1. Chi Squared Tests Hypothesis Tests for Linear Regression
AP Statistics Topic 7
Chi Squared Tests
Hypothesis Tests for Linear Regression

2. These are the last 2 things we’ll study
Chi-squared tests
Goodness of Fit
Independence and Homogeneity
Hypothesis tests for linear regression
The significance of the linear relationship
We’ll spend one week on each

3. Chi-Squared Tests Analysis of categorical data
The tests we’ll study are
Goodness of Fit test
tests for homogeneity and independence
These tests are performed exactly the same way
For homogeneity, we look at two samples and one characteristic
For independence, we look at one sample and two characteristics

4. Goodness of Fit Test
Measures the extent to which some empirical distribution “fits” the distribution expected under the null hypothesis 20 30 40 50 60
Fork length 10
Frequency

5. For example Time % 12-3AM 17 3-6AM 8 6-9AM 8 9AM-noon 6 Noon-3PM 10
A GEICO Direct magazine had an interesting article concerning the percentage of teenage motor vehicle deaths and the time of day. The following percentages were given from a sample.
Time %
12-3AM 17
3-6AM 8
6-9AM 8
9AM-noon 6
Noon-3PM 10
3-6 PM 16
6-9 PM 15
9PM-12AM 19

6. The Distribution and Hypothesis Statements
Is the percentage of teenage motor vehicle deaths the same for each time period? Conduct a hypothesis test at the 1% level.
Ho: The percent of teenage motor vehicle deaths is the same for each time period.
Ha: The percent of teenage motor vehicle deaths is not the same for each time period.

7. Let’s look at this more closely
In this problem, what type of data are we considering?
Categorical data – that is, the time of day
How many classes is our data divided into?
8 different classes

8. Put another way ….
We want to see if the distribution of our data is consistent with the hypothesized distribution
In this case, we want to see if the distribution of accidents is uniform – about 12.5% per period

9. So how can we do this? We have our observed occurrences
Are these consistent with our hypothesis?
What should we compare with these?
Expected values --
Time
12-3
3-6
6-9
9-12
Count 17 8 6 10 16 15 19
Time
12-3
3-6
6-9
9-12
Observed 17 8 6 10 16 15 19
Expected
12.38

10. Test Statistic
Our test statistic is

11. Chi-squared Distribution
Family of curves identified by deg of freedom (k-1)
Mean = degrees of freedom
Variance = 2(degrees of freedom)
As deg of freedom increases, curves approach normal

12. How we’ll use the chi-squared distribution
We’ll use the chi-squared distribution to determine our p-value
If our test statistic is large, then we’ll reject the null hypothesis
P-value

13. How can we find p-values?
Calculate the chi-squared statistic
Use the chi-squared table
Use the chi-squared cdf function on your TI-83

14. The Chi-square table

15. Graphing Calculator
2nd DIST

16. Graphing Calculator STAT Inputs Observed data list Expected data list
Degrees of freedom

17. Assumptions We have 2 assumptions for this test
First, the observed cell counts are based on a random sample (our sample is random)
Our sample is large.
How do we determine large?
Expected cell counts must all be greater than or equal to 5

18. Our conclusions? The same as we’ve always done
We reject or fail to reject the null based on a comparison of the p-value and our significance level
We interpret our conclusion in the context of our alternative hypothesis

19. Let’s summarize Use the same 9 steps for hypothesis testing
Identify the parameter
Null
Alternative
Choose significance level
Test Statistic
Assumptions
Calculate Test Statistic
Determine P-value
Make your conclusion

20. Let’s finish the Geico Problem
Let’s identify the parameter
Proportion of teenage accidents
Null Hypothesis
Ho: The percent of teenage motor vehicle deaths is the same for each time period.
Alternative Hypothesis
Ha: The percent of teenage motor vehicle deaths is not the same for each time period.
Significance level

21. Continuing … Test Statistic Assumptions: The sample is random
The sample is large

22. Continuing … Calculate the Test Statistic Time 12-3 3-6 6-9 9-12
Observed 17 8 6 10 16 15 19
Expected
12.38

23. Conclusion
We fail to reject the null hypothesis because the p-value (.056) is greater than the significance level (.01).
The data does not suggest that the distribution of accidental deaths is not distributed differently among the time periods.

24. Homework 7-1
Read section 12.1 in the textbook
12.10
12.12
12.14

25. Let’s try this for some practice
Using a test, investigate whether it’s reasonable to assume the random number table is random. Use a significance level of .05.

26. Tests for Homogeneity and Independence
In these tests we’ll be taking n samples and looking at one characteristic.
Take samples of 1000 people from 4 different countries and ask how they feel about whether the use of torture against suspected terrorist is justified.
In this case we’d like to see if the responses are distributed equally (homogenous) among the countries.
Or, we’ll take one sample and look at two characteristics.
Take a sample of 300 adults and determine each person’s political philosophy and what television news station they watch
In this case we’d like to see if political philosophy and news station are independent.

27. Let’s do an example of each
First, let’s do a test for independence
Big Office is a chain of large office supply stores that sell an extensive line of desktop
and laptop computers. Company executives want to know whether the demands for
these types of computers are related in any way. They might act as complementary
products or sales may not be related. Big Office randomly selected 250 business days
categorized demand for each type of computer as Low, MedLow, MedHi and Hi.
Desktops
Low
MedLow
MedHi Hi 4 17 5 43 8 23 22 27 80 16 20 14 70 10 19 11 57 38 77 72 63
250
Laptops

28. So how many samples do we have
So how many samples do we have? Is the data we are collecting categorical or numerical? How many characteristics are we investigating? How many classes within those characteristics?

29. Hypotheses Statements
Ho: The two variables are independent.
Ha: The two variables are not independent.

30. How do we test for independence?
Low
MedLow
MedHi Hi 4 17 5 43 8 23 22 27 80 16 20 14 70 10 19 11 57 38 77 72 63
250
Recall that if events A and B are independent
So, we’ll assume the two variables are independent.
Then we’ll determine expected cell counts for each cell.
We’ll look at the differences between the expected and observed counts for out test.

31. Test Statistic

32. Assumptions The sample is random. The sample is large
Each expected cell count is at least 5

33. At this point …
We can calculate the p-value using the chi-squared table
The chi-squared CDF function on our calculator
Or, use the chi-squared test

34. Let’s look at the Chi-Squared Test
This test is a piece of cake ….
First, put your observed matrix into the calculator
STAT – TEST –
Now just select Calculate
The calculator creates the Expected matrix
Output : value of your test statistic and p-value

35. So let’s do this problem using our 9 steps of hypothesis testing

36. Chi-Squared Test for Homogeniety
The paper “No Evidence of Impaired Neurocognitive Performance in Collegiate Soccer Players’ compared collegiate soccer players, athletes in sports other than soccer, and a group of students who were not involved in collegiate sports with respect to head injuries. Three independent random samples were chosen and each person in the sample was asked to complete a medical history survey. The following 2-way contingency table was created based on reported concussions. 1 2 3+
Soccer 45 25 11 10 91
Other 68 15 8 5 96
Non 3 53
158 22
240

37. So how many samples do we have?
Is the data we are collecting categorical or numerical?
How many characteristics are we investigating?
How many classes within those characteristics?

38. Hypotheses Statements
Ho: The populations are homogenous
or, The category proportions are the same for all populations.
Ha: The populations are not homogenous.
or, the category proportions are not the same for all populations.

39. Test Statistic

40. Assumptions The samples are random and independent.
The sample is large
The expected cell counts are at least 5

41. Everything else is the same
Let’s finish this test using our 9 steps. 1 2 3+
Soccer 45 25 11 10 91
Other 68 15 8 5 96
Non 3 53
158 22
240

42. To summarize … Test for Independence Test for Homogeneity One sample
Two characteristics
Assumptions:
Sample is random
Sample is large
Test for Homogeneity
Multiple samples
One characteristic
Samples are independent and random
Samples are large

43. Homework 7-2
Read Section 12.2
12.18
12.22