Chi-Square. All the tests we’ve learned so far assume that our data is normally distributed z-test...
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Transcript of Chi-Square. All the tests we’ve learned so far assume that our data is normally distributed z-test...
Chi-Square
•All the tests we’ve learned so far assume that our data is normally distributed
• z-test• t-test
• We test hypotheses about parameters of these normal distributions• We call these tests “Parametric Tests”
Chi Square
• What if our data is not normally distributed?• Skewed distributions• Nominal or ordinal data• Then we use “Nonparametric Tests”• Tests that do not deal with parameters
Chi Square
Parametric vs. Nonparametric Tests
• Parametric Tests Test hypotheses about some parameter
(e.g. about the mean)
• Nonparametric Tests Test hypotheses about entire distribution
Chi Square
Characteristics of the Chi-Square Distribution
The major characteristics of the chi-square distribution are:
1. It is positively skewed. 2. It is non-negative.3. It is based on degrees of
freedom.4. When the degrees of
freedom change a new distribution is created.
The formula for 2 is:
OR, sometimes written
Where f0 is the observed frequency of
each category in each cell of a table.
efeff )( 0
2
2
E
EO 22 )(
Example(1): Chi-Square Test for Independence
In one large factory, 100 employees were judged to be highly successful and another 100 marginally successful. All workers were asked, “Which do you find more important to you personally, the money you are able to take home or the satisfaction you feel from doing the job?” In the first group, 49% found the money more important, but in the second group 53% responded that way. Test the null hypothesis that job performance and job motivation are independent using the .01 level of significance.
An Example: Chi-Square Test for Independence
State the research hypothesis. Are job performance and job
motivation independent?
State the statistical hypotheses.
t.independennot are motivation job and eperformanc Job:H
t.independen are motivation job and eperformanc Job:H
A
0
An Example:
Set the decision rule.
64.6
1)1)(1()12)(12()1)(1(
01.
2
crit
rcdf
An Example: Chi-Square Test for Independence
Calculate the test statistic.
49200
)100)(98(
totaloverall
)lumn total total)(co(row
51200
)100)(102(
totaloverall
)lumn total total)(co(row
e
e
f
f
High Success Marginal Success Total
Money 49 (51) 53 (51)102
Satisfaction 51 (49) 47 (49)98
Total 100 100 200
An Example: Chi-Square Test for Independence
Calculate the test statistic.
32.0
08.08.08.08.
49
)4947(
51
)5153(
49
)4951(
51
)5149( 22222
High Success Marginal Success Total
Money 49 (51) 53 (51)102
Satisfaction 51 (49) 47 (49)98
Total 100 100 200
e
eo
f
ff 22 )(
An Example: Chi-Square Test for Independence
Decide if your result is significant. Retain H0, 0.32<6.64
Interpret your results. Job performance and job motivation
are independent.
Q2:Is the type of crime independent of whether the criminal is a stranger?
Stranger
Relative
12
39
379
106
727
642
Homicide Robbery Assault
Homicide :Crime of Killing, Robbery: Stealing, Assault: Attacking
Row Total
Column Total
Stranger
Relative
1118
787
1905
12
39
51
379
106
485
727
642
1369
Homicide Robbery Assault
Is the type of crime independent of whether the criminal is a stranger?
Row Total
Column Total
E = (row total) (column total)(grand total)
Stranger
Relative
Homicide Robbery Assault
Is the type of crime independent of whether the criminal is a stranger?
1118
787
1905
12
39
51
379
106
485
727
642
1369
Row Total
(29.93)
Column Total
E = (row total) (column total)(grand total)
E = (1118)(51)
1905= 29.93
Stranger
Relative
Homicide Robbery Assault
Is the type of crime independent of whether the criminal is a stranger?
1118
787
1905
12
39
51
379
106
485
727
642
1369
Row Total
(29.93)
(21.07)
(284.64)
(200.36)
(803.43)
(565.57)
Column Total
E = (row total) (column total)(grand total)
E = (1118)(51)
1905= 29.93 E =
(1118)(485)
1905= 284.64
etc.
Stranger
Relative
Homicide Robbery Assault
Is the type of crime independent of whether the criminal is a stranger?
1118
787
1905
12
39
51
379
106
485
727
642
1369
12
39
379
106
727
642
Homicide Robbery Forgery
(29.93)
(21.07)
(284.64)
(200.36)
(803.43)
(565.57
[10.741]Stranger
Relative
X2 = (O - E )2
E
(O -E )2
EUpper left cell: = = 10.741
(12 -29.93)2
29.93
(E)
(O - E )2
E
Is the type of crime independent of whether the criminal is a stranger?
12
39
379
106
727
642
Homicide Robbery Forgery
(29.93)
(21.07)[15.258]
(284.64)[31.281]
(200.36)[44.439]
(803.43)[7.271]
(565.57)[10.329]
[10.741]Stranger
Acquaintance
or Relative
X2 = (O - E )2
E
(O -E )2
EUpper left cell: = = 10.741
(12 -29.93)2
29.93
(E)
(O - E )2
E
Is the type of crime independent of whether the criminal is a stranger?
12
39
379
106
727
642
Homicide Robbery Forgery
(29.93)
(21.07)[15.258]
(284.64)[31.281]
(200.36)[44.439]
(803.43)[7.271]
(565.57)[10.329]
[10.741]Stranger
Acquaintance
or Relative
X2 = (O - E )2
E
(E)
(O - E )2
E
Is the type of crime independent of whether the criminal is a stranger?
Test Statistic X2 = 10.741 + 31.281 + ... + 10.329 =
119.319
Test Statistic X2 = 119.319
with = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of freedom
Critical Value X2 = 5.991
Critical Values of Chi2
Significance Level
df 0.10 0.05 0.25 0.01 0.005
1 2.7055 3.8415 5.0239 6.6349 7.8794
2 4.6062 5.9915 7.3778 9.2104 10.5965
3 6.2514 7.8147 9.3484 11.3449
12.8381
Test Statistic X2 = 119.319
with = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of freedom
0
= 0.05
X2 = 5.991
RejectIndependence
Critical Value X2 = 5.991
Sample data: X2 =119.319
Fail to RejectIndependence
Test Statistic X2 = 119.319
with = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of freedom
0
= 0.05
X2 = 5.991
RejectIndependence
Critical Value X2 = 5.991
Reject independence
Sample data: X2 =119.319
Fail to RejectIndependence
Test Statistic X2 = 119.319
with = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of freedom
0
= 0.05
X2 = 5.991
RejectIndependence
Critical Value X2 = 5.991
Reject independence
Sample data: X2 =119.319
Fail to RejectIndependence
Claim: The type of crime and knowledge of criminal are independentHo : The type of crime and knowledge of criminal are independent H1 : The type of crime and knowledge of criminal are dependent
Test Statistic X2 = 119.319
with = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of freedom
It appears that the type of crime and knowledge of the criminal are related.
0
= 0.05
X2 = 5.991
RejectIndependence
Critical Value X2 = 5.991
Reject independence
Sample data: X2 =119.319
Fail to RejectIndependence