1. Chi-Square Goodness-of-Fit Test
STAT 250
Dr. Kari Lock Morgan
Chi-Square Goodness-of-Fit Test
SECTION 7.1
Testing the distribution of a single categorical variable : χ2 goodness of fit (7.1)

2. Multiple Categories
So far, we’ve learned how to do inference for categorical variables with only two categories
In lab, you learned how to do inference for a quantitative variable and a categorical variable with multiple categories
Today and next class we’ll learn how to do hypothesis tests for categorical variables with multiple categories

3. Are children with ADHD younger than their peers?
Question of the Day
Are children with ADHD younger than their peers?
pheromones = subconscious chemical signals

4. ADHD or Just Young?
In British Columbia, Canada, the cutoff date for entering school in any year is December 31st, so those born late in the year are almost a year younger than those born early in the year
Is it possible that younger students are being over- diagnosed with ADHD?
Are children diagnosed with ADHD more likely to be born late in the year, and so be younger than their peers?
Data: Sample of kids aged 6-12 in British Columbia who have been diagnosed with ADHD
Morrow, R., et. al., “Influence of relative age on diagnosis and treatment of attention-deficit/hyperactivity disorder in children,” Canadian Medical Association Journal, April 17, 2012; 184(7):

5. Categories For simplicity, we’ll break the year into four categories:
January – March
April – June
July – September
October – December
Are birthdays for kids with ADHD evenly distributed among these four time periods?

6. Hypothesis Testing State hypotheses
Calculate a statistic, based on your sample data
Create a distribution of this statistic, as it would be observed if the null hypothesis were true
Measure how extreme your test statistic from (2) is, as compared to the distribution generated in (3), to get a p-value
Make a conclusion

7. Hypotheses H0: p1 = p2 = p3 = p4 = 1/4 Ha: At least one pi ≠ 1/4
Define parameters as
p1 = Proportion of ADHD births in January – March
p2 = Proportion of ADHD births in April – June
p3 = Proportion of ADHD births in July – Sep
p4 = Proportion of ADHD births in Oct – Dec
H0: p1 = p2 = p3 = p4 = 1/4
Ha: At least one pi ≠ 1/4

8. Observed Counts
The observed counts are the actual counts observed in the study
Data for boys:
Jan-Mar
Apr-Jun
July-Sep
Oct-Dec
Observed
6880
7982
9161
8945

9. Expected Counts
The expected counts are the expected counts if the null hypothesis were true
For each cell, the expected count is the sample size (n) times the null proportion, pi

10. Expected Counts n = 32,968 boys with ADHD H0: p1 = p2 = p3 = p4 = ¼
Expected count for each category, if the null hypothesis were true:
32,968(1/4) = 8242
Jan-Mar
Apr-Jun
July-Sep
Oct-Dec
Observed
6880
7982
9161
8945
Expected
8242

11. Chi-Square Statistic
Jan-Mar
Apr-Jun
July-Sep
Oct-Dec
Observed
6880
7982
9161
8945
Expected
8242
Need a way to measure how far the observed counts are from the expected counts…
Use the chi-square statistic :

12. Chi-Square Statistic Observed Expected Contribution to χ2 Jan-Mar 6880
8242
Apr-Jun
7982
Jul-Sep
9161
Oct-Dec
8945

13. StatKey
More Advanced Randomization Tests: χ2 Goodness-of-Fit

14. A distribution of the test statistic assuming H0 is true
What Next?
We have a test statistic. What else do we need to perform the hypothesis test?
A distribution of the test statistic assuming H0 is true
How do we get this?
Two options:
Simulation
Distributional Theory

15. Upper-Tail p-value
To calculate the p-value for a chi-square test, we always look in the upper tail
Why?
Values of the χ2 are always positive
The higher the χ2 statistic is, the farther the observed counts are from the expected counts, and the stronger the evidence against the null

16. Randomization Distribution
Can we reject the null?
Yes No

17. Chi-Square (χ2) Distribution
If each of the expected counts are at least 5, AND if the null hypothesis is true, then the χ2 statistic follows a χ2 –distribution, with degrees of freedom equal to
df = number of categories – 1
ADHD Birth Month:
df = 4 – 1 = 3

18. Chi-Square Distribution

19. χ2 Null Distribution

20. p-value

21. Conclusion This is a TINY p-value!!!
We have incredibly strong evidence that birthdays of boys with ADHD are not evenly distributed throughout the year.

22. Chi-Square Test for Goodness of Fit
State null hypothesized proportions for each category, pi. Alternative is that at least one of the proportions is different than specified in the null.
Calculate the expected counts for each cell as npi .
Calculate the χ2 statistic:
Compute the p-value as the proportion above the χ2 statistic for either a randomization distribution or a χ2 distribution with df = (# of categories – 1) if expected counts all > 5
Make a conclusion.
n = 1

23. Births Equally Likely?
Maybe all birthdays (not just for boys with ADHD) are not evenly distributed throughout the year
We have the overall proportion of births for all boys during each time period in British Columbia:
January – March: 0.244
April – June: 0.258
July – September: 0.257
October – December: 0.241

24. Hypotheses Define parameters as
p1 = Proportion of ADHD births in January – March
p2 = Proportion of ADHD births in April – June
p3 = Proportion of ADHD births in July – Sep
p4 = Proportion of ADHD births in Oct – Dec
H0: p1 = 0.244, p2 = 0.258, p3 = 0.257, p4 = 0.241
Ha: At least one pi is not as specified in H0

25. ADHD or Just Young? (BOYS)
Birthday
Proportion of Births
ADHD
Expected Counts
Jan-Mar
0.244
6880
Apr-Jun
0.258
7982
Jul-Sep
0.257
9161
Oct-Dec
0.241
8945
n = 32,968
The expected count for the Jan-Mar cell is
8044.2
8505.7
8472.8
7945.3

26. ADHD or Just Young? (BOYS)
Birthday
Proportion of Births
ADHD
Expected Counts
Contribution to χ2
Jan-Mar
0.244
6880
Apr-Jun
0.258
7982
8505.7
Jul-Sep
0.257
9161
8472.8
Oct-Dec
0.241
8945
7945.3
The contribution to χ2 for the Oct-Dec cell is
32.2
55.9
125.8
168.5

27. ADHD or Just Young? (BOYS)
Birth Date
Proportion of Births
ADHD Diagnoses
Expected Counts
Contribution to χ2
Jan-Mar
0.244
6880
168.5
Apr-Jun
0.258
7982
8505.7
32.2
Jul-Sep
0.257
9161
8472.8
55.9
Oct-Dec
0.241
8945
7945.3
n = 32,968
Expected: , , ,
Chi-square: = 382.4

28. ADHD or Just Young (Boys)
p-value ≈ 0
We have VERY (!!!) strong evidence that boys with ADHD do not have the same distribution of birthdays as all boys in British Columbia.

29. A Deeper Understanding
The p-value only tells you that the counts provide evidence against the null distribution
If you want to know more, look at
Which cells contribute most to the χ2 statistic?
Are the observed or expected counts higher?

30. ADHD or Just Young? (BOYS)
Birth Date
Proportion of Births
ADHD Diagnoses
Expected Counts
Contribution to χ2
Jan-Mar
0.244
6880
8044.2
168.5
Apr-Jun
0.258
7982
8505.7
32.2
Jul-Sep
0.257
9161
8472.8
55.9
Oct-Dec
0.241
8945
7945.3
125.8
n = 32,968
Expected: , , ,
Chi-square: = 382.4
Big contributions to χ2 : beginning and end of the year
Jan-Feb: Many fewer ADHD than would be expected
Oct-Dec: Many more ADHD cases than would be expected

31. ADHD and Birth Year Stop and think about the implications of this!
In British Columbia, 6.9% of all boys 6-12 are diagnosed with ADHD
5.5% of boys 6-12 receive medication for ADHD
How many of these diagnoses are simply due to the fact that these kids are younger than their peers???
ARE YOU CONCERNED?

32. ADHD or Just Young? (Girls)
Want more practice?
Here is the data for girls. (χ2 = 236.8)
Birth Date
Proportion of Births
ADHD Diagnoses
Jan-Mar
0.243
1960
Apr-Jun
0.258
2358
Jul-Sep
0.257
2859
Oct-Dec
0.242
2904

33. To Do
HW 7.1 (due Wednesday, 4/12)

34. Mendel’s Pea Experiment
In 1866, Gregor Mendel, the “father of genetics” published the results of his experiments on peas
He found that his experimental distribution of peas closely matched the theoretical distribution predicted by his theory of genetics (involving alleles, and dominant and recessive genes)
Source: Mendel, Gregor. (1866). Versuche über Pflanzen-Hybriden. Verh. Naturforsch. Ver. Brünn 4: 3–47 (in English in 1901, Experiments in Plant Hybridization, J. R. Hortic. Soc. 26: 1–32)

35. Mendel’s Pea Experiment
Mate SSYY with ssyy:
1st Generation: all Ss Yy
Mate 1st Generation:
=> 2nd Generation
 Second Generation
Phenotype
Theoretical Proportion
Round, Yellow
9/16
Round, Green
3/16
Wrinkled, Yellow
Wrinkled, Green
1/16
S, Y: Dominant
s, y: Recessive

36. Mendel’s Pea Experiment
Phenotype
Theoretical
Proportion
Observed
Counts
Round, Yellow
9/16
315
Round, Green
3/16
101
Wrinkled, Yellow
108
Wrinkled, Green
1/16 32
Let’s test this data against the null hypothesis of each pi equal to the theoretical value, based on genetics

37. Mendel’s Pea Experiment
Phenotype
Null pi
Observed
Counts
Expected Counts
Round, Yellow
9/16
315
Round, Green
3/16
101
Wrinkled, Yellow
108
Wrinkled, Green
1/16 32
The expected count for the round, yellow phenotype is
177.2
310.5
312.75
318.25
n = 556
Expected = , , , 34.75
chisquare = 0.47

38. Mendel’s Pea Experiment
Phenotype
Null pi
Observed
Counts
Expected Counts
Contribution to χ2
Round, Yellow
9/16
315
312.75
Round, Green
3/16
101
104.25
Wrinkled, Yellow
108
Wrinkled, Green
1/16 32
34.75
Phenotype
Null pi
Observed
Counts
Expected Counts
Contribution to χ2
Round, Yellow
9/16
315
312.75
0.016
Round, Green
3/16
101
104.25
0.101
Wrinkled, Yellow
108
0.135
Wrinkled, Green
1/16 32
34.75
0.1218
The contribution to the χ2 statistic for the round, yellow phenotype is
0.012
0.014
0.016
0.018
n = 556
Expected = , , , 34.75
chisquare = 0.47

39. Mendel’s Pea Experiment
χ2 = 0.47
Two options:
Simulate a randomization distribution
Compare to a χ2 distribution with 4 – 1 = 3 df

40. Mendel’s Pea Experiment
p-value = 0.925
Does this prove Mendel’s theory of genetics? Or at least prove that his theoretical proportions for pea phenotypes were correct?
Yes No

41. To Do
HW 7.1 (due Wednesday, 4/12)