MGT 2120: Chapter 12
Tests of Goodness of Fit and Independence
§12.1 Goodness of Fit: A Multinomial Population
A population with each observation having only one of just a few possible values (or categories) is called Multinomial population.
Examples of Multinomial populations

Grade students receive in a class (A, B, C, D, or F)
 Opinion about an issue (Favor, Oppose, Don’t know)

Member of a political party (D, R, or Independent)
Using Goodness of fit tests, we test whether or not a given set of probabilities for the possible outcomes is valid. In other words, we test whether a given probability distribution for a multinomial population is valid or not.
H o
: The given probability distribution is valid (state the values for p
1
, p
1
,…, p k
)
H a
: The given probability distribution is not valid
Random sample: n = Sample size k = number of categories f i
= observed frequency for category i in the sample data
 f i
= n e i
= expected frequency for category i in the sample data e i
= n x p i
, where n = sample size
 2
Calc
Population value
Probability as in H o
Value 1 p
1
Value 2 p
2
…. ….
Value k p k
 p i
= 1
Observed frequency (f i
) f
1 f
2
… f k n =
 f i
Expected frequency (e i
) e
1
= n x p
1 e
2
= n x p
2
…. e k
= n x p k
(f
1
– e
(f
2
– e
2
) 2 /e
2
….
1
)
(f k
– e k
)
2
 2
Calc
=
2
/e
/e k k  i

1
1
( f i
 e i
)
2 e i
NOTE: All expected frequencies (e i
) must be > 5 for the test to be valid.
Test statistic =
 2
𝐶𝑎𝑙𝑐
= ∑ 𝑘 𝑖=1
(𝑓 𝑖
−𝑒 𝑖
)
2 𝑒 𝑖
Then, p-value = CHISQ.DIST.RT(
 2
Calc
,df), where df = k-1

§12.2 Test of Independence
 2
test is used to test whether or not two variables are independent.
Examples of test of independence

Is wage rate independent of gender?

Is smoking independent of age group?

Is the choice of major independent of race?
H o
: The two variables are independent
H a
: The two variables are NOT independent
Prepare a two-way table called Contingency Table for the two variables.
Observed frequencies (f ij
= observed frequency for row i and column j in the sample data)
Variable 1 values
Value 1
Value 2
….
Value n f f f
Value 1
11
21 n1
 f i1
= T
.1
f f f
12
22 n2
 f i2
Variable 2 values
Value 2
= T
.2
….
….
….
….
…. f f f
Value m
1m
2m nm
 f im
= T
.m

 f f
1j
2j
= T
= T
1.
2.
….
 f

T nj
= T n.
i.
= T n = Number of categories of the row variable, i.e. number of rows m = Number of categories of the column variable, i.e. number of columns e ij
=
T i.
∗T
.j
where, T i.
= Sum of row i; T
.j
= Sum of column j; and T = Grand sum
T
Expected frequencies (e ij
= expected frequency for category row i and column j)
Variable 1 values
Variable 2 values
Value 1 Value 2
….
Value 1 e
11
= (T
1.
*T
.1
)/T e
12
= (T
1.
*T
.2
)/T ….
Value 2 e
21
= (T
2.
*T
.1
)/T e
22
= (T
2.
*T
.2
)/T ….
…. ….
 e i1
= T
.1
….
 e i2
= T
.2
….
Value n e n1
= (T n.
*T
.1
)/T e n2
= (Tn
.
*T
.2
)/T ….
….
Value m e
1m
= (T
1.
*T
.m
)/T  e
1j
= T
1.
e
2m
= (T
2.
*T
.m
)/T
 e
2j
…. ….
= T
2.
e nm
= (T n.
*T
.m
)/T
 e nj
= T n.
 e im
= T
.m
NOTE: All expected frequencies (e ij
) must be > 5 for the test to be valid.
Test statistic =
 2
Calc
𝑛 𝑖=1
𝑚 𝑗=1
(𝑓 𝑖𝑗
−𝑒 𝑖𝑗
)
2
, where, df = (n-1)(m-1) 𝑒 𝑖𝑗 p-value = CHISQ.DIST.RT(
 2
Calc
,df)

§12.3 Goodness of Fit for Poisson and Normal distributions

2 test procedure for Poisson distribution :
H o
: The population follows Poisson distribution
H a
: The population does NOT follow Poisson distribution
H
0
: The number of customers entering the store during 5-minute intervals has a Poisson probability distribution
H a
: The number of customers entering the store during 5-minute intervals does not have a
Poisson distribution
Take a random sample and compute the sample mean 𝑥̅ and use it as the estimate for the mean of the Poisson distribution.
Mean = SUMPRODUCT(Range1, Range2)/SUM(Range 2)
Range 1 = Cell range that contains the X-values
Range 2 = Cell range that contains the Frequencies
The Poisson probability can be determined using the Excel function POISSON.DIST(X,Mean,0)
Table of observed and expected frequencies:
X Poisson probability with 𝑥̅ as the mean
Observed frequency (f i
)
Expected frequency (e i
)
 2
Calc
0 p
1
1 p
2 p k
 p
…. …. i
= 1 f
1 f
2
… f k n =
 f i e e
2
= n x p
2
…. e
1 k
= n x p
= n x p
1 k
(f
1
– e
1
)
2
/e
1
(f
2
– e
2
)
2
….
/e
2
(f k
– e k
)
2
/e k
 2
Calc
= k  i

1
( f i
 e i
)
2 e i
Make sure all expected frequencies (e i
) are > 5. If not, combine rows.
 p-value for this test is given by = CHISQ.DIST.RT(
 2
Calc
,df), where df = k-2


 2 test procedure for Normal distribution :
Step 1:
Set up the hypotheses:
Step 2:
H o
: The random variable (population) follows Normal distribution
H a
: The random variable (population) does NOT follow Normal distribution
Take a random sample of size n, and use:
Sample mean 𝑥̅ = Estimate for the population mean

Sample standard deviation S = Estimate for the population standard deviation

Step 3:
Set up the table of observed frequencies and expected frequencies. a.
Find number of classes, k = n/5 (round down) b.
Determine the probability per class interval (1/k) c.
Set up a column of cumulative probabilities for the class intervals (The cumulative probability for the last class must be = 1) d.
Using NORM.INV(Cumulative probability, 𝑥̅ , S) function determine the upper limit for each class.
Step 4: a.
Using the “Array Function” FREQUENCY, determine the observed frequencies. (Check to see the sum = n) b.
Compute expected frequency (e i
) = n/k for all classes. c.
Calculate the test statistic =
 2
𝐶𝑎𝑙𝑐
= ∑ 𝑘 𝑖=1
(𝑓 𝑖
−𝑒 𝑖
) 2 𝑒 𝑖 d.
Find the p-vale = CHISQ.DIST.RT(
 2
Calc
,df), with df = k-3 e.
Make a conclusion
Cumulative probability
Upper class limit Observed frequency
(f i
)
Expected frequency (e i
)
 2
Calc
1/k
2/k
…. k/k
NORM.INV(1/k,
NORM.INV(2/k,
NORM.INV(1, 𝑥̅ 𝑥̅ 𝑥̅
, S)
, S)
, S)
Use “Array”
FREQUENCY n = function
 f i f f
1 k e e
1 k
= n/k f
2 e
2
= n/k
… ….
= n/k
(f
1
– e
1
)
2
/e
1
(f
2
– e
2
)
2
….
/e
2
(f k
– e k
)
2
/e k
 2
Calc
= k  i

1
( f i
 e i
)
2 e i