11/1/2023

Characteristics of the F Distribution

 There is a family of F distributions. Each time the degrees
of freedom in either the numerator or the denominator
change, a new distribution is created
 The F distribution is continuous
 The F statistic cannot be negative
 The F distribution is positively skewed
Analysis of Variance  The F distribution is asymptotic

Chapter 12

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Compare Two Population Variances

Comparing Two Population Variances Example
 The value of F is computed using the following equation
Lammers Limos offers limousine service from Government Center in downtown
Toledo, Ohio, to Metro Airport in Detroit. The president of the company is
considering two routes. One is via U.S. 25 and the other via I-75. He wants to study
the time it takes to get to the airport using each route and compare the results. He
collected the following sample data. Using the .10 significance level, is there a
 The larger of the two sample variances is placed in the numerator, difference in the variation in the driving times for the two routes?
forcing the ratio to be at least 1.00
 We calculate the standard deviation, s, and square the standard
deviations to get the variance, s2,for each population
Example
 A health services corporation manages two hospitals in Knoxville: St.
Mary’s North and St. Mary’s South.The mean waiting time in both
Emergency Departments is 42 minutes. The hospital administrator
believes St. Mary’s North has more variation than St. Mary’s South.

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Compare Two Population Variances Compare Two Population Variances

Example (2 of 3) Example (3 of 3)
Step 1: State the null and alternate hypothesis Step 5: Compute the ratio of the two sample variances; it’s 4.23, so we reject H0
H0:
H1:
Step 2: Select the level of significance; we decide to use .10
Step 3: Determine the test statistic; we’ll use F
Step 4: State the decision rule, reject H0 if the ratio of the sample variances > 3.87

The decision is to reject the null hypothesis because the test statistic of
4.23 is larger than the critical value of 3.87.

Step 6: We conclude there is a difference in the variation in the time to travel the
two routes.

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ANOVA: Analysis of Variance ANOVA Example

 A one-way ANOVA is used to compare two or more Joyce Kuhlman manages a regional financial
treatment means center. She wishes to compare the productivity,
as measured by the number of customers
 ANOVA was first developed for use in agriculture; the served, among three employees. Four days are
randomly selected and the number of
term treatment was used to identify how different plots customers served by each employee is
of land were treated with different fertilizers recorded. Is there a difference in the mean
number of customers served?
 A treatment is a source of variation
 The assumptions underlying ANOVA are:
 The samples are from populations that follow the
normal distribution
 The populations have equal standard deviations
 The populations are independent

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The ANOVA Test Finding the Value of F

First, find the overall mean of the 12  The formula for the sum of the squares total, SS total, is
observations. It is 58.
Next, find the difference between each
particular value and the overall mean. Square
these differences and sum up. This result is the
total variation, here 1,082.  The formula for the sum of the squares error, SSE, is
TOTAL VARIATION The sum of the squared differences between each
observation and the overall mean.

Now, break this total variation in two components: variation due to  The formula for the sum of the squares treatment, SST, is
treatment variation and random variation.

TREATMENT VARIATION The sum of the squared differences between  This information is summarized in the ANOVA table
each treatment mean and the grand or overall mean.

RANDOM VARIATION The sum of the squared differences between

each observation and its treatment mean.
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The ANOVA Test Continued Finding the Value of F Example

Recall, the overall mean is 58 and the total variation is 1,082. A group of four airlines hired Brunner Marketing Research Inc. to survey passengers
Now, break this total variation in two components: variation due regarding their level of satisfaction with a recent flight. Twenty-five questions offered a
to treatment variation and random variation. range of possible answers: excellent (4), good (3), fair (2), poor (1), so the highest
• The variation due to treatments is 992, found by squaring the possible score was 100. Brunner randomly selected and surveyed passengers from the
difference between each treatment mean and the overall mean four airlines. Is there a difference in the mean satisfaction level among the four airlines?
and then multiplying each squared difference by the number of
observations in each treatment. Step 1: State the null and the alternate hypothesis
4(56-58)2 + 4(70-58)2 + 4(48-58)2 = 992 H0: = = =
H1: The mean scores are not all equal
• The random variation is 90, found by summing the squared
Step 2: Select the level of significance; we’ll use .01
differences between each value and the mean for each
treatment. Step 3: Determine the test statistic; the test statistic follows the F distribution
Step 4: Formulate the decision rule, reject H0 is F > 5.09
(55-56)2 + (54-56)2 + … + (48-48)2 = 90
Step 5: Select the sample, calculate F (8.99), and make a decision; we reject H0
• Calculate the test statistic, F Step 6: Interpret the result; we conclude the populations are not all equal
/
F = / = 49.6
This ratio is quite different from 1, we can conclude there is a
difference in the mean number of customers served by the three
employees.
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Pairs of Means Pairs of Means Analysis Example

 If a null hypothesis of equal treatment means is rejected, Recall in the previous example of airline
satisfaction, we rejected the null
we can identify the pairs of means that differ with the hypothesis that the population means
following confidence interval were equal; at least one of the airline’s
mean level of satisfaction is different
from the others. But we do not know
which pairs.
Use formula 12-5 to construct a
confidence interval with the mean
scores of Northern and Branson. Using
a 95% level of confidence, we find the
endpoints are 10.457 and 26.043.
Zero is not in the interval; so
 If the confidence interval includes zero, there is not a passengers on Northern rated service
significantly different from those on
difference between the treatment means Branson Airlines.

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A Two-Way ANOVA A Two-Way ANOVA (2 of 2)

 The SSE term, or sum of squares error, is found with the
 In a two-way ANOVA, we consider a second treatment following equation
variable
BLOCKING VARIABLE A second treatment variable that when included
in the ANOVA analysis will have the effect of reducing the SSE term.  The two-way ANOVA table includes an additional row for
 This reduces the amount of error variance the blocking variable
 The blocking variable determined using equation below

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ANOVA Test Example ANOVA Test Example Continued

WARTA, the Warren Area Regional Transit Authority, is expanding bus service Step 1: State the null and alternate hypothesis
from the suburb of Starbrick to the business district of Warren. There are four H0: = = =
routes being considered, U.S. 6, West End, Hickory St. , and Rte. 59. WARTA H1: Not all treatment means are the same
conducted tests to determine whether there is a difference in the mean travel Step 2: Select the level of significance; we decide to use .05
times along the four routes; each driver drove each route. See the travel times in Step 3: Select the test statistic; we use F
minutes for each driver-route combination below.
Step 4: State the decision rule, Reject H0 if F > 3.24
At the .05 significance level, is there a difference in the mean travel time along the Step 5: Make decision; F = 2.483, we do not reject the null hypothesis
four routes? If we remove the effects of the drivers, is there a difference in the Step 6: Interpret; there is no reason to conclude that any one of the routes is faster
mean travel time? than any other.

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Two-Way Analysis of Variance Two-Way Analysis of Variance (2 of 2)

 In the WARTA example, we only considered the variation Including the variance of the drivers, here is a table of the drivers’ respective
due to routes and took all other variables to be random means, with an overall mean of 22.8 minutes.

 Now, we’ll include the variance due to the drivers by

letting the drivers be the blocking variable

 Substituting the information into formula (12-6) we determine SSB, the sum
of squares due to the drivers.

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A Second Treatment Variable Continued Hypothesis Test of Equal Treatment Means

 Determine the F statistics for the treatment variable and Step 1: State the null hypothesis and the alternate hypotheses,
H0: The treatment means are equal (1 = 2= 3= 4)
the blocking variable from the following ANOVA table H1: At least one treatment mean is different
Step 2: Select the level of significance; we’ll use .05
Step 3: Select the test statistic; we use F
Step 4: State the decision rule for the first set of hypotheses, reject H0 if F > 3.49
Step 5: Make decision; the computed F ratio is 7.93 so we reject the null hypothesis
that all treatment means are equal
F = = = 7.93
Step 6: Interpret; we conclude that at least one of the routes mean travel time is
different from the other routes

Next, we test to find if the travel times for the various drivers are equal.

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Hypothesis Test of Equal Block Means (2 of

Hypothesis Test of Equal Block Means 2)

Step 1: State the null hypothesis and the alternate hypotheses, Excel has a two-factor ANOVA procedure. The output for the WARTA example
H0: The block means are equal (D= S= O= Z= F) just completed is shown.
H1: At least one block mean is different
Step 2: Select the level of significance; we’ll use .05
Step 3: Select the test statistic; we use F
Step 4: State the decision rule for the first set of hypotheses, reject H0 if F > 3.26
Step 5: Make decision; the computed F ratio is 9.78 so we reject the null hypothesis
that all block means are equal
F = 9.78
Step 6: Interpret; we conclude at least one driver’s mean travel time is different from
the others. WARTA management can conclude, based on the sample results that there
is a difference in the mean travel times of drivers.

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Interaction Plot Interaction Plot (2 of 2)

 An interaction plot illustrates the interaction of the two  To test for interaction, the sample data must be replicated
factors, route and driver for each route. In this case, each driver drives each route
 Travel time is the response variable three times, so there are three observed times for each
route/driver combination. This information is summarized
INTERACTION The effect of one factor on a response variable differs
depending on the value of another factor. in the following Excel spreadsheet.

12-25 12-26

Hypothesis Tests for Interaction ANOVA Table including Interactions

 The next step is to investigate the interaction effects  The complete ANOVA table including interactions
 Is there an interaction between drivers and routes?
 Are the mean travel times for drivers the same?
 Are the mean travel times for the routes the same?
 Test three sets of hypotheses
 Interaction
 H0: There is no interaction between drivers and routes
 H1: There is interaction between drivers and routes
 Blocks
 H0: The driver means are equal
 H1: At least one driver travel time mean is different
 Treatments
 H0: The route means are equal
 H1: At least one route travel time mean is different

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A One-Way ANOVA to Test a Hypothesis

 We will continue the
analysis by conducting
a one-way ANOVA for
each route by testing
the hypothesis
H0: Driver times are equal Chapter 12 Practice Problems
The results show there are
significant differences in the mean
travel times among the drivers
for every route, except Route 59
which has a p-value of .06.

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Question 5 LO12-1 Question 9 LO12-2

Arbitron Media Research Inc. conducted a study of the iPod listening A real estate developer is considering investing in a shopping mall on
habits of men and women. One facet of the study involved the mean the outskirts of Atlanta, Georgia. Three parcels of land are being
listening time. It was discovered that the mean listening time for a evaluated. Of particular importance is the income in the area
sample of 10 men was 35 minutes per day. The standard deviation was surrounding the proposed mall. A random sample of four families is
10 minutes per day. The mean listening time for a sample of 12 women selected near each proposed mall. Following are the sample results. At
was also 35 minutes, but the standard deviation of the sample was 12 the .05 significance level, can the developer conclude there is a
minutes. At the .10 significance level, can we conclude that there is a difference in the mean income?
difference in the variation in the listening times for men and women?
a. What are the null and
a. State the null hypothesis and the alternate hypothesis alternate hypotheses?
b. State the decision rule b. What is the critical value?
c. Compute the value of the test statistic c. Compute the test statistic.
d. What is the p-value d. Compute the p-value.
e. What is your decision regarding H0 e. What is your decision
f. Interpret the result regarding the null hypothesis?
f. Interpret the result.
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Question 11 LO12-3 Question 17 LO12-4

The following are three observations collected from treatment 1, five Chapin Manufacturing Company operates 24 hours a day, 5 days a
observations collected from treatment 2, and four observations week.The workers rotate shifts each week. Management is
collected from treatment 3. Test the hypothesis that the treatment
interested in whether there is a difference in the number of units
means are equal at the .05 significance level.
produced when the employees work on various shifts. A sample of
five workers is selected and their output recorded on each shift. At
a. State the null hypothesis and the .05 significance level, can we conclude there is a difference in
the alternate hypothesis.
the mean production rate by shift or by employee?
b. What is the decision rule?
c. Compute SST, SSE, and SS total.
d. Complete an ANOVA table.
e. Based on the value of the test statistic, state your decision
regarding the null hypothesis.
f. If H0 is rejected, can we conclude that treatment 1 and treatment
2 differ? Use the 95% level of confidence.

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Question 19 LO12-5

Consider the following sample data for a two-factor ANOVA analysis.

There are two levels (heavy and light) of factor A (weight), and three
levels (small, medium, and large) of factor B (size). For each
combination of size and weight, there are three observations.

Compute an ANOVA with statistical

software, and use the .05 significance
level to answer the following
questions.
a. Is there a difference in the Size means?
b. Is there a difference in the Weight means?
c. Is there a significant interaction between Weight and Size?

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