1. CHAPTER 14 Chi- square goodness of fit Test “GOF”
Chi-square Test for Homogeneity
Chi-square Test for Independence
Symbol for Chi-Square is χ2
DONE SAME WAY

2. What’s the difference?
Goodness of Fit…..how well does the observed data match the expected….2 rows or 2 columns
Homogeneity…..More than one sample is taken with one categorical variable in mind
(2+ Samples, 1 category)
Independence/Association…..Only one sample is taken and there are two or more categories.
(1 sample, 2+ categories)

3. Chi-Square Curve It is not a NORMAL CURVE!!!
It is always skewed to the right some

4. Is your die fair—one more time.
Roll your die 60 times. Write down the number for every roll.

5. Chi-Square GOF Test
If your die is fair you would expect to get 10 of each number in 60 rolls.
In this test we compare the EXPECTED results vs the OBSERVED results.

6. Hypotheses
Ho: The proportion of each number that occurs on my die is 1/6
Ha: The proportion of each number that occurs on my die is different than 1/6
There are no symbols for the chi-square. However, it is always one-sided, even though the word “different” is used.

7. Normal condition for χ2
80% of the expected cells are greater than or equal to 5.
(not observed cells—expected cells!)

8. Formula

9. Degrees of Freedom (df)
For all chi-square tests use the following:
df = (r – 1)(c – 1)
r is the row and c is the column

11. Calculator steps TI-83+ Calculator Then hit 2nd and Stat
Put your observed counts in L1 and Expected in L2
Tyoe the observed in L1 and expected in L2. Then click on the L3 heading and type the formula(then click enter), then quit out to the main screen
Find the sum of L3, your answer is the chi-square statistic

12. Calculator steps X2 , UB , df
After your get the sum you need to obtain the p-value
X2 , UB , df
This will give you the p-value

13. Calculator steps 5 TI-84 calculator does most of the work for you
Make sure you have typed your observed counts in L1 and expected counts in L2 5

16. Demographics.
Rancho is approximately 53.6% Hispanic, 29.2% Asian, 12.7% white and 4.5% other.(data as of school year)
Does Mr. Pines’ AP stats classes reflect this diversity? Run the appropriate test, verify your requirements, and write a conclusion.

17. Demographics.
2015
Ho: The diversity in Mr. Pines classes is the same as Rancho’s diversity
Ha: The diversity in Mr. Pines classes is different than Rancho’s diversity
Assumptions: We have an independent random sample of 159 students ethnicity. We can assume that there have been at least 1590 students in Mr. Pines classes. 100% of our expected cells are 5 or more.
Chi-square GOF test
This p-value is low enough to reject Ho at the 1% level
This is strong evidence to suggest that Mr. Pines class diversity may be different than Rancho’s diversity.
X2 = 12.38
P-value =.0062
df = 3

18. Is there a difference……
Do boys and girls prefer types of social media?
Please choose your favorite of the three below.

19. Instagram Snapchat Twitter Girls Boys
One Hundred Fifty Seven students were surveyed………….
We took a sample of 83 girls and a sample of 74 boys.
Instagram
Snapchat
Twitter
Girls 36 25 22
Boys 19

20. What’s the difference?
Goodness of Fit…..how well does the observed data match the expected….2 rows or 2 columns
Homogeneity…..More than one sample is taken with one categorical variable in mind
(2+ Samples, 1 category)
Independence/Association…..Only one sample is taken and there are two or more categories.
(1 sample, 2+ categories)

21. One Hundred Fifty Seven students were surveyed………….
Instagram
Snapchat
Twitter
Girls
36(29.076)
25(32.248)
22(21.675)
Boys
19(25.924)
36(28.752)
19(19.325)

22. Hypotheses for Chi-Square Test for Homogeneity
Ho: There is no difference between gender and social media preference Ha: There is a difference between gender and social media preference OR
Ho: The proportions of boys and girls who prefer each type of social media are the same
Ha: The proportions of boys and girls who prefer each type of social media are different

23. Ho: There is no difference between gender and social media preference Ha: There is a difference between gender and social media preference
Assumptions: We have two independent random samples of students social media preferences(83 girls and 74 boys). There are obviously more than 830 girls and 740 boys in the population sampled from who use social media. 100% of our expected cells are 5 or more.
Chi-square test of homogeneity
This p-value is low enough to reject Ho at the 5% level
This is evidence to suggest that there may be a difference between gender and social media preference.
X2 = 6.96
P-value =.0307
df = 2

24. Period 1

25. Period 3

26. Referrals vs Days of week
Monday
Tuesday
Wednesday
Thursday
Friday 12 5 9 4 15
The table shows the number of students referred for disciplinary reasons to the principals office, broken down by day of the week.
Are referrals related to the day of the week?

28. BIRTHDAYS.
Are Mr. Pines students birth months distributed in proportion to the number of days in each month?
We can run a chi-square GOF based on the # of days in each month
n = 153

29. C/O 2015 per 1 only
H0:
Mr. Pines students birthday months are in proportion to the number of days in each month.
Mr. Pines students birthday months are different than the proportion of the number of days in each month.
Ha:
We have an independent sample of 53 students birth months. All births is obviously more than 10x our sample. 100% of our expected counts are 5 or more.
X2 = 9.81
P-value = .5475
df = 11
Chi-square GOF Test
This p-value is too high to reject Ho
Based on this sample, there is NOT enough evidence to suggest that Mr. Pines students birthday months are different than the proportion of the number of days in each month.

30. BIRTHDAYS O E 9 7 12 11 14 16 20 17
To figure out the expected we need to think about the number of days in each month.
Jan July 31
Feb Aug 31
Mar Sep 30 Total = 365 days
Apr Oct 31
May Nov 30
June Dec 31
n = 153

31. Births are uniformly distributed by the # of days in each month
C/O 2014
H0:
Births are uniformly distributed by the # of days in each month O E 12
10.956 5
9.896
10.603 13 7 9 14 11 8 15
Ha:
Births are not uniformly distributed by the # of days in each month
CONDITIONS
We have an independent sample of 129 students birth months. All births is obviously more than 10x our sample. 100% of our expected counts are 5 or more.
df = 11
X2 = 7.75
P = .7355
This p-value is too high to reject Ho.
There is not enough evidence to suggest that births are not uniformly distributed by the # of days in each month.

32. Chi-Square Test for Homogeneity
Data is given in a 2-way table
Expected counts are found by using a matrix on your calculator or by multiplying the (ROW TOTAL)(COLUMN TOTAL)/GRAND TOTAL
Conditions and df are the same as GOF test

33. Is there a difference……
Do boys and girls prefer different video game consoles?
Please choose your favorite console out of the 3?

34. Hypotheses for Chi-Square Test for Homogeneity
Remember Ha is always means different! Ho: There is no difference between gender and video game console preference Ha: There is a difference between gender and video game preference OR
Ho: The proportions of boys and girls who prefer each type of console are the same
Ha: The proportions of boys and girls who prefer each type of console are different

35. Why is this a Homogeneity Test?
Two samples were taken separately
Boys console preference
Girls console preference
There is ONE category of interest.

36. VIDEO GAME CONSOLE PREFERENCE
Two-Way Table
2013
VIDEO GAME CONSOLE PREFERENCE
GENDER
Wii
Xbox 360
PS3
Girls 27 16 20 63
Boys 3 33 32 68
Total 30 49 52
131

37. VIDEO GAME CONSOLE PREFERENCE
Two-Way Table
2014
VIDEO GAME CONSOLE PREFERENCE
GENDER
Wii
Xbox
Playstation
Girls 19 10 32 61
Boys 6 21 34
Total 25 31 66
122

38. Quit to home screen, go to test menu
Hit 2nd Matrix, go to EDIT
Should be already setup if you used A and B
Set the appropriate matrix size
Enter observed counts in matrix

39. REMEMBER!....EXPECTED COUNTS MUST BE ON YOUR PAPER!
You need the expected counts….so go back to 2nd matrix. Use NAMES and go down to Matrix B, calculator generates them after you run the test
You will most likely have to scroll to the right to see all of the expected counts
REMEMBER!....EXPECTED COUNTS MUST BE ON YOUR PAPER!

41. College Students’ Drinking
In 1987, a random sample of undergraduate students at Rutgers University was sent a questionnaire that asked about their alcohol drinking habits. Here are the results displayed in a two-way table.
Live on Campus
Live Off Campus(Not with Parents)
Live Off Campus with Parents
Total
Abstain 46 17 43
106
Light Drinker 71 38 42
151
Light-Moderate 55 34 26
115
Moderate - Heavy 63 24 15
102
Heavy 67 28
112
302
141
143
586

42. Chi-Square Test for Independence
There was one sample taken and then data was broken down into different categories.
When only one sample is taken we are doing a Chi-Square Test for Independence/Association

43. Hypotheses
This is a chi-square test for Independence/Association, you have a few options for writing the hypotheses
Ho: There is no association between students’ residence type and drinking habits.
Ha There is an association between students’ residence type and drinking habits. OR
Ho: Student drinking habits and residence type are independent
Ha Student drinking habits and residence type are not independent

44. Full Moon This is a chi-square test for Homogeneity
Some people believe that a full moon elicits unusual behavior in people. The table shows the number of arrests made in a small town during the weeks of six full moons and six other randomly selected weeks in the same year. Is there evidence of a difference in the types of illegal activity that takes place.
This is a chi-square test for Homogeneity
Full Moon
Not Full
Violent (murder,assault,rape, etc.) 2 3
Property(burglary, vandalism, etc.) 17 21
Drugs/Alcohol 27 19
Domestic Abuse 11 14
Other offenses 9 6

45. Thanks To:
Grace Montgomery

46. THANKS TO: Amy Nguyen

47. Testing M&M’s
The Mars company has always claimed that the color distribution of their M&M’s follow a certain proportion as follows:
Brown
Red
Yellow
Green
Orange
Blue
13%
14%
16%
20%
24%
Check the M&M’s that were given to you. How many of each color do you have? We will run a Chi-Square GOF test to see if their claim is accurate.
Do not eat your M&M’s until we have all observed and expected counts completed!

48. Hypotheses
Ho: My bag of M&M’s follow the same color distribution as the Mars company claim.
Ha: My bag of M&M’s follows a different color distribution as the Mars company claim.

49. Assumptions/Conditions
___% of expected counts >5
My bag of M&M’s can be considered an independent random sample of M&M’s

50. M&M Combined Results There were a total of 5606 M&M’s sampled.
Colors
Brown
Red
Yellow
Green
Orange
Blue
Claim %
13%
14%
16%
20%
24%
Expected
728.78
784.84
896.96
1121.2
Observed
606
586
803
1101
1335
1175
There were a total of 5606 M&M’s sampled.
We have a chi-square statistic of which gives a P-value of 0.

51. Mr. Pines Poker Chips 44 white chips = 20pts 5 blue chips = 30pts
1 red chip = 50pts

52. Mr. Pines Poker Chips 44 white chips = 20pts 5 blue chips = 30pts
1 red chip = 50pts
There have been 135 attempts at randomly choosing poker chips out of the bag. A White chip has been pulled 301 times, a Blue chip 47 times, and the Red chip 16 times. Has this followed the expected probabilities?
Run a chi-square GOF Test.

53. Baseball Bats
There have been some major bat changes for the 2011 season. Aluminum baseball bats have been regulated so that they meet certain safety standards. After 5 games this season, coach Pines has noticed significant reductions in power numbers such as 2B’s, 3B’s, and HR’s…..Of course he would like to test his hypothesis.

54. Baseball Bats
Run a Chi-Squared two-way table test to see if there is an association between the power numbers and types of bats.
Also run a 2-Prop. Z Test between the types of bats used.
If these are done correctly, Z2 = X2

55. Hypotheses
Ho: There is no association between type of bats and extra base hits
Ha: There is an association between type of bats and extra base hits

56. Assumptions/Conditions
E---All expected counts > 5
S----We have a random sample of 23 schools hitting stats for the first 5 games of the 2010 and 2011 baseball seasons
I----We can assume that all stats are independent of other teams stats

57. Observed and Expected Counts
2010(BESR Bats)
2011(BBCOR Bats)
Singles
(695.26)
(679.74)
Extra Base Hits
(289.74)
(283.26)

58. X2 = 5.35
P-value = .0207
This p-value is low enough to reject at the 5% level.
There is evidence to suggest that there may be an association between the types of bats and extra base hits

59. Tootsie Pop Wrappers
We are interested in whether or not the designs on the wrappings on Tootsie Roll Pops are independent of the flavor of the pop.

60. Hypotheses
This is a chi-square test for Independence/Association, you have a few options for writing the hypotheses
Ho: There is no association between pop flavor and designs on the wrapper.
Ha There is an association between pop flavor and designs on the wrapper. OR
Ho: Pop flavor and wrapper designs are independent.
Ha Pop flavor and wrapper designs are not independent.