056_Hypothesis Testing - Chi Square

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Hypothesis Testing -
Chi Square

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Two Chi-Square Tests?
• Chi-Square ‘Goodness of Fit Test’
Used to test if a particular distribution (model)
is a good fit for a population. Use with 1
sample.
• Chi-Square ‘Test for Association’
Used to test if a relationship between two
attribute variables exists. Use with 2 samples.
Both of these tests use the
Chi-Square distribution, where
fo is the observed frequency and
fe is the expected frequency.
Note: the sum of the expected
frequencies is always equal to
the sum of the observed
frequencies.
( )
∑=
−
=
g
1j
e
2
eo
f
ff2
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When to Use Goodness-Of-Fit Test?
The Chi-Square Goodness of fit test compares an
observed frequency to an expected frequency.
Example
A coin is flipped 100 times and 63 heads and 37 tails
are recorded. Could this result have occurred by
chance, or is the coin biased?
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1. State the practical problem:
Is the coin biased?
2. State the Null & Alternate Hypothesis
H0: Probability of Heads = Tails = 0.5
Ha: Probability of Heads ≠ Tails ≠ 0.5
3. Select α = 0.05
4. Create a contingency table
5. Calculate degrees of freedom (df), (# Rows-1) *
(# Columns-1) = (2-1)*(2-1)=1
Goodness of Fit - Example
Observed
( fo )
Expected
( fe )
Head
s 63 50
Tails 37 50

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Test for Association

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Test For Association - Example
1000 Adults were surveyed to determine their political
views:
246 were liberal
405 were moderate
349 were conservative
The split of males and females was also recorded
447 were male
553 were female
Is gender and political view related?

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Liberal Moderate Conservative
Male 73 213 161 447
Female 173 192 188 553
246 405 349 1000
4. Create a contingency table continued:
5. Calculate degrees of freedom (df), (# Rows-1) *
(# Columns-1) = (2-1)*(3-1)=2
6. Calculate the χ2 test statistic - χ2 (calculated), recall:
fo
( )
∑=
−
=
g
1j
e
2
eo
f
ff2
χ

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9. Practical solution:
• Proportion of liberal males is lower than expected; proportion
of females is higher
• Proportion of moderate males somewhat higher; proportion of
moderate females somewhat lower
• Proportion of male to female conservatives are consistent with
expected frequencies
Liberal Moderate Conservative
Male - Obs 73 213 161 447
- Exp 109.96 181.04 156
Female - Obs 173 192 188 553
- Exp 136.04 223.96 193
246 405 349 1000

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Multiple Proportion Comparison
Practical Question : Are all populations’ proportions statistically different?
P1
P2 P3
vs. vs.

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1. Practical problem:
Are some shifts producing more defectives than others?
2. Statistical problem:
H0: P1 = P2 = P3
Ha: P1 ≠ P2 ≠ P3
3. α = 0.05:
Test For Multiple Proportions - Example

1st Shift 2nd Shift 3rd Shift Total
1 43 56 68 167
60.31 55.89 50.80
4.970 0.000 5.824
2 965 878 781 2624
947.69 878.11 798.20
0.316 0.000 0.371
Total 1008 934 849 2791
Chi-Sq = 11.481, DF = 2, P-Value = 0.003 < α = 0.05, Reject H0
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Test For Association – Example
Minitab Solution
• Create a table in Minitab:
• Stat > Tables > χ2 Chi Square Test
• Analyze Results:
1st Shift 2nd Shift 3rd Shift
Defectives 43 56 68
Non-Defectives 965 878 781

Questions
1. Good of Fit and Test for Association are the same test.
2. Chi Square is used to test multiple proportions.
3. Binomial is the underlying distribution.
4. Chi Square critical values are dependant on the degrees of freedom.
5. If Calc > Crit, reject H0.
6. If P < alpha , reject H0.
7. Test for Association can be used to test multiple supplier locations.
8. Reject H0 = accept Ha.
9. Fail to Reject H0 = reject Ha.
10. Confidence Level = 1 – alpha = Confidence Interval.
T F
T F
T F
T F
T F
T F
T F
T F
T F
T F
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056_Hypothesis Testing - Chi Square