# Statistics Chi Square Data Analysis Discussion Questions

SPSS required to complete. please see for ASSIGNMENT attachments and Discussion questions below (attachment are the guides/ example).

1. During this week’s lesson, you learned that the chi-square test is a nonparametric test. When

compared to parametric tests such as t or F, nonparametric tests tend to have less statistical

power than parametric tests. Explain why this is the case. Given this distinction, would a test

that has less statistical power increase the risk of a Type I error or a Type II error? Explain.

2. Is the chi-square goodness-of-fit test a univariate test or a bivariate test? In other words, does

it involve one variable or does it involve two variables? Explain.

3. Present a research question in an aviation context—other than the ones presented in the

guided example and graded assignment—that would be a good application of the chi-square

goodness-of-fit test. As part of your RQ include any corresponding operational definitions

and the targeted variable.

Runway Incursions

Situational Awareness

Situational Awareness

Situational Awareness

Situational Awareness

Situational Awareness

Situational Awareness

Situational Awareness

Situational Awareness

Situational Awareness

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Situational Awareness

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Situational Awareness

Situational Awareness

Miscommunication

Miscommunication

Miscommunication

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Distraction

Distraction

Distraction

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Distraction

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Airport Markings

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Complex Taxiways

Seat Preferences

Aisle Seat

Aisle Seat

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Window Seat

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Delta Comfort Plus

Delta Comfort Plus

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Week 13

One-Way Chi-Square

Graded Assignment

Let’s assume that Delta Airlines reports the following information about seat preferences among

U.S. travelers who took a Delta flight to/from at least one domestic airport during the past 3 months:

• 30% prefer an aisle seat

• 20% prefer a window seat

• 20% prefer an exit row seat

• 10% prefer a middle seat

• 10% prefer to sit in Delta Comfort Plus

• 10% prefer to sit in the main cabin

Let’s further assume that you randomly surveyed 400 Delta passengers with similar flight

experiences and asked them to specify their seat preferences. A copy of these data is contained in

the Excel file titled “Week 13 Graded Assignment Data.” For the reader’s convenience these

data are summarized in the table below.

Seat Preference

Aisle

Window

Exit Row

Middle

Delta Comfort Plus

Main Cabin

f

125

77

90

31

42

35

A. Pre-Data Analysis

1. What is the research question and corresponding operational definitions?

2. What is the research methodology/design and why is this methodology appropriate?

3. Conduct an a priori power analysis to determine the minimum sample size needed. Is the

given data set sufficient relative to this minimum sample size? In what way do you think

the size of the given sample will impact the results?

B. Data Analysis

Using the data from the Excel file (or the table above), conduct a hypothesis test as follows:

1. Formulate the null and alternative hypotheses in words.

2. Determine the test criteria.

3. Confirm the data set is compliant with the two primary assumptions for the goodness-offit test.

4. Run the analysis and record the results, or calculate the chi-square goodness-of-fit test

statistic by hand similar to what was done in the guided example.

5. Make a decision to reject or fail to reject the null hypothesis and write a concluding

statement.

C. Post-Data Analysis

1. Determine and interpret the effect size.

2. Determine and interpret the power of the study.

3. Present at least three plausible explanations for the results.

4. Interpret the findings from a practical perspective.

Week 13

One-Way Chi-Square

This handout material supplements the information about one-way chi-square that is

presented in Chapter 5 of the assigned textbook by Wilson and Joye. You are encouraged to read

this chapter before reviewing the information contained here. You also should review the Week

13 Overview to ensure that you have a general understanding of the need for chi-square tests.

The Concept of the Chi-Square Test for Goodness of Fit

In the most general sense, the chi-square test is a statistical procedure that tests whether

sets of frequencies follow certain patterns. For example, Torres, Metscher, and Smith (2011),

examined the relationship between various human factor errors and the occurrence of runway

incursions. As part of their study, Torres et al. reviewed reports of runway incursions submitted to

the Aviation Safety Reporting System (ASRS) and NTSB between January 2005 and

March 2009. From these reports they recorded the number of runway incursions that were

attributed to various human factors errors such as situational awareness, miscommunications, and

airport markings. In the context of chi-square, each contributing factor is considered a category

and the number of incursions associated with each category represents the corresponding

frequency. These data usually are placed in a frequency distribution table similar to what we

presented in Week 2’s lesson. Table 13.1 contains a distribution table of the top five categories

from the Torres et al. study, but the frequencies are fictitious for instructional purposes.

When examined from a hypothesis test perspective, Torres et al. (2011) applied a oneway chi-square strategy to test the null hypothesis that the frequencies of “scores” falling into the

different categories do not differ significantly from the frequencies of scores that have been

hypothesized from theory or the literature. In the Torres et al.’s study, they hypothesized there

would be an even distribution among the runway incursion factors. In other words, they posited

Table 13.1

Number of Runway Incursions

per Human Factors Errors

Category

Situational Awareness

Miscommunication

Distraction

Airport Markings

Complex Taxiways

f

90

70

40

30

20

Note. N = 250.

Michael A. Gallo © 2018

Week 13: Gallo Supplement: The Concept of One-Way Chi-Square Page 1

Table 13.2

Observed and Expected Number of Runway Incursions per Human

Factors Errors Based on A Hypothesized Even Distribution

Category

Situational Awareness

Miscommunication

Distraction

Airport Markings

Complex Taxiways

Observed

Frequencies

(O)

Expected

Proportion

90

70

40

30

20

.2

.2

.2

.2

.2

Expected

Frequencies

(E)

.2 × 250 = 50

.2 × 250 = 50

.2 × 250 = 50

.2 × 250 = 50

.2 × 250 = 50

Note. N = 250.

that the proportion of incursions would be the same across all categories. With five categories,

this mean each category is expected to have 100% / 5 = 20%. Based on the data given in Table

13.1 and a sample size of N = 250, this means that each category is expected to have .20 × 250 =

50 incursions. Thus, to determine the expected frequencies, we multiplied the hypothesized

proportions by the total sample size. This is shown in Table 13.2. Note also from Table 13.2 the

following:

• The scores obtained from sample data are called observed frequencies, denoted O.

• The scores of the claimed or established distribution are called expected frequencies,

denoted E.

In short, the chi-square test for goodness of fit is used to determine if a frequency distribution

obtained using sample data is consistent with a claimed or established distribution. In other

words, it is used to compare observed data to a theoretical model. (Note: Recall from the Week

13 Overview that one-way chi-square also is called the chi-square goodness-of-fit test.)

The Chi-Square Distribution

The chi-square test is based on the chi-square distribution, which is illustrated in Figure

13.1. You will note that this distribution is not symmetrical, but instead its shape depends on the

degrees of freedom similar to the t and F distributions. Unlike the F distribution, though, but

similar to the t distribution, the chi-square distribution has only one degree of freedom. As df

increases, the chi-square distribution becomes more symmetrical. The values of chi-square—

denoted as χ2 —are nonnegative (≥ 0), and df = C − 1 where C = the number of categories. The

critical values of the chi-square distribution are provided in Table 13.3.

Michael A. Gallo © 2018

Week 13: Gallo Supplement: The Concept of One-Way Chi-Square Page 2

Relative Frequency

df = 1

df = 5

df = 9

χ2

Figure 13.1. The chi-square distribution.

From Table 13.3 note that as df increases so too does the critical value. Observe that this

pattern is the complete opposite of the other statistical tests we have discussed in which as df

increases the corresponding critical values decrease. This is because the degrees of freedom for

chi-square are independent of sample size—they are not related to the number of scores in a

sample. Instead, the degrees of freedom are related to the number of categories or possible scores

instead of sample size. For example, the degrees of freedom for the runway incursion example

are df = 5 – 1 = 4 because there are five categories.

Assumptions of Chi-Square

As a nonparametric test, chi-square has fewer assumptions than parametric tests. For

example, the chi-square test does not assume that: (a) data are measured on an interval or ratio

scale, (b) the scores form a normal distribution, and (c) the data fit any particular shape.

Furthermore, there is no homogeneity of variance assumption as there is with the independent

samples t test, and outliers usually are not a problem with nominal or ordinal data. In addition to

the data type requirement the chi-square test has two primary assumptions:

• Independence. Each “score” or observation must be independent of all the other

scores. In other words, there must be one score per participant.

• Sample size. A sufficiently large sample size is needed so that the expected frequency

of each cell is at least 5. Furthermore, all individual cell frequencies must be at least 1.

In other words, each table cell must have a frequency of 1 or more, and each cell

associated with the expected frequencies must have a frequency that is 5 or greater.

Michael A. Gallo © 2018

Week 13: Gallo Supplement: The Concept of One-Way Chi-Square Page 3

Table 13.3

Critical Values for Chi-Square

df

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

40

50

60

70

80

90

100

.995

–0.010

0.072

0.207

0.412

0.676

0.989

1.344

1.735

2.156

2.603

3.074

3.565

4.075

4.601

5.142

5.697

6.265

6.844

7.434

8.034

8.643

9.260

9.886

10.520

11.160

11.808

12.461

13.121

13.787

20.707

27.991

35.534

43.275

51.172

59.196

67.328

.99

–0.020

0.115

0.297

0.554

0.872

1.239

1.646

2.088

2.558

3.053

3.571

4.107

4.660

5.229

5.812

6.408

7.015

7.633

8.260

8.897

9.542

10.196

10.856

11.524

12.198

12.879

13.565

14.256

14.953

22.164

29.707

37.485

45.442

53.540

61.754

70.065

p

0

χ2

Proportion in Critical Region

.975

.95

.90

.10

.05

.025

.01

0.001 0.004

0.016

2.706 3.841

5.024

6.635

0.051 0.103

0.211

4.605 5.991

7.378

9.210

0.216 0.352

0.584

6.251 7.815

9.348 11.345

0.484 0.711

1.064

7.779 9.488 11.143 13.277

0.831 1.145

1.610

9.236 11.070 12.833 15.086

1.237 1.635

2.204 10.645 12.592 14.449 16.812

1.690 2.167

2.833 12.017 14.067 16.013 18.475

2.180 2.733

3.490 13.362 15.507 17.535 20.090

2.700 3.325

4.168 14.684 16.919 19.023 21.666

3.247 3.940

4.865 15.987 18.307 20.483 23.209

3.816 4.575

5.578 17.275 19.675 21.920 24.725

4.404 5.226

6.304 18.549 21.026 23.337 26.217

5.009 5.892

7.042 19.812 22.362 24.736 27.688

5.629 6.571

7.790 21.064 23.685 26.119 29.141

6.262 7.261

8.547 22.307 24.996 27.488 30.578

6.908 7.962

9.312 23.542 26.296 28.845 32.000

7.564 8.672 10.085 24.769 27.587 30.191 33.409

8.231 9.390 10.865 25.989 28.869 31.526 34.805

8.907 10.117 11.651 27.204 30.144 32.852 36.191

9.591 10.851 12.443 28.412 31.410 34.170 37.566

10.283 11.591 13.240 29.615 32.671 35.479 38.932

10.982 12.338 14.041 30.813 33.924 36.781 40.289

11.689 13.091 14.848 32.007 35.172 38.076 41.638

12.401 13.848 15.659 33.196 36.415 39.364 42.980

13.120 14.611 16.473 34.382 37.652 40.646 44.314

13.844 15.379 17.292 35.563 38.885 41.923 45.642

14.573 16.151 18.114 36.741 40.113 43.195 46.963

15.308 16.928 18.939 37.916 41.337 44.461 48.278

16.047 17.708 19.768 39.087 42.557 45.722 49.588

16.791 18.493 20.599 40.256 43.773 46.979 50.892

24.433 26.509 29.051 51.805 55.758 59.342 63.691

32.357 34.764 37.689 63.167 67.505 71.420 76.154

40.482 43.188 46.459 74.397 79.082 83.298 88.379

48.758 51.739 55.329 85.527 90.531 95.023 100.425

57.153 60.391 64.278 96.578 101.879 106.629 112.329

65.647 69.126 73.291 107.565 113.145 118.136 124.116

74.222 77.929 82.358 118.498 124.342 129.561 135.807

.005

7.879

10.597

12.838

14.860

16.750

18.548

20.278

21.955

23.589

25.188

26.757

28.300

29.819

31.319

32.801

34.267

35.718

37.156

38.582

39.997

41.401

42.796

44.181

45.559

46.928

48.290

49.645

50.993

52.336

53.672

66.766

79.490

91.952

104.215

116.321

128.299

140.169

Note. The proportions represent the areas to the right of the critical value. To look up an area to the

left of the critical value, subtract the proportion from 1 and then use the corresponding column.

Michael A. Gallo © 2018

Week 13: Gallo Supplement: The Concept of One-Way Chi-Square Page 4

The Test Statistic for the Chi-Square Goodness-of-Fit Test

Although we will use a statistical software program to calculate the chi-square test

statistic, it nevertheless is helpful to know the formula for this test statistic:

χ2 = Σ

(Oi − Ei ) 2

Ei

From the formula we: (a) calculate the difference between the observed (O) and expected (E)

frequencies for each category, (b)€square each difference and divide these squared values by the

corresponding expected frequency, and (c) add the individual quotients across all categories.

To illustrate how to apply this formula, we will use the data from Table 3.2 for the

runway incursion example by expanding the table to include columns associated with the various

parts of the formula. As shown in Table 13.4, we created three new columns: one for the

difference between the observed and expected frequencies (O – E), one for the squared

difference (O – E)2, and one for the quotient between the squared difference and expected

frequency (O – E)2 / E. We then summed the quotients. Thus, χ2 = 68.0.

The Effect Size for the Chi-Square Goodness-of-Fit Test

The effect size for the chi-square goodness-of-fit test is Cohen’s w:

w=

χ2

N

Similar to the Pearson r, Cohen’s w can vary between 0 and 1. As w approaches 1, the relationship

becomes stronger, and as w approaches 0 the relationship becomes weaker. Cohen also suggested

€

2

the following for χ : (a) ES = .10 is small, (b) ES = .30 is medium, and (c) ES = .50 is large.

Table 13.4

Observed and Expected Number of Runway Incursions per Human Factors Errors Based on A Hypothesized Even

Distribution and Configured to Follow the Chi-Square Test Statistic Formula

Category

Situational Awareness

Miscommunication

Distraction

Airport Markings

Complex Taxiways

Observed

Frequencies

(O)

Expected

Proportion

90

70

40

30

20

.2

.2

.2

.2

.2

Expected

Frequencies

(E)

.2 × 250 = 50

.2 × 250 = 50

.2 × 250 = 50

.2 × 250 = 50

.2 × 250 = 50

(O – E)

90 – 50 = 40

70 – 50 = 20

40 – 50 = -10

30 – 50 = -20

20 – 50 = -30

2

(O – E)

402 = 1600

202 = 400

-102 =€100

-202 = 400

-302 = 900

Sum (Σ)

(O − E) 2

E

1600/50 = 32

400/50 = 8

100/50 = 2

400/50 = 8

900/50 = 18

68.0

Note. N = 250.

Michael A. Gallo © 2018

Week 13: Gallo Supplement: The Concept of One-Way Chi-Square Page 5

Week 13

One-Way Chi-Square: A Guided Example

This handout material provides a guided example of one-way chi-square, which also is

called the chi-square goodness-of-fit test. The example is similar in structure to the other guided

examples that have been presented in previous weeks. Prior to working through this example you

are encouraged to review Chapter 5 of the assigned textbook by Wilson and Joye as well as the

Gallo supplement on one-way chi-square.

Guided Example Context

The context of this guided example is from Torres et al. (2011), which examined the

relationship between various human factors errors and runway incursions. The data set we will use

is summarized in Table 13.1 from the Gallo supplement on one-way chi-square, and is replicated

here for the reader’s convenience and labeled Table 1. A copy of the data also is given in the Excel

file, “Week 13 Guided Example Data.” As shown in Table 1, five categories of human factors

errors are being considered—situational awareness, miscommunication, distraction, airport

markings, and complex taxiways—and their corresponding frequencies are provided.

Pre-data analysis. Before we begin data collection, we first must pose the research

question, identify the correct research methodology to answer the RQ, and conduct an a priori

power analysis to determine the minimum sample size needed.

What is the RQ? The overriding research question for the current example is: “What is

the relationship between the observed frequencies of runway incursions for the targeted five

categories of human factors errors and their corresponding expected frequencies?” As noted in

Torres et al. (2011), runway incursions were defined as those involving pilot deviations or

operational errors; vehicle/pedestrian deviations were not included. Furthermore, the runway

incursions data were collected between January 2005 and March 2009.

Table 1

Number of Runway Incursions

per Human Factors Errors

Category

Situational Awareness

Miscommunication

Distraction

Airport Markings

Complex Taxiways

f

90

70

40

30

20

Note. N = 250.

Michael A. Gallo © 2018

Week 13: Gallo Guided Example: Applying One-Way Chi-Square Page 1

What is the research methodology? The research methodology that would best answer

this question is correlational because we are examining a relationship between two entities

(observed vs. expected frequencies).

What is the minimum sample size needed? To determine the minimum sample size, we

consult G•Power using the following parameters:

• Test family = χ2 tests.

• Statistical test = Goodness of-fit tests: Contingency tables.

• Type of power analysis = A priori: Compute required sample size—given α, power,

and effect size.

• Input parameters are: Effect size w = 0.3 (this is a medium effect), α error prob = .05,

Power = .80, and df = C – 1 = 5 – 1 = 4.

These parameters result in a minimum sample size of N = 133. A copy of the G*Power output is

given in Figure 1.

Figure 1. G•Power output for chi-square test for goodness of fit (a priori power analysis).

Michael A. Gallo © 2018

Week 13: Gallo Guided Example: Applying One-Way Chi-Square Page 2

Data analysis. We now direct our attention to hypothesis testing. Following is a

summary of the steps associated with the corresponding hypothesis test. Before doing so, though,

observe that the data do indeed satisfy the two assumptions for the chi-square goodness-of-fit

test: (a) there is independence of the observations, and (b) each cell for the expected frequencies

is 5 or greater with each cell in Table 3.1 having a frequency of at least 1.

Step 1: Formulate the null and alternative hypotheses.

H0: There is an even distribution of frequencies across all five human factors errors

categories. In other words: There is no significant difference between the observed

frequencies from the sample data and the expected frequencies.

H1: There is a significant difference between the observed frequencies from the sample

data and the expected frequencies.

Step 2: Determine the test criteria. The test statistic is χ2, the level of significance is α =

.05, and the boundary of the critical region is determined from Table 13.3, which was given in

the Week 13 Gallo supplement on one-way chi-square. Based on df = 4 and α = .05, the

corresponding critical χ2 value is 9.488 as illustrated in Figure 2.

α = .05

2

χ 4 = 9.488

Figure 2. Critical region for

χ 2 with df = 4 and α = .05.

Step 3: Collect data and compute sample statistics. The computation of the chi-square

test statistic was done in Table 13.4 of the Week 13 Gallo supplement on one-way chi-square.

Following is an alternative approach to the calculations given in Table 13.4.

χ2 = Σ

(Oi − Ei ) 2

(90 − 50) 2

(70 − 50) 2

(40 − 50) 2

(30 − 50) 2

(20 − 50) 2

=

+

+

+

+

Ei

50

50

50

50

50

=

€

€

(40) 2

(20) 2

(−10) 2

(−20) 2

(−30) 2

+

+

+

+

50€

50

50

50 €

€ 50

€

=

€

€

€

Michael A. Gallo © 2018

1600

400

100

400

900

+

+

+

+

50

50

50

50

50

€

€

€

= 32.0 + 8.0 + 2.0 + 8.0 + 18.0

€

€

=€68.0 €

Week 13: Gallo Guided Example: Applying One-Way Chi-Square Page 3

Step 4: Make a decision: Either reject or fail to reject the null hypothesis. The

calculated χ2 = 68.0 is greater than the χ2 critical boundary of 9.488 for df = 4 and α = .05, and

hence lies in the critical region. Therefore, the decision is to reject the null hypothesis and

conclude there is a significant relationship between the observed frequencies and the expected

frequencies. The number of runway incursions is not uniform across the targeted five categories

of human factors errors.

Post-data analysis. After completing the hypothesis test, we now perform various postdata analysis activities. For one-way chi-square these include determining and reporting the

corresponding effect size and power, and then discussing some plausible explanations for the

results.

What is the effect size? The effect size for the chi-square goodness-of-fit test is Cohen’s

w, which is given by the following formula:

w=

χ2

N

Applying this formula to our results:

w=

χ2

€=

N

68.0

=

250

0.272 ≈ 0.52

Based on Cohen’s guidance, this is a large effect size. As a result, the relationship between the

€

observed and expected frequencies

of runway

incursions yielded a large effect.

€

€

What is the power of this study? Recall that power refers to the probability that the effect

found in the sample truly exists in the parent population. To determine the actual power of this

study we consult G*Power again, but this time we make the following changes:

• The type of power analysis gets changed to Post hoc: Compute achieved power—given

α, sample size, and effect size.

• The input parameters are changed to reflect the actual effect size (ES = 0.52) and Total

sample size (N = 250).

When these changes are made, the power of the study is 1.0, which means there is a greater than

99% probability that the large effect found in the sample truly exists in the population.

What are some plausible explanations for the result? The results indicate there is a

statistically significant relationship between the observed and expected frequencies of runway

incursions. Of the targeted five factors, situational awareness was the most common, followed by

miscommunications and distractions, which led to either a pilot deviation or operational error. A

Michael A. Gallo © 2018

Week 13: Gallo Guided Example: Applying One-Way Chi-Square Page 4

plausible explanation for these findings is that these top three factors relate to the infallible

nature of human beings. It is conceivable that pilots can (and do) lose sight of the actual location

of their aircraft at an airport, particularly if an airport is undergoing construction, which can lead

to blocked signage or closed taxiways and runways. It also is conceivable that ATCs could lose

sight of the big picture by not knowing where the aircraft under their control are located at the

airport. With respect to miscommunications and distractions, it certainly is conceivable that

pilots either do not communicate their position to other pilots in the area or do not monitor this

information from other pilots due to being distracted by various conditions. A plausible

explanation for the last three factors, which are airport related, is the lack of consistency across

airports with respect to their markings, taxiways, and signage. Can you think if other plausible

explanations?

Michael A. Gallo © 2018

Week 13: Gallo Guided Example: Applying One-Way Chi-Square Page 5

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