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 chisquare test (meaning the distribution for the hypothesis test is chisquare) 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 goodnessoffit test is:
where:
 O = observed values (data)
 E = expected values (from theory)
 k = 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 n terms of the form $\frac{{(OE)}^{2}}{E}$.
The number of degrees of freedom is df = (number of categories – 1).
The goodnessoffit test is almost always righttailed. 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 chisquare curve.
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 [link].
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 in [link] displays the results of that survey.
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 goodnessoffit test.
H_{0}: Student absenteeism fits faculty perception.
The alternative hypothesis is the opposite of the null hypothesis.
H_{a}: Student absenteeism does not fit faculty perception.
a. Can you use the information as it appears in the charts to conduct the goodnessoffit test?
a. 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 in [link] and [link].
Number of absences per term  Expected number of students 
0–2  50 
3–5  30 
6–8  12 
9+  8 
Number of absences per term  Actual number of students 
0–2  35 
3–5  40 
6–8  20 
9+  5 
b. What is the number of degrees of freedom (df)?
b. There are four "cells" or categories in each of the new tables.
df = number of cells – 1 = 4 – 1 = 3
Employers want to know which days of the week employees are absent in a fiveday 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 in [link]. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a fiveday work week? Test at a 5% significance level.
Monday  Tuesday  Wednesday  Thursday  Friday  
Number of Absences  15  12  9  9  15 
The null and alternative hypotheses are:
 H_{0}: The absent days occur with equal frequencies, that is, they fit a uniform distribution.
 H_{a}: 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: 15 + 12 + 9 + 9 + 15 = 60), 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 (E) values. The values in the table are the observed (O) values or data.
This time, calculate the χ^{2} test statistic by hand. Make a chart with the following headings and fill in the columns:
 Expected (E) values (12, 12, 12, 12, 12)
 Observed (O) values (15, 12, 9, 9, 15)
 (O – E)
 (O – E)^{2}
 $\frac{{(O\text{\u2013}E)}^{2}}{E}$
Now add (sum) the last column. The sum is three. This is the χ^{2} test statistic.
To find the pvalue, calculate P(χ^{2} > 3). This test is righttailed. (Use a computer or calculator to find the pvalue. You should get pvalue = 0.5578.)
The dfs are the number of cells – 1 = 5 – 1 = 4
Press
2nd DISTR
. Arrow down to χ^{2}cdf
. Press ENTER
. Enter (3,10^99,4)
. Rounded to four decimal places, you should see 0.5578, which is the pvalue.Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.)
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.
TI83+ and some TI84 calculators do not have a special program for the test statistic for the goodnessoffit test. The next example [link] has the calculator instructions. The newer TI84 calculators have in
STAT TESTS
the test Chi2 GOF
. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS
and Chi2 GOF
. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate
or draw
. Make sure you clear any lists before you start. To Clear Lists in the calculators: Go into STAT EDIT
and arrow up to the list name area of the particular list. Press CLEAR
and then arrow down. The list will be cleared. Alternatively, you can press STAT
and press 4 (for ClrList
). Enter the list name and press ENTER
.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 [link].
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?
df = 6
pvalue = 0.6093
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.
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 [link].
Number of Televisions  Percent 
0  10 
1  16 
2  55 
3  11 
4+  8 
The table contains expected (E) percents.
A random sample of 600 families in the far western United States resulted in the data in [link].
Number of Televisions  Frequency 
Total = 600  
0  66 
1  119 
2  340 
3  60 
4+  15 
The table contains observed (O) 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?
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 righttailed.
The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are shown in [link].
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. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter 0.10*600.
H_{0}: The "number of televisions" distribution of far western United States families is the same as the "number of televisions" distribution of the American population.
H_{a}: 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: ${\chi}_{4}^{2}$ where df = (the number of cells) – 1 = 5 – 1 = 4.
df ≠ 600 – 1
Calculate the test statistic: χ2 = 29.65
Graph:
Probability statement: pvalue = P(χ^{2} > 29.65) = 0.000006
Compare α and the pvalue:
 α = 0.01
 pvalue = 0.000006
Make a decision: Since α > pvalue, reject H_{o}.
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.
Press
STAT
and ENTER
. Make sure to clear lists L1
, L2
, and L3
if they have data in them (see the note at the end of [link]). Into L1
, put the observed frequencies 66
, 119
, 349
, 60
, 15
. Into L2
, put the expected frequencies .10*600, .16*600
, .55*600
, .11*600
, .08*600
. Arrow over to list L3
and up to the name area "L3"
. Enter (L1L2)^2/L2
and ENTER
. Press 2nd QUIT
. Press 2nd LIST
and arrow over to MATH
. Press 5
. You should see "sum" (Enter L3)
. Rounded to 2 decimal places, you should see 29.65
. Press 2nd DISTR
. Press 7
or Arrow down to 7:χ2cdf
and press ENTER
. Enter (29.65,1E99,4)
. Rounded to four places, you should see 5.77E6 = .000006
(rounded to six decimal places), which is the pvalue.The newer TI84 calculators have in
STAT TESTS
the test Chi2 GOF
. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS
and Chi2 GOF
. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate
or draw
. Make sure you clear any lists before you start.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 [link].
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 [link].
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 pvalue?
pvalue = 0.0036
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.
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.
This problem can be set up as a goodnessoffit 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 X = 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 X = 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 righttailed.
H_{0}: The coins are fair.
H_{a}: The coins are not fair.
Distribution for the test: ${\chi}_{2}^{2}$ where df = 3 – 1 = 2.
Calculate the test statistic: χ^{2} = 2.14
Graph:
Probability statement: pvalue = P(χ^{2} > 2.14) = 0.3430
Compare α and the pvalue:
 α = 0.05
 pvalue = 0.3430
Make a decision: Since α < pvalue, do not reject H_{0}.
Conclusion: There is insufficient evidence to conclude that the coins are not fair.
Press
STAT
and ENTER
. Make sure you
clear lists L1
, L2
, and L3
if they have data in them. Into L1
, put the observed
frequencies 20
, 57
, 23
. Into L2
, put the expected frequencies 25
, 50
, 25
. Arrow
over to list L3
and up to the name area "L3"
. Enter (L1L2)^2/L2
and
ENTER
. Press 2nd QUIT
. Press 2nd LIST
and arrow over to MATH
. Press
5
. You should see "sum"
.Enter L3
. Rounded to two decimal places, you
should see 2.14
. Press 2nd DISTR
. Arrow down to 7:χ2cdf
(or press 7
). Press
ENTER
. Enter 2.14,1E99,2)
. Rounded to four places, you should see .3430
, which
is the pvalue.The newer TI84 calculators have in
STAT TESTS
the test Chi2 GOF
. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS
and Chi2 GOF
. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate
or draw
. Make
sure you clear any lists before you start.Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. [link] shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom?
MDG Region  Adult Literacy Rate (%) 
Developed Regions  99.0 
Commonwealth of Independent States  99.5 
Northern Africa  67.3 
SubSaharan Africa  62.5 
Latin America and the Caribbean  91.0 
Eastern Asia  93.8 
Southern Asia  61.9 
SouthEastern Asia  91.9 
Western Asia  84.5 
Oceania  66.4 
degrees of freedom = 9
chi^{2} test statistic = 26.38
Press
STAT
and ENTER
. Make sure you clear lists L1, L2,
and L3
if they have data in them. Into L1, put the observed frequencies 99, 99.5, 67.3, 62.5, 91, 93.8, 61.9, 91.9, 84.5, 66.4
. Into L2
, put the expected frequencies 82, 82, 82, 82, 82, 82, 82, 82, 82, 82
. Arrow over to list L3
and up to the name area "L3"
. Enter (L1L2)^2/L2
and ENTER
. Press 2nd QUIT
. Press 2nd LIST
and arrow over to MATH
. Press 5
. You should see "sum"
. Enter L3
. Rounded to two decimal places, you should see 26.38
. Press 2nd DISTR
. Arrow down to 7:χ2cdf
(or press 7
). Press ENTER
. Enter 26.38,1E99,9)
. Rounded to four places, you should see .0018
, which is the pvalue.The newer TI84 calculators have in
STAT TESTS
the test Chi2 GOF
. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS
and Chi2 GOF
. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate
or draw
. Make sure you clear any lists before you start.
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 Freeliving US Adult Population.” American Journal of Epidemiology volume 158, no. 1, pages 8592.
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., “Multifamily 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).
Chapter Review
To assess whether a data set fits a specific distribution, you can apply the goodnessoffit hypothesis test that uses the chisquare 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 righttailed. Each observation or cell category must have an expected value of at least five.
Formula Review
$\sum _{k}\frac{{(OE)}^{2}}{E}$
goodnessoffit test statistic where:
O: observed values
E: expected values
k: number of different data cells or categories
df = k − 1 degrees of freedom
Determine the appropriate test to be used in the next three exercises.
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.
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.
a goodnessoffit test
A personal trainer is putting together a weightlifting program for her clients. For a 90day 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.
Use the following information to answer the next five exercises: A teacher predicts that the distribution of grades on the final exam will be and they are recorded in [link].
Grade  Proportion 
A  0.25 
B  0.30 
C  0.35 
D  0.10 
The actual distribution for a class of 20 is in [link].
Grade  Frequency 
A  7 
B  7 
C  5 
D  1 
$df=$ ______
3
State the null and alternative hypotheses.
χ^{2} test statistic = ______
2.04
pvalue = ______
At the 5% significance level, what can you conclude?
We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.
Use the following information to answer the next nine exercises: The following data are real. The cumulative number of AIDS cases reported for Santa Clara County is broken down by ethnicity as in [link].
Ethnicity  Number of Cases 
White  2,229 
Hispanic  1,157 
Black/AfricanAmerican  457 
Asian, Pacific Islander  232 
Total = 4,075 
The percentage of each ethnic group in Santa Clara County is as in [link].
Ethnicity  Percentage of total county population  Number expected (round to two decimal places) 
White  42.9%  1748.18 
Hispanic  26.7%  
Black/AfricanAmerican  2.6%  
Asian, Pacific Islander  27.8%  
Total = 100% 
If the ethnicities of AIDS victims followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group.
Perform a goodnessoffit test to determine whether the occurrence of AIDS cases follows the ethnicities of the general population of Santa Clara County.
H_{0}: _______
H_{0}: the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County.
H_{a}: _______
Is this a righttailed, lefttailed, or twotailed test?
righttailed
degrees of freedom = _______
χ^{2} test statistic = _______
88,621
pvalue = _______
Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the pvalue.
Let α = 0.05
Decision: ________________
Reason for the Decision: ________________
Conclusion (write out in complete sentences): ________________
Graph: Check student’s solution.
Decision: Reject the null hypothesis.
Reason for the Decision: pvalue < alpha
Conclusion (write out in complete sentences): The makeup of AIDS cases does not fit the ethnicities of the general population of Santa Clara County.
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?
Homework
For each problem, use a solution sheet to solve the hypothesis test problem. Go to [link] for the chisquare solution sheet. Round expected frequency to two decimal places.
A sixsided 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 [link] are the result of the 120 rolls.
Face Value  Frequency  Expected Frequency 
1  15  
2  29  
3  16  
4  15  
5  30  
6  15 
The marital status distribution of the U.S. male population, ages 15 and older, is as shown in [link].
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 [link], rounding to two decimal places.
Marital Status  Frequency 
never married  140 
married  238 
widowed  2 
divorced/separated  20 
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 
 The data fits the distribution.
 The data does not fit the distribution.
 3
 chisquare distribution with df = 3
 19.27
 0.0002
 Check student’s solution.

 Alpha = 0.05
 Decision: Reject null
 Reason for decision: pvalue < alpha
 Conclusion: Data does not fit the distribution.
Use the following information to answer the next two exercises: The columns in [link] contain the Race/Ethnicity of U.S. Public Schools for a recent year, the percentages for the Advanced Placement Examinee Population for that class, and the Overall Student Population. Suppose the right column contains the result of a survey of 1,000 local students from that year who took an AP Exam.
Race/Ethnicity  AP Examinee Population  Overall Student Population  Survey Frequency 
Asian, Asian American, or Pacific Islander  10.2%  5.4%  113 
Black or AfricanAmerican  8.2%  14.5%  94 
Hispanic or Latino  15.5%  15.9%  136 
American Indian or Alaska Native  0.6%  1.2%  10 
White  59.4%  61.6%  604 
Not reported/other  6.1%  1.4%  43 
Perform a goodnessoffit test to determine whether the local results follow the distribution of the U.S. overall student population based on ethnicity.
Perform a goodnessoffit test to determine whether the local results follow the distribution of U.S. AP examinee population, based on ethnicity.
 H_{0}: The local results follow the distribution of the U.S. AP examinee population
 H_{a}: The local results do not follow the distribution of the U.S. AP examinee population
 df = 5
 chisquare distribution with df = 5
 chisquare test statistic = 13.4
 pvalue = 0.0199
 Check student’s solution.
 Alpha = 0.05
 Decision: Reject null when a = 0.05
 Reason for Decision: pvalue < alpha
 Conclusion: Local data do not fit the AP Examinee Distribution.
 Decision: Do not reject null when a = 0.01
 Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee 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 selfreported Asians in the Manhattan, NY, area yielded the data in [link]. Conduct a goodnessoffit test to determine if the selfreported subgroups of Asians in the Manhattan area fit that of the Lake Tahoe area.
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 following information to answer the next two exercises: 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
Conduct a goodnessoffit test to determine if the actual college majors of graduating females fit the distribution of their expected majors.
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 
 H_{0}: The actual college majors of graduating females fit the distribution of their expected majors
 H_{a}: The actual college majors of graduating females do not fit the distribution of their expected majors
 df = 10
 chisquare distribution with df = 10
 test statistic = 11.48
 pvalue = 0.3211
 Check student’s solution.

 Alpha = 0.05
 Decision: Do not reject null when a = 0.05 and a = 0.01
 Reason for decision: pvalue > 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.
Conduct a goodnessoffit test to determine if the actual college majors of graduating males fit the distribution of their expected majors.
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 
Read the statement and decide whether it is true or false.
In a goodnessoffit test, the expected values are the values we would expect if the null hypothesis were true.
true
In general, if the observed values and expected values of a goodnessoffit 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.
Use a goodnessoffit test to determine if high school principals believe that students are absent equally during the week or not.
true
The test to use to determine if a sixsided die is fair is a goodnessoffit test.
In a goodnessof fit test, if the pvalue is 0.0113, in general, do not reject the null hypothesis.
false
A sample of 212 commercial businesses was surveyed for recycling one commodity; a commodity here means any one type of recyclable material such as plastic or aluminum. [link] shows the business categories in the survey, the sample size of each category, and the number of businesses in each category that recycle one commodity. Based on the study, on average half of the businesses were expected to be recycling one commodity. As a result, the last column shows the expected number of businesses in each category that recycle one commodity. At the 5% significance level, perform a hypothesis test to determine if the observed number of businesses that recycle one commodity follows the uniform distribution of the expected values.
Business Type  Number in class  Observed Number that recycle one commodity  Expected number that recycle one commodity 
Office  35  19  17.5 
Retail/Wholesale  48  27  24 
Food/Restaurants  53  35  26.5 
Manufacturing/Medical  52  21  26 
Hotel/Mixed  24  9  12 
[link] 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.
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 
 H_{0}: Surveyed obese fit the distribution of expected obese
 H_{a}: Surveyed obese do not fit the distribution of expected obese
 df = 4
 chisquare distribution with df = 4
 test statistic = 54.01
 pvalue = 0
 Check student’s solution.

 Alpha: 0.05
 Decision: Reject the null hypothesis.
 Reason for decision: pvalue < 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.
 Introductory Statistics
 Preface
 Sampling and Data
 Descriptive Statistics
 Introduction
 StemandLeaf Graphs (Stemplots), Line Graphs, and Bar Graphs
 Histograms, Frequency Polygons, and Time Series Graphs
 Measures of the Location of the Data
 Box Plots
 Measures of the Center of the Data
 Skewness and the Mean, Median, and Mode
 Measures of the Spread of the Data
 Descriptive Statistics
 Probability Topics
 Discrete Random Variables
 Introduction
 Probability Distribution Function (PDF) for a Discrete Random Variable
 Mean or Expected Value and Standard Deviation
 Binomial Distribution
 Geometric Distribution
 Hypergeometric Distribution
 Poisson Distribution
 Discrete Distribution (Playing Card Experiment)
 Discrete Distribution (Lucky Dice Experiment)
 Continuous Random Variables
 The Normal Distribution
 The Central Limit Theorem
 Confidence Intervals
 Hypothesis Testing with One Sample
 Hypothesis Testing with Two Samples
 The ChiSquare Distribution
 Linear Regression and Correlation
 F Distribution and OneWay ANOVA
 Appendix A: Review Exercises (Ch 313)
 Appendix B: Practice Tests (14) and Final Exams
 Appendix C: Data Sets
 Appendix D: Group and Partner Projects
 Appendix E: Solution Sheets
 Appendix F: Mathematical Phrases, Symbols, and Formulas
 Appendix G: Notes for the TI83, 83+, 84, 84+ Calculators
 Appendix H: Tables