistics by Weiers
Prepared by:
JOHN MELVIN V. BARANDA, LPT