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Chi-squared Tests

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Title: Chi-squared Tests


1
Chi-squared Tests
2
We want to test the goodness of fit of a
particular theoretical distribution to an
observed distribution. The procedure is
  • 1. Set up the null and alternative hypotheses
    and select the significance level.
  • 2. Draw a random sample of observations from
    a population or process.
  • 3. Derive expected frequencies under the
    assumption that the null hypothesis is true.
  • 4. Compare the observed frequencies and the
    expected frequencies.
  • 5. If the discrepancy between the observed
    and expected frequencies is too great to
    attribute to chance fluctuations at the selected
    significance level, reject the null hypothesis.

3
Example 1 Five brands of coffee are
taste-tested by 1000 people with the results
below. Test at the 5 level the hypothesis that,
in the general population, there is no difference
in the proportions preferring each brand (i.e.
H0 pA pB pC pD pE versus H1 not all the
proportions are the same).
Brand preference Observed frequency fo Theoretical frequency ft fo-ft (fo-ft)2
A 210
B 312
C 170
D 85
E 223
1000
4
If all the proportions were the same, wed expect
about 200 people in each group, if we have a
total of 1000 people.
Brand preference Observed frequency fo Theoretical frequency ft fo-ft (fo-ft)2
A 210 200
B 312 200
C 170 200
D 85 200
E 223 200
1000 1000
5
We next compute the differences in the observed
and theoretical frequencies.
Brand preference Observed frequency fo Theoretical frequency ft fo-ft (fo-ft)2
A 210 200 10
B 312 200 112
C 170 200 -30
D 85 200 -115
E 223 200 23
1000 1000
6
Then we square each of those differences.
Brand preference Observed frequency fo Theoretical frequency ft fo-ft (fo-ft)2
A 210 200 10 100
B 312 200 112 12544
C 170 200 -30 900
D 85 200 -115 13225
E 223 200 23 539
1000 1000
7
Then we divide each of the squares by the
expected frequency and add the quotients. The
resulting statistic has a chi-squared (?2)
distribution.
Brand preference Observed frequency fo Theoretical frequency ft fo-ft (fo-ft)2
A 210 200 10 100 0.500
B 312 200 112 12544 62.720
C 170 200 -30 900 4.500
D 85 200 -115 13225 66.125
E 223 200 23 539 2.645
1000 1000 136.49
8
The chi-squared (?2) distribution
The chi-squared distribution is skewed to the
right. (i.e. It has the bump on the left and
the tail on the right.)
9
In these goodness of fit problems, the number of
degrees of freedom is
In the current problem, we have 5 categories (the
5 brands). We have 1 restriction. When we
determined our expected frequencies, we
restricted our numbers so that the total would be
the same total as for the observed frequencies
(1000). We didnt estimate any parameters in
this particular problem. So dof 5 1 0 4 .
10
Large values of the ?2 statistic indicate big
discrepancies between the observed and
theoretical frequencies.
So when the ?2 statistic is large, we reject the
hypothesis that the theoretical distribution is a
good fit. That means the critical region consists
of the large values, the right tail.
acceptance region
crit. reg.
11
From the ?2 table, we see that for a 5 test with
4 degrees of freedom, the cut-off point is 9.488.
In the current problem, our ?2 statistic had a
value of 136.49. So we reject the null hypothesis
and conclude that the proportions preferring each
brand were not the same.
f(?2)
acceptance region
0.05
crit. reg.
9.488
136.49
12
Example 2 A diagnostic test of mathematics is
given to a group of 1000 students. The
administrator analyzing the results wants to know
if the scores of this group differ significantly
from those of the past. Test at the 10 level.
Grade Historical Rel. freq. Expected Abs. freq. ft CurrentObs. freq. fo fo-ft (fo-ft)2
90-100 0.10 50
80-89 0.20 100
70-79 0.40 500
60-69 0.20 200
lt60 0.10 150
1000
13
Grade Historical Rel. freq. Expected Abs. freq. ft CurrentObs. freq. fo fo-ft (fo-ft)2
90-100 0.10 50
80-89 0.20 100
70-79 0.40 500
60-69 0.20 200
lt60 0.10 150
1000
14
Based on the historical relative frequency, we
determine the expected absolute frequency,
restricting the total to the total for the
current observed frequency.
Grade Historical Rel. freq. Expected Abs. freq. ft CurrentObs. freq. fo fo-ft (fo-ft)2
90-100 0.10 100 50
80-89 0.20 200 100
70-79 0.40 400 500
60-69 0.20 200 200
lt60 0.10 100 150
1000 1000
15
We subtract the theoretical frequency from the
observed frequency.
Grade Historical Rel. freq. Expected Abs. freq. ft CurrentObs. freq. fo fo-ft (fo-ft)2
90-100 0.10 100 50 -50
80-89 0.20 200 100 -100
70-79 0.40 400 500 100
60-69 0.20 200 200 0
lt60 0.10 100 150 50
1000 1000
16
We square those differences.
Grade Historical Rel. freq. Expected Abs. freq. ft CurrentObs. freq. fo fo-ft (fo-ft)2
90-100 0.10 100 50 -50 2500
80-89 0.20 200 100 -100 10,000
70-79 0.40 400 500 100 10,000
60-69 0.20 200 200 0 0
lt60 0.10 100 150 50 2500
1000 1000
17
We divide the square by the theoretical frequency
and sum up.
Grade Historical Rel. freq. Expected Abs. freq. ft CurrentObs. freq. fo fo-ft (fo-ft)2
90-100 0.10 100 50 -50 2500 25
80-89 0.20 200 100 -100 10,000 50
70-79 0.40 400 500 100 10,000 25
60-69 0.20 200 200 0 0 0
lt60 0.10 100 150 50 2500 25
1000 1000 125
18
We have 5 categories (the 5 grade groups). We
have 1 restriction. We restricted our expected
frequencies so that the total would be the same
total as for the observed frequencies (1000).
We didnt estimate any parameters in this
particular problem. So dof 5 1 0 4 .
19
From the ?2 table, we see that for a 10 test
with 4 degrees of freedom, the cut-off point is
7.779.
In the current problem, our ?2 statistic had a
value of 125. So we reject the null hypothesis
and conclude that the grade distribution is NOT
the same as it was historically.
f(?2)
acceptance region
0.10
crit. reg.
7.779
125
20
Example 3 Test at the 5 level whether the
demand for a particular product as listed below
has a Poisson distribution.
of units demanded per day x Observed of days fo xfo prob. f(x) Expected of days ft fo-ft (fo-ft)2
0 11
1 28
2 43
3 47
4 32
5 28
6 7
7 0
8 2
9 1
10 1
200
21
Multiplying the number of days on which each
amount was sold by the amount sold on that day,
and then adding those products, we find that the
total number of units sold on the 200 days is
600. So the mean number of units sold per day is
3.
of units demanded per day x Observed of days fo xfo prob. f(x) Expected of days ft fo-ft (fo-ft)2
0 11 0
1 28 28
2 43 86
3 47 141
4 32 128
5 28 140
6 7 42
7 0 0
8 2 16
9 1 9
10 1 10
200 600
22
We use the 3 as the estimated mean for the
Poisson distribution. Then using the Poisson
table, we determine the probabilities for each x
value.
of units demanded per day x Observed of days fo xfo prob. f(x) Expected of days ft fo-ft (fo-ft)2
0 11 0 0.050
1 28 28 0.149
2 43 86 0.224
3 47 141 0.224
4 32 128 0.168
5 28 140 0.101
6 7 42 0.050
7 0 0 0.022
8 2 16 0.008
9 1 9 0.003
10 1 10 0.001
200 600 1.
23
Then we multiply the probabilities by 200 to
compute ft, the expected number of days on which
each number of units would be sold. By
multiplying by 200, we restrict the ft total to
be the same as the fo total.
of units demanded per day x Observed of days fo xfo prob. f(x) Expected of days ft fo-ft (fo-ft)2
0 11 0 0.050 10.0
1 28 28 0.149 29.8
2 43 86 0.224 44.8
3 47 141 0.224 44.8
4 32 128 0.168 33.6
5 28 140 0.101 20.2
6 7 42 0.050 10.0
7 0 0 0.022 4.4
8 2 16 0.008 1.6
9 1 9 0.003 0.6
10 1 10 0.001 0.2
200 600 1. 200
24
When the fts are small (less than 5), the test
is not reliable. So we group small ft values.
In this example, we group the last 4 categories.
of units demanded per day x Observed of days fo Observed of days fo xfo prob. f(x) Expected of days ft Expected of days ft fo-ft (fo-ft)2
0 11 11 0 0.050 10.0 10.0
1 28 28 28 0.149 29.8 29.8
2 43 43 86 0.224 44.8 44.8
3 47 47 141 0.224 44.8 44.8
4 32 32 128 0.168 33.6 33.6
5 28 28 140 0.101 20.2 20.2
6 7 7 42 0.050 10.0 10.0
7 0 4 0 0.022 4.4 6.8
8 2 4 16 0.008 1.6 6.8
9 1 4 9 0.003 0.6 6.8
10 1 4 10 0.001 0.2 6.8
200 200 600 200 200
25
Next we subtract the theoretical frequencies ft
from the observed frequencies fo.
of units demanded per day x Observed of days fo Observed of days fo xfo prob. f(x) Expected of days ft Expected of days ft fo-ft (fo-ft)2
0 11 11 0 0.050 10.0 10.0 1
1 28 28 28 0.149 29.8 29.8 1.8
2 43 43 86 0.224 44.8 44.8 -1.8
3 47 47 141 0.224 44.8 44.8 2.2
4 32 32 128 0.168 33.6 33.6 -1.6
5 28 28 140 0.101 20.2 20.2 7.8
6 7 7 42 0.050 10.0 10.0 -3
7 0 4 0 0.022 4.4 6.8 -2.8
8 2 4 16 0.008 1.6 6.8 -2.8
9 1 4 9 0.003 0.6 6.8 -2.8
10 1 4 10 0.001 0.2 6.8 -2.8
200 200 600 200 200
26
Then we square the differences
of units demanded per day x Observed of days fo Observed of days fo xfo prob. f(x) Expected of days ft Expected of days ft fo-ft (fo-ft)2
0 11 11 0 0.050 10.0 10.0 1 1
1 28 28 28 0.149 29.8 29.8 1.8 3.24
2 43 43 86 0.224 44.8 44.8 -1.8 3.24
3 47 47 141 0.224 44.8 44.8 2.2 4.84
4 32 32 128 0.168 33.6 33.6 -1.6 2.5
5 28 28 140 0.101 20.2 20.2 7.8 60.84
6 7 7 42 0.050 10.0 10.0 -3 9
7 0 4 0 0.022 4.4 6.8 -2.8 7.84
8 2 4 16 0.008 1.6 6.8 -2.8 7.84
9 1 4 9 0.003 0.6 6.8 -2.8 7.84
10 1 4 10 0.001 0.2 6.8 -2.8 7.84
200 200 600 200 200
27
divide by the theoretical frequencies, and sum
up.
of units demanded per day x Observed of days fo Observed of days fo xfo prob. f(x) Expected of days ft Expected of days ft fo-ft (fo-ft)2
0 11 11 0 0.050 10.0 10.0 1 1 0.10
1 28 28 28 0.149 29.8 29.8 1.8 3.24 0.11
2 43 43 86 0.224 44.8 44.8 -1.8 3.24 0.07
3 47 47 141 0.224 44.8 44.8 2.2 4.84 0.11
4 32 32 128 0.168 33.6 33.6 -1.6 2.5 0.08
5 28 28 140 0.101 20.2 20.2 7.8 60.84 3.01
6 7 7 42 0.050 10.0 10.0 -3 9 0.90
7 0 4 0 0.022 4.4 6.8 -2.8 7.84 1.15
8 2 4 16 0.008 1.6 6.8 -2.8 7.84 1.15
9 1 4 9 0.003 0.6 6.8 -2.8 7.84 1.15
10 1 4 10 0.001 0.2 6.8 -2.8 7.84 1.15
200 200 600 200 200 5.53
28
We have 8 categories (after grouping the small
ones). We have 1 restriction. We restricted our
expected frequencies so that the total would be
the same total as for the observed frequencies
(200). We estimated 1 parameter, the mean for the
Poisson distribution. So dof 8 1 1 6 .
29
From the ?2 table, we see that for a 5 test with
6 degrees of freedom, the cut-off point is 12.592.
In the current problem, our ?2 statistic had a
value of 5.53. So we accept the null hypothesis
that the Poisson distribution is a reasonable fit
for the product demand.
f(?2)
acceptance region
0.05
crit. reg.
12.592
5.53
30
Example 4 Test at the 10 level whether the
following exam grades are from a normal
distribution. Note This is a very long problem.

grade intervals midpoint X fo X fo
50, 60) 14
60,70) 18
70,80) 36
80.90) 18
90,100 14
100
31
If the distribution is normal, we need to
estimate its mean and standard deviation.
grade intervals midpoint X fo X fo
50, 60) 14
60,70) 18
70,80) 36
80.90) 18
90,100 14
100
32
To estimate the mean, we first determine the
midpoints of the grade intervals.
grade intervals midpoint X fo X fo
50, 60) 55 14
60,70) 65 18
70,80) 75 36
80.90) 85 18
90,100 95 14
100
33
We then multiple these midpoints by the observed
frequencies of the intervals, add the products,
and divide the sum by the number of
observations. The resulting mean is 7500/100 75.
grade intervals midpoint X fo X fo
50, 60) 55 14 770
60,70) 65 18 1170
70,80) 75 36 2700
80.90) 85 18 1530
90,100 95 14 1330
100 7500
34
Next we need to calculate the standard deviation
We begin by subtracting the mean of 75 from
each midpoint, and squaring the differences.
grade intervals midpoint X fo X fo
50, 60) 55 14 770 -20 400
60,70) 65 18 1170 -10 100
70,80) 75 36 2700 0 0
80.90) 85 18 1530 10 100
90,100 95 14 1330 20 400
100 7500
35
We multiply by the observed frequencies and sum
up. Dividing by n 1 or 99, the sample variance
s2 149.49495. The square root is the sample
standard deviation s 12.2268.
grade intervals midpoint X fo X fo
50, 60) 55 14 770 -20 400 5600
60,70) 65 18 1170 -10 100 1800
70,80) 75 36 2700 0 0 0
80.90) 85 18 1530 10 100 1800
90,100 95 14 1330 20 400 5600
100 7500 14,800
36
We will use the 75 and 12.2268 as the mean ? and
the standard deviation ? of our proposed normal
distribution. We now need to determine what the
expected frequencies would be if the grades were
from that normal distribution.
37
Start with our lowest grade category, under 60.
We then expect that 10.93 of our 100
observations, or about 11 grades, would be in the
lowest grade category. So 11 will be one of our
ft values. We need to do similar calculations for
our other grade categories.
38
The next grade category is 60,70).
So 23.16 of our 100 observations, or about 23
grades, are expected to be in that grade category.
39
The next grade category is 70,80).
So 31.82 of our 100 observations, or about 32
grades, are expected to be in that grade category.
40
The next grade category is 80,90).
So 23.16 of our 100 observations, or about 23
grades, are expected to be in that grade category.
41
The highest grade category is 90 and over.
So 10.93 of our 100 observations, or about 11
grades, are expected to be in that grade category.
42
Now we can finally compute our ?2 statistic. We
put in the observed frequencies that we were
given and the theoretical frequencies that we
just calculated.
grade category fo ft
under 60 14 11
60,70) 18 23
70,80) 36 32
80.90) 18 23
90 and up 14 11

43
We subtract the theoretical frequencies from the
observed frequencies, square the differences,
divide by the theoretical frequencies, and sum
up. The resulting ?2 statistic is 4.3104.
grade category fo ft
under 60 14 11 0.8182
60,70) 18 23 1.0870
70,80) 36 32 0.5000
80.90) 18 23 1.0870
90 and up 14 11 0.8182
4.3104
44
We have 5 categories (the 5 grade groups). We
have 1 restrictions. We restricted our expected
frequencies so that the total would be the same
total as for the observed frequencies (100). We
estimated two parameters, the mean and the
standard deviation. So dof 5 1 2 2 .
45
From the ?2 table, we see that for a 10 test
with 2 degrees of freedom, the cut-off point is
4.605.
In the current problem, our ?2 statistic had a
value of 4.31. So we accept the null hypothesis
that the normal distribution is a reasonable fit
for the grades.
f(?2)
acceptance region
0.10
crit. reg.
4.605
4.31
46
We can also use the ?2 statistic to test whether
two variables are independent of each other.
47
Example 5 Given the following frequencies for a
sample of 10,000 households, test at the 1 level
whether the number of phones and the number of
cars for a household are independent of each
other.
of cars of cars of cars
0 1 2
of phones 0 1,000 900 100
of phones 1 1500 2600 500
of phones 2 or more 500 2500 400
10,000

48
We first compute the row and column totals,
of cars of cars of cars
0 1 2 row total
of phones 0 1,000 900 100 2000
of phones 1 1500 2600 500 4600
of phones 2 or more 500 2500 400 3400
column total 3,000 6,000 1,000 10,000

49
and the row and column percentages (marginal
probabilities).
of cars of cars of cars
0 1 2 row total
of phones 0 1,000 900 100 2000 0.20
of phones 1 1500 2600 500 4600 0.46
of phones 2 or more 500 2500 400 3400 0.34
column total 3,000 6,000 1,000 10,000 1.00
0.30 0.60 0.10 1.00
50
Recall that if 2 variables X and Y are
independent of each other, then Pr(Xx and Yy)
Pr(Xx) Pr(Yy)
51
We can use our row and column percentages as
marginal probabilities, and multiply to determine
the probabilities and numbers of households we
would expect to see in the center of the table if
the numbers of phones and cars were independent
of each other.
of cars of cars of cars
0 1 2 row total
of phones 0 0.20
of phones 1 0.46
of phones 2 or more 0.34
column total 1.00
0.30 0.60 0.10 1.00
52
First calculate the expected probability. For
example, Pr(0 phones 0 cars) Pr(0 phones)
Pr(0 cars) (0.20)(0.30) 0.06. So we expect 6
of our 10,000 households, or 600 households to
have 0 phones and 0 cars.
of cars of cars of cars
0 1 2 row total
of phones 0 600 0.20
of phones 1 0.46
of phones 2 or more 0.34
column total 10,000 1.00
0.30 0.60 0.10 1.00
53
Pr(0 phones 1 car) Pr(0 phones) Pr(1 car)
(0.20)(0.60) 0.12. So we expect 12 of our
10,000 households, or 1200 households to have 0
phones and 1 car.
of cars of cars of cars
0 1 2 row total
of phones 0 600 1200 0.20
of phones 1 0.46
of phones 2 or more 0.34
column total 10,000 1.00
0.30 0.60 0.10 1.00
54
Pr(0 phones 2 cars) Pr(0 phones) Pr(2 cars)
(0.20)(0.10) 0.02. So we expect 2 of our
10,000 households, or 200 households to have 0
phones and 2 cars.
of cars of cars of cars
0 1 2 row total
of phones 0 600 1200 200 0.20
of phones 1 0.46
of phones 2 or more 0.34
column total 10,000 1.00
0.30 0.60 0.10 1.00
55
Notice that when we add the 3 numbers that we
just calculated, we get the same total for the
row (2,000) that we had observed. The row and
column totals should be the same for the observed
and expected tables.
of cars of cars of cars
0 1 2 row total
of phones 0 600 1200 200 2,000 0.20
of phones 1 0.46
of phones 2 or more 0.34
column total 10,000 1.00
0.30 0.60 0.10 1.00
56
Continuing, we get the following numbers for the
2nd and 3rd rows.
of cars of cars of cars
0 1 2 row total
of phones 0 600 1200 200 2,000 0.20
of phones 1 1380 2760 460 4600 0.46
of phones 2 or more 1020 2040 340 3400 0.34
column total 10,000 1.00
0.30 0.60 0.10 1.00
57
The column totals are the same as for the
observed table.
of cars of cars of cars
0 1 2 row total
of phones 0 600 1200 200 2,000 0.20
of phones 1 1380 2760 460 4600 0.46
of phones 2 or more 1020 2040 340 3400 0.34
column total 3000 6000 1000 10,000 1.00
0.30 0.60 0.10 1.00
58
Now we set up the same type of table that we did
for our earlier ?2 goodness-of-fit tests. We
put in the fo column the observed frequencies and
in the ft column the expected frequencies that we
calculated.
of cars of phones fo ft
0 0 1000 600
0 1 1500 1380
0 2 or more 500 1020
1 0 900 1200
1 1 2600 2760
1 2 or more 2500 2040
2 0 100 200
2 1 500 460
2 2 or more 400 340

59
Then we subtract the theoretical frequencies from
the observed frequencies, square the differences,
divide by the theoretical frequencies, and sum to
get our ?2 statistic.
of cars of phones fo ft
0 0 1000 600 266.67
0 1 1500 1380 10.43
0 2 or more 500 1020 265.10
1 0 900 1200 75.00
1 1 2600 2760 9.28
1 2 or more 2500 2040 103.73
2 0 100 200 50.00
2 1 500 460 3.48
2 2 or more 400 340 10.59
794.28
60
In these tests of independence, the number of
degrees of freedom is
In our example, we have 3 rows and 3 columns. So
dof (3 1)( 3 1) (2)(2) 4 .
61
From the ?2 table, we see that for a 1 test with
4 degrees of freedom, the cut-off point is 13.277.
In the current problem, our ?2 statistic had a
value of 794.28. So we reject the null hypothesis
and conclude that the number of phones and the
number of cars in a household are not independent.
f(?2)
acceptance region
0.01
crit. reg.
13.277
794.28
62
Yates Correction
In testing for independence in 2x2 tables, the
chi-square statistic has only (r-1)(c-1) 1
degree of freedom. In these cases, it is often
recommended that the value of the statistic be
corrected so that its discrete distribution
will be better approximated by the continuous
chi-square distribution.
63
The Hypothesis Test for the Variance or Standard
Deviation
This test is another one that uses the
chi-squared distribution.
64
Sometimes it is important to know the variance or
standard deviation of a variable.
For example, medication often needs to be
extremely close to the specified dosage. If the
dosage is too low, the medication may be
ineffective and a patient may die from inadequate
treatment. If the dosage is too high, the
patient may die from an overdose. So you may
want to make sure that the variance is a very
small amount.
65
If the data are normally distributed, the
chi-squared test for the variance or standard
deviation is appropriate.
The statistic is
n is the sample size, and s2 is the
hypothesized population variance. The number of
degrees of freedom is n-1.
66
Example Suppose you want to test at the 5
level whether the population standard deviation
for a particular medication is 0.5 mg. Based on
a sample of 25 capsules, you determine the sample
standard deviation to be 0.6 mg. Perform the
test.
Now we need to determine the critical region for
the test.
67
Because the chi-squared distribution is not
symmetric, you need to look up the two critical
values for a two-tailed test separately.
You can find the two numbers either by looking
under Cumulative Probabilities 0.025 and
1-0.0250.975 or under Upper-Tail Areas 0.975
and 0.025 .
Recall that the value of the test statistic was
34.56, which is in the acceptance region. So we
can not rule out the null hypothesis and
therefore we conclude that the population
standard deviation is 0.5 mg.
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