Section 10.2 Independence. Section 10.2 Objectives Use a chi-square distribution to test whether two...
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Transcript of Section 10.2 Independence. Section 10.2 Objectives Use a chi-square distribution to test whether two...
Section 10.2
Independence
Section 10.2 Objectives
• Use a chi-square distribution to test whether two variables are independent
• Use a contingency table to find expected frequencies
Contingency Tables
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
Example:• The contingency table shows the number of times a
random sample of former smokers tried to quit smoking before they were habit free. They are classified by gender.
Number of times tried to quit before habit-free
Gender1 2-3 4 or
more
Male 271 257 149
Female 146 139 80
Finding the Expected Frequency
• 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
Find the expected frequency for each cell in the contingency table. Assume that the variables, favorite way to eat ice cream and gender, are independent.
Number of times tried to quit
Gender1 2-3 4 or
more Total
Male 271 257 149 677
Female 146 139 80 365
Total 417 396 229 1042
marginal totals
Solution: Finding Expected Frequencies
Number of times tried to quit
Gender1 2-3 4 or
more Total
Male 271 257 149 677
Female 146 139 80 365
Total 417 496 229 1042
,(Sum of row ) (Sum of column )
Sample sizer cr cE
E1,1= 270.93
Solution: Finding Expected Frequencies
Number of times tried to quit
Gender1 2-3 4 or
more Total
Male 271 257 149 677
Female 146 139 80 365
Total 417 496 229 1042
,(Sum of row ) (Sum of column )
Sample sizer cr cE
= 322.26 = 148.78
Solution: Finding Expected Frequencies
Number of times tried to quit
Gender1 2-3 4 or
more Total
Male 271 257 149 677
Female 146 139 80 365
Total 417 496 229 1042
,(Sum of row ) (Sum of column )
Sample sizer cr cE
= 322.26 = 173.74 = 80.22
Example: Finding Expected Frequencies
Find the expected frequency for each cell in the contingency table. Assume that the variables, favorite way to eat ice cream and gender, are independent.
Favorite way to eat ice cream
GenderCup Cone Sundae Sandwich Other Total
Male 600 288 204 24 84 1200
Female 410 340 180 20 50 1000
Total 1010 628 384 44 134 2200
marginal totals
Solution: Finding Expected Frequencies
Favorite way to eat ice cream
GenderCup Cone Sundae Sandwich Other Total
Male 600 288 204 24 84 1200
Female 410 340 180 20 50 1000
Total 1010 628 384 44 134 2200
,(Sum of row ) (Sum of column )
Sample sizer cr cE
1,1
1200 1010550.91
2200E
Solution: Finding Expected Frequencies
Favorite way to eat ice cream
GenderCup Cone Sundae
Sandwich
Other Total
Male600 288 204 24 84 1200
Female 410 340 180 20 50 1000
Total 1010 628 384 44 134 2200
1,2
1200 628342.55
2200E
1,3
1200 384209.45
2200E
1,4
1200 4424
2200E
1,5
1200 13473.09
2200E
Solution: Finding Expected Frequencies
Favorite way to eat ice cream
GenderCup Cone Sundae Sandwich Other Total
Male 600 288 204 24 84 1200
Female 410 340 180 20 50 1000
Total 1010 628 384 44 134 2200
2,2
1000 628285.45
2200E
2,4
1000 4420
2200E
2,5
1000 13460.91
2200E
2,1
1000 1010459.09
2200E
2,3
1000 384174.55
2200E
Chi-Square Independence Test
Chi-square independence test• Used 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
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
• 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
1. Identify the claim. State the null and alternative hypotheses.
2. Specify the level of significance.
3. Determine the degrees of freedom.
4. Determine the critical value.
State H0 and Ha.
Identify α.
Use Table 6 in Appendix B.
d.f. = (r – 1)(c – 1)
In Words In Symbols
Chi-Square Independence Test
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.
22 ( )O E
E
In Words In Symbols
Example: Performing a χ2 Independence Test
Using the gender/favorite way to eat ice cream contingency table, can you conclude that the adults favorite ways to eat ice cream are related to gender? Use α = 0.01. Expected frequencies are shown in parentheses.
Favorite way to eat ice cream
GenderCup Cone Sundae Sandwich Other Total
Male600
(550.91)288
(342.55)204
(209.45)24
(24)84
(73.09)1200
Female410
(459.09)340
(285.45)180
(174.55)20
(20)50
(60.91)1000
Total 1010 628 384 44 134 2200
Solution: Performing a Goodness of Fit Test
• H0:
• Ha:
• α =• d.f. = • Rejection Region
• Test Statistic: • Decision:
0.01(2 – 1)(5 – 1) = 4
The adults’ favorite ways to eat ice cream are independent of gender. The adults’ favorite ways to eat ice cream are dependent on gender. (Claim)
Solution: Performing a Goodness of Fit Test
O E (O-E) (O-E)2 (O-E)2/E
600 550.91 49.09 2409.8281 4.3743
288 342.55 -54.55 2975.7025 8.6869
204 209.45 -5.45 29.7025 0.1418
24 24 0 0 0
84 73.09 10.91 119.0281 1.6285
410 459.09 -49.09 2409.8281 5.2491
340 285.45 54.55 2975.7025 10.4246
180 174.55 5.45 29.7025 0.1702
20 20 0 0 0
50 60.91 -10.91 119.0281 1.95422
2 ( )O E
E
4.3743+8.6869+.1418+0+1.6285+5.2491+10.4246+.1702+0+1.9542 = 32.630
Solution: Performing a Goodness of Fit Test
• H0:
• Ha:
• α =
• d.f. =
• Rejection Region
• Test Statistic:
• Decision:
0.01
(2 – 1)(5 – 1) = 4
The adults’ favorite ways to eat ice cream are independent of gender.The adults’ favorite ways to eat ice cream are dependent on gender. (Claim)
χ2 ≈ 32.630
There is enough evidence at the 1% level of significance to conclude that the adults’ favorite ways to eat ice cream and gender are dependent.
Reject H0
Section 10.2 Summary
• Used a contingency table to find expected frequencies• Used a chi-square distribution to test whether two
variables are independent
Contingency Tables
Example:• The contingency table shows the results of a random
sample of 2200 adults classified by their favorite way to eat ice cream and gender.
Favorite way to eat ice cream
GenderCup Cone Sundae Sandwich Other
Male 600 288 204 24 84
Female 410 340 180 20 50
Solution: Performing a Goodness of Fit Test
2 2 2 2 2
2 2 2 2 2
(600 550.91) (288 342.55) (204 209.45) (24 24) (84 73.09)
550.91 342.55 209.45 24 73.09
(410 459.09) (340 285.45) (180 174.55) (20 20) (50 60.91)
459.09 285.45 174.55 20 60.9132.630
22 ( )O E
E
Example: Performing a χ2 Independence Test
Using the gender/times to quit contingency table, can you conclude that the number of times they tried to quit are related to gender? Use α = 0.05. Expected frequencies are shown in parentheses.
