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Chapter 11: The Chi-Square Distribution

11.2 Goodness-of-Fit Test

Learning Objectives

By the end of this section, the student should be able to:

  • Conduct and interpret chi-square goodness-of-fit hypothesis tests.

In this type of hypothesis test, you determine whether the data "fit" a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities.

The test statistic for a goodness-of-fit test is: [latex]\underset{k}{\Sigma }\frac{{\left(O-E\right)}^{2}}{E}[/latex]

where:

  • [latex]O =[/latex] observed values (data)
  • [latex]E =[/latex] expected values (from theory)
  • [latex]k =[/latex] the number of different data cells or categories

The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true. There are [latex]n[/latex] terms of the form [latex]\frac{{\left(O-E\right)}^{2}}{E}[/latex].

The number of degrees of freedom is [latex]df = (\text{number of categories} – 1)[/latex].

The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve.

Note

The expected value for each cell needs to be at least five in order for you to use this test.

Example

Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to the table below.

Table 1: Number of Absences, Expected Number of Students
Number of absences per term Expected number of students
0–2 50
3–5 30
6–8 12
9–11 6
12+ 2

A random survey across all mathematics courses was then done to determine the actual number (observed) of absences in a course. The chart below displays the results of that survey.

Table 2: Number of Absences, Actual Number of Students
Number of absences per term Actual number of students
0–2 35
3–5 40
6–8 20
9–11 1
12+ 4

Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test.

[latex]H_0:[/latex] Student absenteeism fits faculty perception.

The alternative hypothesis is the opposite of the null hypothesis.

[latex]H_a:[/latex] Student absenteeism does not fit faculty perception.

  1. Can you use the information as it appears in the charts to conduct the goodness-of-fit test?
  2. What is the number of degrees of freedom ([latex]df[/latex])?
Solution
  1. No. Notice that the expected number of absences for the "12+" entry is less than five (it is two). Combine that group with the "9–11" group to create new tables where the number of students for each entry are at least five. The new results are shown below.
  2. There are four "cells" or categories in each of the new tables. [latex]df = \text{number of cells} – 1 = 4 – 1 = 3[/latex]
Table 3: Number of Absences Edited, Expected Number of Students
Number of absences per term Expected number of students
0–2 50
3–5 30
6–8 12
9+ 8
Table 4: Number of Absences Edited, Actual Number of Students
Number of absences per term Actual number of students
0–2 35
3–5 40
6–8 20
9+ 5

Your Turn!

A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed below.

Table 5: Number of Products, Expected Number Defective
Number produced Number defective
0–100 5
101–200 6
201–300 7
301–400 8
401–500 10

A random sample was taken to determine the actual number of defects. The following table shows the results of the survey.

Table 6: Number of Products, Actual Number Defective
Number produced Number defective
0–100 5
101–200 7
201–300 8
301–400 9
401–500 11

State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.

Solution

[latex]H_0:[/latex] The number of defaults fits expectations.

[latex]H_a:[/latex] The number of defaults does not fit expectations.

[latex]df = 4[/latex]

Example

Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as shown in the table below. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5% significance level.

Table 7: Day of the Week Employees Were Most Absent
Monday Tuesday Wednesday Thursday Friday
Number of Absences 15 12 9 9 15
Solution

The null and alternative hypotheses are:

  • [latex]H_0:[/latex] The absent days occur with equal frequencies, that is, they fit a uniform distribution.
  • [latex]H_a:[/latex] The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution.

If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: [latex]15 + 12 + 9 + 9 + 15 = 60[/latex]), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the expected ([latex]E[/latex]) values. The values in the table are the observed ([latex]O[/latex]) values or data.

Calculate the [latex]\chi^2[/latex] test statistic. Make a chart with the following headings and fill in the columns:

  • Expected ([latex]E[/latex]) values (12, 12, 12, 12, 12)
  • Observed ([latex]O[/latex]) values (15, 12, 9, 9, 15)
  • ([latex]O – E[/latex])
  • [latex](O – E)^2[/latex]
  • [latex]\frac{{(O – E)}^{2}}{E}[/latex]

Now add (sum) the last column. The sum is three. This is the [latex]\chi^2[/latex] test statistic.

To find the p-value, calculate [latex]P(\chi^2 > 3)[/latex]. This test is right-tailed. You should get [latex]\text{p-value} = 0.5578[/latex].

The [latex]df \text{s}[/latex] are [latex]\text{the number of cells} – 1 = 5 – 1 = 4[/latex].

Use the probability table found in the Back Matter - Statistics Tables where needed.

Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.)

This is a blank nonsymmetrical chi-square curve for the test statistic of the days of the week absent.
Figure 1. Chi-Square Distribution Example

 

The decision is not to reject the null hypothesis.

Conclusion: At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies.

Your Turn!

Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 49 students were asked on which night of the week they did the most homework. The results were distributed as in the chart below.

Table 8: Day of the Week Students Did Homework
Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Number of Students 11 8 10 7 10 5 5

From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use?

Solution

[latex]df = 6[/latex]

[latex]\text{p-value} = 0.6093[/latex]

We decline to reject the null hypothesis. There is not enough evidence to support that students do not do the majority of their homework equally throughout the week.

Example

One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population) as in the following table.

Table 9: Number of Televisions and Percent
Number of Televisions Percent
0 10
1 16
2 55
3 11
4+ 8

The table contains expected ([latex]E[/latex]) percentages.

A random sample of 600 families in the far western United States resulted in the data in the following table.

Table 10: Number of Televisions and Frequency
Number of Televisions Frequency
Total = 600
0 66
1 119
2 340
3 60
4+ 15

The table contains observed ([latex]O[/latex]) frequency values.

At the 1% significance level, does it appear that the distribution "number of televisions" of far western United States families is different from the distribution for the American population as a whole?

Solution

This problem asks you to test whether the far western United States families distribution fits the distribution of the American families. This test is always right-tailed.

The first table contains expected percentages. To get expected ([latex]E[/latex]) frequencies, multiply the percentage by 600. The expected frequencies are shown in the following table.

Table 11: Number of Televisions, Percent, and Expected Frequency
Number of Televisions Percent Expected Frequency
0 10 (0.10)(600) = 60
1 16 (0.16)(600) = 96
2 55 (0.55)(600) = 330
3 11 (0.11)(600) = 66
over 3 8 (0.08)(600) = 48

Therefore, the expected frequencies are 60, 96, 330, 66, and 48. You can let the calculator do the math. For example, instead of 60, enter 0.10*600.

[latex]H_0[/latex]: The "number of televisions" distribution of far western United States families is the same as the "number of televisions" distribution of the American population.

[latex]H_a[/latex]: The "number of televisions" distribution of far western United States families is different from the "number of televisions" distribution of the American population.

Distribution for the test: [latex]{\chi }_{4}^{2}[/latex] where [latex]df = \text{the number of cells} – 1 = 5 – 1 = 4[/latex].

Note: [latex]df \neq 600 – 1[/latex]

Calculate the test statistic: [latex]\chi^2 = 29.65[/latex]

Graph:

A nonsymmetric chi-square curve with a value of 4 at the peak of the curve. A value of 29.65 is on the axis with a region to the right shaded.
Figure 2. A nonsymmetric chi-square curve with 4 at the peak. The shaded area is equal to the p-value of 0.000006.

 

Probability statement: [latex]\text{p-value} = P(\chi^2 > 29.65) = 0.000006[/latex]

Compare α and the p-value: [latex]\alpha = 0.01[/latex]; [latex]\text{p-value} = 0.000006[/latex]; So, [latex]\alpha > \text{p-value}[/latex].

Make a decision: Since [latex]\alpha > \text{p-value}[/latex], reject [latex]H_0[/latex].

This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole.

Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude that the "number of televisions" distribution for the far western United States is different from the "number of televisions" distribution for the American population as a whole.

Your Turn!

The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in the table below.

Table 12: Number of Pets and Percent
Number of Pets Percent
0 18
1 25
2 30
3 18
4+ 9

A random sample of 1,000 students from the Eastern United States resulted in the data in the table below.

Table 13: Number of Pets and Frequency
Number of Pets Frequency
0 210
1 240
2 320
3 140
4+ 90

At the 1% significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the p-value?

Solution

[latex]\text{p-value} = 0.0036[/latex]

We reject the null hypothesis that the distributions are the same. There is sufficient evidence to conclude that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole.

Example

Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are the coins fair? Test at a 5% significance level.

Solution

This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is {HH, HT, TH, TT}. Out of 100 flips, you would expect 25 HH, 25 HT, 25 TH, and 25 TT. This is the expected distribution. The question, "Are the coins fair?" is the same as saying, "Does the distribution of the coins (20 HH, 27 HT, 30 TH, 23 TT) fit the expected distribution?"

Random Variable: Let [latex]X =[/latex] the number of heads in one flip of the two coins. X takes on the values 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the number of cells is three. Since [latex]X =[/latex] the number of heads, the observed frequencies are 20 (for two heads), 57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (for two heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed.

[latex]H_0:[/latex] The coins are fair.

[latex]H_a:[/latex] The coins are not fair.

Distribution for the test: [latex]{\chi }_{2}^{2}[/latex] where [latex]df = 3 – 1 = 2[/latex].

Calculate the test statistic: [latex]\chi^2 = 2.14[/latex]

Graph:

A nonsymmetrical chi-square curve with no labeled value at the peak, but a value of 2.14 is on the axis with a region to the right shaded.
Figure 3. A nonsymmetric chi-square curve with a shaded area equal to the p-value of 0.3430.

 

Probability statement: p-value = [latex]P(\chi^2 > 2.14) = 0.3430[/latex]

Compare [latex]\alpha[/latex] and the p-value: Since [latex]\alpha = 0.05[/latex] and [latex]\text{p-value} = 0.3430[/latex], [latex]\alpha \lt \text{p-value}[/latex].

Make a decision: Since [latex]\alpha \lt \text{p-value}[/latex], do not reject [latex]H_0[/latex].

Conclusion: There is insufficient evidence to conclude that the coins are not fair.

Your Turn!

Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. The following table shows the actual literacy rates across the world broken down by region. What are the test statistics and the degrees of freedom?

Table 14: MDG Region and Adult Literacy Rate %
MDG Region Adult Literacy Rate (%)
Developed Regions 99.0
Commonwealth of Independent States 99.5
Northern Africa 67.3
Sub-Saharan Africa 62.5
Latin America and the Caribbean 91.0
Eastern Asia 93.8
Southern Asia 61.9
South-Eastern Asia 91.9
Western Asia 84.5
Oceania 66.4
Solution

[latex]df = 9[/latex]

[latex]\chi^2 \text{test statistic} = 26.38[/latex]

A nonsymmetric chi-square curve, df=9, with a value of 9 at the peak of the curve. A value of 26.38 is on the axis with a region to the right shaded.
Figure 4. A nonsymmetric chi-square curve with 9 at the peak. The shaded area is equal to the p-value of 0.0018

Videos

Below are helpful videos for the content covered in this Section. Videos are provided from YouTube.

Section 11.2 Review

To assess whether a data set fits a specific distribution, you can apply the goodness-of-fit hypothesis test that uses the chi-square distribution. The null hypothesis for this test states that the data come from the assumed distribution. The test compares observed values against the values you would expect to have if your data followed the assumed distribution. The test is almost always right-tailed. Each observation or cell category must have an expected value of at least five.

Formula Review

  • Goodness-of-fit test statistic: [latex]\sum _{k}\frac{{\left(O-E\right)}^{2}}{E}[/latex], where:
    • [latex]O:[/latex] observed values
    • [latex]E:[/latex] expected values
    • [latex]k:[/latex] number of different data cells or categories
  • Degrees of freedom: [latex]df = k − 1[/latex]

Section 11.2 Practice

Determine the appropriate test to be used:

  1. An archeologist is calculating the distribution of the frequency of the number of artifacts she finds in a dig site. Based on previous digs, the archeologist creates an expected distribution broken down by grid sections in the dig site. Once the site has been fully excavated, she compares the actual number of artifacts found in each grid section to see if her expectation was accurate.
  2. An economist is deriving a model to predict outcomes on the stock market. He creates a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, he records the actual points on the index. He wants to see how well his model matched what actually happened.
  3. A personal trainer is putting together a weight-lifting program for her clients. For a 90-day program, she expects each client to lift a specific maximum weight each week. As she goes along, she records the actual maximum weights her clients lifted. She wants to know how well her expectations met with what was observed.

A teacher predicts that the distribution of grades on the final exam will be and they are recorded in the following table.

Table 15: Grade and Proportion
Grade Proportion
A 0.25
B 0.30
C 0.35
D 0.10

The actual distribution for a class of 20 is in the following table.

Table 16: Grade and Percent
Grade Frequency
A 7
B 7
C 5
D 1
  1. [latex]df= \underline{\hspace{2cm}}[/latex]
  2. State the null and alternative hypotheses.
  3. [latex]\chi^2 \text{ test statistic} = \underline{\hspace{2cm}}[/latex]
  4. [latex]\text{p-value} = \underline{\hspace{2cm}}[/latex]
  5. At the 5% significance level, what can you conclude?

The following data are real. The cumulative number of AIDS cases reported for Santa Clara County is broken down by ethnicity as in the following table.

Table 17: Ethnicity and Number of Cases
Ethnicity Number of Cases
White 2,229
Hispanic 1,157
Black/African-American 457
Asian, Pacific Islander 232
Total = 4,075

The percentage of each ethnic group in Santa Clara County is as in the following table.

Table 18: Ethnicity and Percentage of Total Population
Ethnicity Percentage of total county population Number
expected (round to two decimal places)
White 42.9% 1748.18
Hispanic 26.7%
Black/African-American 2.6%
Asian, Pacific Islander 27.8%
Total = 100%
  1. If the ethnicities of AIDS victims followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group.
  2. Perform a goodness-of-fit test to determine whether the occurrence of AIDS cases follows the ethnicities of the general population of Santa Clara County.
    • [latex]H_0: \underline{\hspace{2cm}}[/latex]
    • [latex]H_a: \underline{\hspace{2cm}}[/latex]
    • Is this a right-tailed, left-tailed, or two-tailed test?
    • [latex]df = \underline{\hspace{2cm}}[/latex]
    • [latex]\chi^2 \text{ test statistic} = \underline{\hspace{2cm}}[/latex]
    • [latex]\text{p-value} = \underline{\hspace{2cm}}[/latex]
    • Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p-value.
    • Let [latex]\alpha = 0.05[/latex].
      • Decision: [latex]\underline{\hspace{2cm}}[/latex]
      • Reason for the Decision: [latex]\underline{\hspace{2cm}}[/latex]
      • Conclusion (write out in complete sentences): [latex]\underline{\hspace{2cm}}[/latex]
    • Does it appear that the pattern of AIDS cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not?

A six-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct a hypothesis test to determine if the die is fair. The data in the table below are the result of the 120 rolls.

Table 19: Face Value of Dice and Frequency, Blank Expected Frequency
Face Value Frequency Expected Frequency
1 15
2 29
3 16
4 15
5 30
6 15

Use the Chi-Square Distribution - Solution Sheet on the Introduction to Chapter 11 page. (Round expected frequency to two decimal places.)

The marital status distribution of the U.S. male population, ages 15 and older, is as shown in the following table.

Table 20: Marital Status and Percent, Blank Expected Frequency
Marital Status Percent Expected Frequency
never married 31.3
married 56.1
widowed 2.5
divorced/separated 10.1

Suppose that a random sample of 400 U.S. young adult males, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group of males fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying 400 people. Fill in the following table, rounding to two decimal places.

Table 21: Marital Status and Frequency
Marital Status Frequency
never married 140
married 238
widowed 2
divorced/separated 20

Use the Chi-Square Distribution - Solution Sheet on the Introduction to Chapter 11 page. (Round expected frequency to two decimal places.)

Solution
Table 22: Marital Status, Percent, and Expected Frequency
Marital Status Percent Expected Frequency
never married 31.3 125.2
married 56.1 224.4
widowed 2.5 10
divorced/separated 10.1 40.4
  1. [latex]H_0:[/latex] The data fits the distribution.
  2. [latex]H_a:[/latex] The data does not fit the distribution.
  3. [latex]df = 3[/latex]
  4. chi-square distribution with [latex]df = 3[/latex]
  5. chi-square test statistic = 19.27
  6. p-value = 0.0002
  7. Check student’s solution.
  8. Correct decision (“reject” or “do not reject” the null hypothesis) and appropriate conclusions:
    • Alpha = 0.05
    • Decision: Reject null
    • Reason for decision: p-value < alpha
    • Conclusion: Data does not fit the distribution.

The City of South Lake Tahoe, CA, has an Asian population of 1,419 people, out of a total population of 23,609. Suppose that a survey of 1,419 self-reported Asians in the Manhattan, NY, area yielded the data in the table below. Conduct a goodness-of-fit test to determine if the self-reported sub-groups of Asians in the Manhattan area fit that of the Lake Tahoe area.

Table 23: Race, Lake Tahoe and Manhattan Frequencies
Race Lake Tahoe Frequency Manhattan Frequency
Asian Indian 131 174
Chinese 118 557
Filipino 1,045 518
Japanese 80 54
Korean 12 29
Vietnamese 9 21
Other 24 66

Use the Chi-Square Distribution - Solution Sheet on the Introduction to Chapter 11 page. (Round expected frequency to two decimal places.)

UCLA conducted a survey of more than 263,000 college freshmen from 385 colleges in fall 2005. The results of students' expected majors by gender were reported in The Chronicle of Higher Education (2/2/2006). Suppose a survey of 5,000 graduating females and 5,000 graduating males was done as a follow-up last year to determine what their actual majors were. The results are shown in the tables below. The second column in each table does not add to 100% because of rounding. Conduct a goodness-of-fit test to determine if the actual college majors of graduating females fit the distribution of their expected majors.

Table 24: Major, Women Expected and Actual
Major Women - Expected Major Women - Actual Major
Arts & Humanities 14.0% 670
Biological Sciences 8.4% 410
Business 13.1% 685
Education 13.0% 650
Engineering 2.6% 145
Physical Sciences 2.6% 125
Professional 18.9% 975
Social Sciences 13.0% 605
Technical 0.4% 15
Other 5.8% 300
Undecided 8.0% 420
Table 25: Major, Men Expected and Actual
Major Men - Expected Major Men - Actual Major
Arts & Humanities 11.0% 600
Biological Sciences 6.7% 330
Business 22.7% 1130
Education 5.8% 305
Engineering 15.6% 800
Physical Sciences 3.6% 175
Professional 9.3% 460
Social Sciences 7.6% 370
Technical 1.8% 90
Other 8.2% 400
Undecided 6.6% 340

Use the Chi-Square Distribution - Solution Sheet on the Introduction to Chapter 11 page. (Round expected frequency to two decimal places.)

Solution
  1. [latex]H_0:[/latex] The actual college majors of graduating females fit the distribution of their expected majors
  2. [latex]H_a:[/latex] The actual college majors of graduating females do not fit the distribution of their expected majors
  3. [latex]df = 10[/latex]
  4. chi-square distribution with [latex]df = 10[/latex]
  5. chi-square test statistic = 11.48
  6. p-value = 0.3211
  7. Check student’s solution.
  8. Correct decision (“reject” or “do not reject” the null hypothesis) and appropriate conclusions:
    • Alpha = 0.05
    • Decision: Do not reject null when [latex]\alpha = 0.05[/latex] and [latex]\alpha = 0.01[/latex]
    • Reason for decision: p-value > alpha
    • Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females fits the distribution of their expected majors.

Read the statement and decide whether it is true or false.

  1. In a goodness-of-fit test, the expected values are the values we would expect if the null hypothesis were true.
  2. In general, if the observed values and expected values of a goodness-of-fit test are not close together, then the test statistic can get very large and on a graph will be way out in the right tail.
  3. Use a goodness-of-fit test to determine if high school principals believe that students are absent equally during the week or not.
  4. The test to use to determine if a six-sided die is fair is a goodness-of-fit test.
  5. In a goodness-of fit test, if the p-value is 0.0113, in general, do not reject the null hypothesis.

The table below contains information from a survey among 499 participants classified according to their age groups. The second column shows the percentage of obese people per age class among the study participants. The last column comes from a different study at the national level that shows the corresponding percentages of obese people in the same age classes in the USA. Perform a hypothesis test at the 5% significance level to determine whether the survey participants are a representative sample of the USA obese population.

Table 26: Age, Obese Percentage and Expected Percentage
Age Class (Years) Obese (Percentage) Expected USA average (Percentage)
20–30 75.0 32.6
31–40 26.5 32.6
41–50 13.6 36.6
51–60 21.9 36.6
61–70 21.0 39.7

Use the Chi-Square Distribution - Solution Sheet on the Introduction to Chapter 11 page. (Round expected frequency to two decimal places.)

Solution
  1. [latex]H_0:[/latex] Surveyed obese fit the distribution of expected obese
  2. [latex]H_a:[/latex] Surveyed obese do not fit the distribution of expected obese
  3. [latex]df = 4[/latex]
  4. chi-square distribution with [latex]df = 4[/latex]
  5. chi- square test statistic = 54.01
  6. p-value = 0
  7. Check student’s solution.
  8. Correct decision (“reject” or “do not reject” the null hypothesis) and appropriate conclusions:
    • Alpha: 0.05
    • Decision: Reject the null hypothesis.
    • Reason for decision: p-value < alpha
    • Conclusion: At the 5% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese.

References

Data from the U.S. Census Bureau

Data from the College Board. Available online at http://www.collegeboard.com.

Data from the U.S. Census Bureau, Current Population Reports.

Ma, Y., E.R. Bertone, E.J. Stanek III, G.W. Reed, J.R. Hebert, N.L. Cohen, P.A. Merriam, I. S. Ockene, “Association between Eating Patterns and Obesity in a Free-living US Adult Population.” American Journal of Epidemiology volume 158, no. 1, pages 85-92.

Ogden, Cynthia L., Margaret D. Carroll, Brian K. Kit, Katherine M. Flegal, “Prevalence of Obesity in the United States, 2009–2010.” NCHS Data Brief no. 82, January 2012. Available online at http://www.cdc.gov/nchs/data/databriefs/db82.pdf (accessed May 24, 2013).

Stevens, Barbara J., “Multi-family and Commercial Solid Waste and Recycling Survey.” Arlington Count, VA. Available online at http://www.arlingtonva.us/departments/EnvironmentalServices/SW/file84429.pdf (accessed May 24,2013).

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