Independence 1 Section 10.2
Section 10.2
Feb 06, 2016
Section 10.2. Independence. Section 10.2 Objectives. Use a contingency table to find expected frequencies Use a chi-square distribution to test whether two variables are independent. Contingency Tables. r c contingency table Shows the observed frequencies for two variables. - PowerPoint PPT Presentation
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Independence
1
Section 10.2
Section 10.2 Objectives
2
Use a contingency table to find expected frequencies
Use a chi-square distribution to test whether two variables are independent
Contingency Tables
3
r c contingency table Shows the observed frequencies for two
variables. The observed frequencies are arranged in r
rows and c columns. The intersection of a row and a column is
called a cell.
Contingency Tables
4
Example:The contingency table shows the results of a
random sample of 550 company CEOs classified by age and size of company.(Adapted from Grant Thornton LLP, The Segal Company)
Age
Company size
39 and
under
40 - 49
50 - 59
60 - 69
70 and over
Small / Midsize
42 69 108 60 21
Large 5 18 85 120 22
Finding the Expected Frequency
5
Assuming the two variables are independent, you can use the contingency table to find the expected frequency for each cell.
The expected frequency for a cell Er,c in a contingency table is
,(Sum of row ) (Sum of column )Expected frequency
Sample sizer cr cE
Example: Finding Expected Frequencies
6
Find the expected frequency for each cell in the contingency table. Assume that the variables, age and company size, are independent.
Age
Company size
39 and
under
40 - 49
50 - 59
60 - 69
70 and over
Total
Small / Midsize
42 69 108 60 21 300
Large 5 18 85 120 22 250
Total 47 87 193 180 43 550marginal totals
Solution: Finding Expected Frequencies
7
Age
Company size
39 and
under
40 - 49
50 - 59
60 - 69
70 and over
Total
Small / Midsize
42 69 108 60 21 300
Large 5 18 85 120 22 250
Total 47 87 193 180 43 550
,(Sum of row ) (Sum of column )
Sample sizer cr cE
1,1
300 4725.64
550E
Solution: Finding Expected Frequencies
8
Age
Company size
39 and
under
40 - 49
50 - 59
60 - 69
70 and over
Total
Small / Midsize
42 69 108 60 21 300
Large 5 18 85 120 22 250
Total 47 87 193 180 43 550
1,2
300 8747.45
550E
1,3
300 193105.27
550E
1,4
300 18098.18
550E
1,5
300 4323.45
550E
1,2
300 8747.45
550E
1,3
300 193105.27
550E
1,4
300 18098.18
550E
Solution: Finding Expected Frequencies
9
Age
Company size
39 and
under
40 - 49
50 - 59
60 - 69
70 and over
Total
Small / Midsize
42 69 108 60 21 300
Large 5 18 85 120 22 250
Total 47 87 193 180 43 550
2,2
250 8739.55
550E
2,4
250 18081.82
550E
2,5
250 4319.55
550E
2,1
250 4721.36
550E
2,3
250 19387.73
550E
Chi-Square Independence Test
10
Chi-square independence testUsed to test the independence of two
variables. Can determine whether the occurrence of
one variable affects the probability of the occurrence of the other variable.
Chi-Square Independence Test
11
For the chi-square independence test to be used, the following must be true.1.The observed frequencies must be obtained by using a random sample.2.Each expected frequency must be greater than or equal to 5.
Chi-Square Independence Test
12
If these conditions are satisfied, then the sampling distribution for the chi-square independence test is approximated by a chi-square distribution with (r – 1)(c – 1) degrees of freedom, where r and c are the number of rows and columns, respectively, of a contingency table.
The test statistic for the chi-square independence test is
where O represents the observed frequencies and E represents the expected frequencies.
22 ( )O E
E The test is always a
right-tailed test.
Chi-Square Independence Test
13
1. Identify the claim. State the null and alternative hypotheses.
2. Specify the level of significance.
3. Identify the degrees of freedom.
4. Determine the critical value.
State H0 and Ha.
Identify .
Use Table 6 in Appendix B.
d.f. = (r – 1)(c – 1)
In Words In Symbols
Chi-Square Independence Test
14
22 ( )O E
E
If χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0.
5. Determine the rejection region.
6. Calculate the test statistic.
7. Make a decision to reject or fail to reject the null hypothesis.
8. Interpret the decision in the context of the original claim.
In Words In Symbols
Example: Performing a χ2 Independence Test
15
Using the age/company size contingency table, can you conclude that the CEOs ages are related to company size? Use α = 0.01. Expected frequencies are shown in parentheses.
Age
Company size
39 and
under
40 - 49
50 - 59
60 - 69
70 and over
Total
Small / Midsize
42(25.64
)
69(47.45
)
108(105.2
7)
60(98.18
)
21(23.45
)
300
Large5
(21.36)
18(39.55
)
85(87.73
)
120(81.82
)
22(19.55
)
250
Total 47 87 193 180 43 550
Solution: Performing a Goodness of Fit Test
16
• H0:
• Ha:
• α =
• d.f. =
• Rejection Region
• Test Statistic:
• Decision:
0.01
(2 – 1)(5 – 1) = 4
0.01
χ2
0 13.277
CEOs’ ages are independent of company size
CEOs’ ages are dependent on company size
Solution: Performing a Goodness of Fit Test
17
2 2 2 2 2
2 2 2 2 2
(42 25.64) (69 47.45) (108 105.27) (60 98.18) (21 23.45)
25.64 47.45 105.27 98.18 23.45
(5 21.36) (18 39.55) (85 87.73) (120 81.82) (22 19.55)
21.36 39.55 87.73 81.82 19.5577.9
22 ( )O E
E
Solution: Performing a Goodness of Fit Test
18
• H0:
• Ha:
• α =
• d.f. =
• Rejection Region
• Test Statistic:
• Decision:
0.01
(2 – 1)(5 – 1) = 4
0.01
χ2
0 13.277
CEOs’ ages are independent of company size
CEOs’ ages are dependent on company size
χ2 = 77.9
There is enough evidence to conclude CEOs’ ages are dependent on company size.
77.9
Reject H0
Section 10.2 Summary
19
Used a contingency table to find expected frequencies
Used a chi-square distribution to test whether two variables are independent
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