<< Chapter < Page Chapter >> Page >

The goodness–of–fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity , can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence.

Note

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

Hypotheses


H 0 : The distributions of the two populations are the same.

H a : The distributions of the two populations are not the same.

Test statistic

Use a χ 2 test statistic. It is computed in the same way as the test for independence.

Degrees of freedom ( df )

df = number of columns - 1

Requirements

All values in the table must be greater than or equal to five.

Common uses

Comparing two populations. For example: men vs. women, before vs. after, east vs. west. The variable is categorical with more than two possible response values.

Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are shown in [link] . Do male and female college students have the same distribution of living arrangements?

Distribution of living arragements for college males and college females
Dormitory Apartment With Parents Other
Males 72 84 49 45
Females 91 86 88 35

H 0 : The distribution of living arrangements for male college students is the same as the distribution of living arrangements for female college students.

H a : The distribution of living arrangements for male college students is not the same as the distribution of living arrangements for female college students.

Degrees of Freedom ( df ):
df = number of columns – 1 = 4 – 1 = 3

Distribution for the test: χ 3 2

Calculate the test statistic: χ c 2 = 10.129

...


The graph of the Chi-square shows the distribution and marks the critical value with three degrees of freedom at 95% level of confidence, α = 0.05, 7.815. The graph also marks the calculated χ 2 test statistic of 10.129. Comparing the test statistic with the critical value, as we have done with all other hypothesis tests, we reach the conclusion.

Make a decision: Because the calculated test statistic is in the tail we cannot accept H 0 . This means that the distributions are not the same.

Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the distributions of living arrangements for male and female college students are not the same.

Notice that the conclusion is only that the distributions are not the same. We cannot use the test for homogeneity to draw any conclusions about how they differ.

Try it

Do families and singles have the same distribution of cars? Use a level of significance of 0.05. Suppose that 100 randomly selected families and 200 randomly selected singles were asked what type of car they drove: sport, sedan, hatchback, truck, van/SUV. The results are shown in [link] . Do families and singles have the same distribution of cars? Test at a level of significance of 0.05.

Sport Sedan Hatchback Truck Van/SUV
Family 5 15 35 17 28
Single 45 65 37 46 7

With a p -value of almost zero, we reject the null hypothesis. The data show that the distribution of cars is not the same for families and singles.

Try it

Ivy League schools receive many applications, but only some can be accepted. At the schools listed in [link] , two types of applications are accepted: regular and early decision.

Application Type Accepted Brown Columbia Cornell Dartmouth Penn Yale
Regular 2,115 1,792 5,306 1,734 2,685 1,245
Early Decision 577 627 1,228 444 1,195 761

We want to know if the number of regular applications accepted follows the same distribution as the number of early applications accepted. State the null and alternative hypotheses, the degrees of freedom and the test statistic, sketch the graph of the χ 2 distribution and show the critical value and the calculated value of the test statistic, and draw a conclusion about the test of homogeneity.

H 0 : The distribution of regular applications accepted is the same as the distribution of early applications accepted.

H a : The distribution of regular applications accepted is not the same as the distribution of early applications accepted.
df = 5
χ 2 test statistic = 430.06

This is a nonsymmetric chi-square curve with df = 5. The values 0, 5, and 430.06 are labeled on the horizontal axis. The value 5 coincides with the peak of the curve. A vertical upward line extends from 430.06 to the curve, and the region to the right of this line is shaded. The shaded area is equal to the p-value.

Press the MATRX key and arrow over to EDIT . Press 1:[A] . Press 3 ENTER 3 ENTER . Enter the table values by row. Press ENTER after each. Press 2nd QUIT . Press STAT and arrow over to TESTS . Arrow down to C:χ2-TEST . Press ENTER . You should see Observed:[A] and Expected:[B] . Arrow down to Calculate . Press ENTER . The test statistic is 430.06 and the p -value = 9.80E-91. Do the procedure a second time but arrow down to Draw instead of calculate .

References

Data from the Insurance Institute for Highway Safety, 2013. Available online at www.iihs.org/iihs/ratings (accessed May 24, 2013).

“Energy use (kg of oil equivalent per capita).” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/EG.USE.PCAP.KG.OE/countries (accessed May 24, 2013).

“Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubsearch/pubsinfo.asp?pubid=2009030 (accessed May 24, 2013).

“Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubs2009/2009030_sup.pdf (accessed May 24, 2013).

Chapter review

To assess whether two data sets are derived from the same distribution—which need not be known, you can apply the test for homogeneity that uses the chi-square distribution. The null hypothesis for this test states that the populations of the two data sets come from the same distribution. The test compares the observed values against the expected values if the two populations followed the same distribution. The test is right-tailed. Each observation or cell category must have an expected value of at least five.

Formula review

i j ( O E ) 2 E Homogeneity test statistic where: O = observed values
E = expected values
i = number of rows in data contingency table
j = number of columns in data contingency table

df = ( i −1)( j −1) Degrees of freedom

A math teacher wants to see if two of her classes have the same distribution of test scores. What test should she use?

test for homogeneity

What are the null and alternative hypotheses for [link] ?

A market researcher wants to see if two different stores have the same distribution of sales throughout the year. What type of test should he use?

test for homogeneity

A meteorologist wants to know if East and West Australia have the same distribution of storms. What type of test should she use?

What condition must be met to use the test for homogeneity?

All values in the table must be greater than or equal to five.

Use the following information to answer the next five exercises: Do private practice doctors and hospital doctors have the same distribution of working hours? Suppose that a sample of 100 private practice doctors and 150 hospital doctors are selected at random and asked about the number of hours a week they work. The results are shown in [link] .

20–30 30–40 40–50 50–60
Private Practice 16 40 38 6
Hospital 8 44 59 39

State the null and alternative hypotheses.

df = _______

3

What is the test statistic?

What can you conclude at the 5% significance level?

Questions & Answers

what is defense mechanism
Chinaza Reply
what is defense mechanisms
Chinaza
I'm interested in biological psychology and cognitive psychology
Tanya Reply
what does preconceived mean
sammie Reply
physiological Psychology
Nwosu Reply
How can I develope my cognitive domain
Amanyire Reply
why is communication effective
Dakolo Reply
Communication is effective because it allows individuals to share ideas, thoughts, and information with others.
effective communication can lead to improved outcomes in various settings, including personal relationships, business environments, and educational settings. By communicating effectively, individuals can negotiate effectively, solve problems collaboratively, and work towards common goals.
it starts up serve and return practice/assessments.it helps find voice talking therapy also assessments through relaxed conversation.
miss
Every time someone flushes a toilet in the apartment building, the person begins to jumb back automatically after hearing the flush, before the water temperature changes. Identify the types of learning, if it is classical conditioning identify the NS, UCS, CS and CR. If it is operant conditioning, identify the type of consequence positive reinforcement, negative reinforcement or punishment
Wekolamo Reply
please i need answer
Wekolamo
because it helps many people around the world to understand how to interact with other people and understand them well, for example at work (job).
Manix Reply
Agreed 👍 There are many parts of our brains and behaviors, we really need to get to know. Blessings for everyone and happy Sunday!
ARC
A child is a member of community not society elucidate ?
JESSY Reply
Isn't practices worldwide, be it psychology, be it science. isn't much just a false belief of control over something the mind cannot truly comprehend?
Simon Reply
compare and contrast skinner's perspective on personality development on freud
namakula Reply
Skinner skipped the whole unconscious phenomenon and rather emphasized on classical conditioning
war
explain how nature and nurture affect the development and later the productivity of an individual.
Amesalu Reply
nature is an hereditary factor while nurture is an environmental factor which constitute an individual personality. so if an individual's parent has a deviant behavior and was also brought up in an deviant environment, observation of the behavior and the inborn trait we make the individual deviant.
Samuel
I am taking this course because I am hoping that I could somehow learn more about my chosen field of interest and due to the fact that being a PsyD really ignites my passion as an individual the more I hope to learn about developing and literally explore the complexity of my critical thinking skills
Zyryn Reply
good👍
Jonathan
and having a good philosophy of the world is like a sandwich and a peanut butter 👍
Jonathan
generally amnesi how long yrs memory loss
Kelu Reply
interpersonal relationships
Abdulfatai Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Introductory statistics. OpenStax CNX. Aug 09, 2016 Download for free at http://legacy.cnx.org/content/col11776/1.26
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Introductory statistics' conversation and receive update notifications?

Ask