1. Chapter 14: Chi-Square Procedures

2. 14.1 – Test for Goodness of Fit

3. (2)
Chi-square test for goodness of fit:
Used to determine if what outcome you expect to happen actually does happen
Observed Counts:
Count of actual results
Expected Counts:
Count of expected results
To find the expected counts multiply the proportion you expect times the sample size

4. Note: Sometimes the probabilities will be expected to be the same and sometimes they will be expected to be different

5. Chi-square test for goodness of fit:P:
Don’t need! H:
Ho: All of the proportions are as expected
HA: One or more of the proportions are different
from expected A:
all expected counts are ≥ 5 N:
GOF test T:
df = k – 1 categories

6. Properties of the Chi-distribution:
Always positive (being squared)
Skewed to the right
Distribution changes as degrees of freedom change

8. Properties of the Chi-distribution:
Always positive (being squared)
Skewed to the right
Distribution changes as degrees of freedom change
Area is shaded to the right

9. Calculator Tip!
Goodness of Fit
L1: Observed
L2: Expected
L3: (L1 – L2)2 / L2
List – Math – Sum – L3

10. Calculator Tip!
P(2 > #)
2nd dist - 2cdf(2, big #, df)

13. Example #1
A study yields a chi-square statistic value of 20 (2 = 20). What is the P value of the test if the study was a goodness-of-fit test with 12 categories?

15. 2nd dist - 2cdf(2, big #, df)
Example #1
A study yields a chi-square statistic value of 20 (2 = 20). What is the P value of the test if the study was a goodness-of-fit test with 12 categories?
0.025
< P <
0.05
Or by calc:
2nd dist - 2cdf(2, big #, df)
2cdf(20, , 11) =

16. Observed Expected 8.57 Example #2
A geneticist claims that four species of fruit flies should appear in the ratio 1:3:3:9. Suppose that a sample of 480 flies contains 25, 92, 68, and 295 flies of each species, respectively. Find the chi-square statistic and probability.
Observed
Expected 25 92 68
295
480/16
480/16  3
480/16  3
480/16  9 30 90 90
270
0.833
0.044
5.378
2.315 =
8.57

17. P(2 > 8.57) =
df =
4 – 1 = 3

19. P(2 > 8.57) =
df =
4 – 1 = 3
.025 < P(2 > 8.57) < 0.05
Or by calc:
2cdf(8.57, , 3) =

20. Example #3
The number of defects from a manufacturing process by day of the week are as follows:
The manufacturer is concerned that the number of defects is greater on Monday and Friday. Test, at the 0.05 level of significance, the claim that the proportion of defects is the same each day of the week.
Monday
Tuesday
Wednesday
Thursday
Friday # 36 23 26 25 40
The proportion of defects from a manufacturing process is the same Mon-Fri H:
Ho:
HA:
The proportion of defects from a manufacturing process is not the same Mon-Fri (on one day or more)

21. A:
Expected Counts

22. Chi-Square Goodness of Fit
Expected Counts
Monday
Tuesday
Wednesday
Thursday
Friday # 36 23 26 25 40
Expected
150 errors total.
150 5 = 30 N:
Chi-Square Goodness of Fit

23.  T: 30 30 30 30 30 (O – E)2 2 = = E 7.533 Monday Tuesday Wednesday
Thursday
Friday # 36 23 26 25 40
Expected
2 = 
(O – E)2 E =
7.533

24. O:
P(2 > 7.533) =
df =
5 – 1 = 4

26. O:
P(2 > 7.533) =
df =
5 – 1 = 4
0.10 < P(2 > 7.533) > 0.15
Or by calc:
2cdf(7.533, , 4) =
0.1102

27. M:
P ___________ 
>
0.1102
0.05
Accept the Null

28. S:
There is not enough evidence to claim the proportion of defects from a manufacturing process is not the same Mon-Fri (on one day or more)

29. 14.2 – Inference for Two-Way Tables

30. To compare two proportions, we use a 2-Proportion Z Test
To compare two proportions, we use a 2-Proportion Z Test. If we want to compare three or more proportions, we need a new procedure.

31. Two – Way Table:
Organize the data for several proportions
R rows and C columns
Dimensions are r x c

32. Row total x Column total
To calculate the expected counts, multiply the row total by the column total, and divide by the table total:
Row total x Column total
table total
Expected count =
Degrees of Freedom:
(r – 1)(c – 1)

33. Chi-Square test for Homogeneity:
Compare two or more populations on one categorical variable
Ho: The proportions are the same among all populations
Ha: The proportions are different among all populations
Conditions:
SRS
Expected Counts are ≥ 5

34. Chi-Square test for Association/Independence:
Two categorical variables collected from a single population
Ho: There is no association between two categorical variables
(independent)
Ha: There is an association (dependent)
Conditions:
SRS
Expected Counts are ≥ 5

35. Test for Homogeneity/Independence
Calculator Tip!
Test for Homogeneity/Independence
2nd – matrx – edit – [A] – rxc – Table info
Then:
Stat – tests – 2–test
Observed: [A]
Expected: [B]
Calculate
Note: Expected: [B] is done automatically!

36. Row total x Column total
Example #1
The table shows the number of people in each grade of high school who preferred a different color of socks. 18 18 20 15 15 12 14 56
a. What is the expected value for the number of 12th graders who prefer red socks?
Row total x Column total
table total
20 x 14 56 = 5
Expected count = =

37. Example #1
The table shows the number of people in each grade of high school who preferred a different color of socks.
b. Find the degrees of freedom.
(r – 1)(c – 1)
(3 – 1)(4 – 1)
(2)(3) 6

38. Row total x Column total
Example #2
An SRS of a group of teens enrolled in alternative schooling programs was asked if they smoked or not. The information is classified by gender in the table. Find the expected counts for each cell, and then find the chi-square statistic, degrees of freedom, and its corresponding probability. 70
147 79
138
217
Expected Counts:
Row total x Column total
table total
70 x 79
217
70 x 138
217 =
25.484 =
44.516
147 x 79
217
147 x 138
217 =
53.516 =
93.484

39. Expected Counts:
Smoker
Non-Smoker
Male
Female
25.484
44.516
53.516
93.484
(O – E)2 E
2 =  =
(23 – )2
25.48
(47 – )2
44.516
(56 – )2
53.516
(91 – )2
93.484 + + +

40.  (O – E)2 2 = = E P(2 > 0.56197) = Expected Counts: Smoker
Non-Smoker
Male
Female
25.484
44.516
53.516
93.484
2 = 
(O – E)2 E =
Degrees of Freedom:
(r – 1)(c – 1) =
(2 – 1)(2 – 1) =
(1)(1) = 1
P(2 > ) =

42.  (O – E)2 2 = = E P(2 > 0.56197) = Expected Counts: Smoker
Non-Smoker
Male
Female
25.484
44.516
53.516
93.484
(O – E)2 E
2 =  =
Degrees of Freedom:
(r – 1)(c – 1) =
(2 – 1)(2 – 1) =
(1)(1) = 1
P(2 > ) =
More than 0.25
OR:
0.4535

43. Example #3
At a school a random sample of 20 male and 16 females were asked to classify which political party they identified with.
Democrat
Republican
Independent
Male 11 7 2
Female 8 1
Are the proportions of Democrats, Republicans, and Independents the same within both populations? Conduct a test of significance at the α = 0.05 level. H:
Ho: The proportions are the same among males and females and their political party
HA: The proportions are different among males and females and their political party

44. Row total x Column total
SRS
(says)
Row total x Column total
table total
Expected Counts  5
Democrat
Republican
Independent
Male 11 7 2
Female 8 1 20 16 36 18 15 3
20 x 18 36
20 x 15 36
20 x 3 36 = 10 =
8.33 =
1.67
16 x 18 36
16 x 15 36
16 x 3 36 = 8 =
6.67 =
1.33
Not all are expected counts are  5, proceed with caution!

45. N:
Chi-Square test for Homogeneity

46. T:
Democrat
Republican
Independent
Male 11 7 2
Female 8 1
Expected
Male 10
8.33
1.67
Female 8
6.67
1.33
(O – E)2 E
2 =  =
0.855
(11 – 10)2 10
(7 – 8)2 8
(7 – 8.33)2
8.33
(8 – 6.67)2
6.67
(2 – 1.67)2
1.67
(1 – 1.33)2
1.33 + + + + +

47. P(2 > 0.855) = O: Degrees of Freedom: (r – 1)(c – 1) =
(2 – 1)(3 – 1) =
(1)(2) = 2

49. P(2 > 0.855) = P(2 > 0.855) = O: More than 0.25
Degrees of Freedom:
(r – 1)(c – 1) =
(2 – 1)(3 – 1) =
(1)(2) = 2
OR:
P(2 > 0.855) =
0.6521

50. M:
P ___________ 
>
0.6521
0.05
Accept the Null

51. S:
There is not enough evidence to claim the proportions are different among males and females and their political party

52. Ho: AP scores and AP test are independent.
Example #4
The following chart represents the score distribution on the AP Exams for different subjects at a certain high school. Is there evidence that the score distribution is dependent from the subject? H:
Ho: AP scores and AP test are independent.
HA: AP scores and AP test are not independent.

53. Row total x Column total
SRS
(says)
Expected Counts  5 34 42 44 18 12 52 68 30
150
Row total x Column total
table total

54. 34 42 44 18 12 52 68 30
150
34 x 52
150
34 x 68
150
34 x 30
150 =
11.787 =
15.413 =
6.8
42 x 52
150
42 x 68
150
42 x 30
150 =
14.56 =
19.04 =
8.4
44 x 52
150
44 x 68
150
44 x 30
150 =
15.253 =
19.947 =
8.8

55. 34 42 44 18 12 52 68 30
150
18 x 52
150
18 x 68
150
18 x 30
150 =
6.24 =
8.16 =
3.6
12 x 52
150
12 x 68
150
12 x 30
150 =
4.16 =
5.44 =
2.4
Not all are expected counts are  5, proceed with caution!

56. N:
Chi-Square test for Independence

57. T: Expected (O – E)2 E 2 =  = 5.227698 5 11.787 15.413 6.8 4 14.56
19.04
8.4 3
15.253
19.947
8.8 2
6.24
8.16
3.6 1
4.16
5.44
2.4
(O – E)2 E
2 =  =

58. P(2 > 5.227698) = O: Degrees of Freedom: (r – 1)(c – 1) =
(5 – 1)(3 – 1) =
(4)(2) = 8

60. P(2 > 5.227698) = P(2 > 5.227698) = O: More than 0.25
Degrees of Freedom:
(r – 1)(c – 1) =
(5 – 1)(3 – 1) =
(4)(2) = 8
OR:
P(2 > ) =

61. M:
P ___________ 
>
0.05
Accept the Null

62. S:
There is not enough evidence to claim the AP scores and AP test are dependent