Section 10.1
Goodness of Fit

Section 10.1 Objectives
•
Use the chi-square distribution to test whether a frequency distribution fits a claimed distribution
“chi” - kie (rhymes with “eye”) [see page 330]
- a distribution
- area under curve = 1
- all values of
χ 2 are greater than or equal to zero
- positively skewed
df = n -1

Multinomial Experiments
Multinomial experiment
•
A probability experiment consisting of a fixed number of independent trials in which there are more than two possible outcomes for each trial.
•
The probability for each outcome is fixed and each outcome is classified into categories .
•
Recall that a binomial experiment had only two possible outcomes.

Multinomial Experiments
Books on a shelf
Fiction
Non-fiction
Reference
60%
30%
10%
3 categories – type of books
If distribution of books is “true” and one had a library of 50 books, how many books of fiction would there be?
60% of 50 is 30 (50*.60= 30)
Non-fiction? 50*.30 = 15
Reference? 50*.10 = 5

Multinomial Experiments
3 categories
– type of books
If distribution of books is “ true ” and one had a library of 50 books, how many books of fiction would there be?
60% of 50 is 30 (50*.60= 30)
Non-fiction? 50*.30 = 15
Reference? 50*.10 = 5
These are Expected Values (given a sample of size n and assuming a given distribution is truth)
Expected value = number in sample size * assumed probability
E = n*p

Multinomial Experiments
•
A tax preparation company wants to determine the proportions of people who used different methods to prepare their taxes.
•
The company can perform a multinomial experiment.
• It wants to test a previous survey’s claim concerning the distribution of proportions of people who use different methods to prepare their taxes.
•
It can compare the distribution of proportions obtained in the multinomial experiment with the previous survey’s specified distribution.
•
It can perform a chi-square goodness-of-fit test.

Chi-Square Goodness-of-Fit Test
Chi-Square Goodness-of-Fit Test
•
Used to test whether a frequency distribution fits an expected distribution.
•
The null hypothesis states that the frequency distribution fits the specified distribution.
•
The alternative hypothesis states that the frequency distribution does not fit the specified distribution.

Multinomial Experiments
•
Results of a survey of tax preparation methods.
Each outcome is classified into categories .
Distribution of tax preparation methods
Accountant 25%
By hand
Computer software
20%
35%
Friend/family
Tax preparation service
5%
15%
The probability for each possible outcome is fixed.

Chi-Square Goodness-of-Fit Test
• To test the previous survey’s claim, a company can perform a chi-square goodness-of-fit test using the following hypotheses.
H
0
: The distribution of tax preparation methods is 25% by accountant, 20% by hand, 35% by computer software, 5% by friend or family, and 15 % by tax preparation service. (claim)
H a
: The distribution of tax preparation methods differs from the claimed or expected distribution.

Chi-Square Goodness-of-Fit Test
•
To calculate the test statistic for the chi-square goodness-of-fit test, the observed frequencies and the expected frequencies are used.
•
The observed frequency O of a category is the frequency for the category observed in the sample data.

Chi-Square Goodness-of-Fit Test
•
The expected frequency E of a category is the calculated frequency for the category.

Expected frequencies are obtained assuming the specified (or hypothesized) distribution. The expected frequency for the i th category is
E i
= np i where n is the number of trials (the sample size) and p i is the assumed probability of the category.
i th

Example: Finding Observed and
Expected Frequencies
A tax preparation company randomly selects 300 adults and asks them how they prepare their taxes. The results are shown at the right. Find the observed frequency and the expected frequency for each tax preparation method.
Survey results
(n = 300)
Accountant
By hand
Computer software
Friend/family
Tax preparation service
71
40
101
35
53

Solution: Finding Observed and
Expected Frequencies
Observed frequency: The number of adults in the survey naming a particular tax preparation method
Survey results
(n = 300)
Accountant
By hand
71
40
101 Computer software
Friend/family
Tax preparation service
35
53 observed frequency

Solution: Finding Observed and
Expected Frequencies
Expected Frequency: E i
= np i
Tax preparation method
% of people
Observed frequency
Expected frequency
Accountant 25% 71 300(0.25) = 75
By hand
Computer Software
Friend/family
Tax preparation service
20%
35%
5%
15%
40
101
35
53
300(0.20) = 60
300(0.35) = 105
300(0.05) = 15
300(0.15) = 45 n = 300

Chi-Square Goodness-of-Fit Test
For the chi-square goodness-of-fit 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 Goodness-of-Fit Test
•
If these conditions are satisfied, then the sampling distribution for the goodness-of-fit test is approximated by a chi-square distribution with k
– 1 degrees of freedom, where k is the number of categories.
•
The test statistic for the chi-square goodness-of-fit test is
 2   ( O

E )
2
E
The test is always a right-tailed test.
where O represents the observed frequency of each category and E represents the expected frequency of each category.

Chi Square Goodness of Fit Test
In Words
1.
Identify the claim. State the null and alternative hypotheses.
2.
Specify the level of significance.
3.
Identify the degrees of freedom.
4.
Determine the critical value.
In Symbols
State H
0 and H a
.
Identify
α
.
d.f. = k
– 1
Use Table 6 in
Appendix B.

Chi Square Goodness of Fit Test
In Symbols In Words
5.
Determine the rejection region.
6.
Calculate the test statistic.
7.
Make a decision to reject or fail to reject the null hypothesis.
 2   ( O

E )
2
E
If
χ 2 is in the rejection region, reject H
0
.
Otherwise, fail to reject H
0
.
8.
Interpret the decision in the context of the original claim.

Example: Performing a Goodness of Fit Test
Use the tax preparation method data to perform a chisquare goodness-of-fit test to test whether the distributions are different. Use α = 0.01.
Distribution of tax preparation methods
Accountant
By hand
Computer software
Friend/family
Tax preparation service
25%
20%
35%
5%
15%
Survey results
(n = 300)
Accountant
By hand
Computer software
Friend/family
Tax preparation service
71
40
101
35
53

Solution: Performing a Goodness of Fit Test
•
H
0
:
The distribution is 25% by accountant, 20% by hand, 35% by computer software, 5% by friend/
•
H a
: family, and 15% by tax preparation service. (Claim)
The distribution of tax preparation methods differs from the claimed or expected distribution.
α = 0.01
• d.f. = 5 – 1 = 4
•
Rejection Region
•
Test Statistic:
•
Decision:
•
Conclusion:

Solution: Performing a Goodness of Fit Test method Observ ed
Expe cted
O-E (O-E) 2 (𝑶 − 𝑬) 𝟐
𝑬
(O) (E)
Accountant
By hand
71
40
75
60
-4
-20
16
400
.213
6.667
Computer hardware
Friend/famil y
Tax prep service
101
35
53
105
15
45
-4
20
8
16
400
64
.152
26.667
1.422
So compute “chi-square:

2   (

)
2
E
= .
213 + 6.667 + .152 + 26.667 + 1.422 = 35.121

Solution: Performing a Goodness of Fit Test
•
H
0
:
The distribution is 25% by accountant, 20% by hand, 35% by computer software, 5% by friend/ family, and 15% by tax preparation service. (Claim)
•
H a
: The distribution of tax preparation methods
• α = differs from the claimed or expected distribution.
0.01
•
Test Statistic:
• d.f. = 5 – 1 = 4
χ 2 ≈ 35.121
•
Rejection Region
•
Decision: Reject H
0
There is enough evidence at the 1% significance level to conclude that the distribution of tax preparation methods differs from the previous survey’s claimed or expected distribution.

Example: Performing a Goodness of Fit Test
A researcher claims that the number of different-colored candies in bags of dark chocolate M&M’s is uniformly distributed. To test this claim, you randomly select a bag that contains 500 dark chocolate M&M’s. The results are shown in the table on the next slide. Using
α
= 0.10, perform a chi-square goodness-of-fit test to test the claimed or expected distribution. What can you conclude?

Example: Performing a Goodness of Fit Test
Color Frequency
Brown 80
Yellow
Red
95
88
Blue
Orange
Green
83
76
78 n = 500
Solution:
•
The claim is that the distribution is uniform, so the expected frequencies of the colors are equal.
•
To find each expected frequency, divide the sample size by the number of colors.
•
E = 500/6 ≈ 83.3

Solution: Performing a Goodness of Fit Test
•
H
0
: Distribution of different-colored candies in bags of dark chocolate M&M’s is uniform. (Claim)
•
H a
: Distribution of different-colored candies in bags of dark chocolate M&M’s is not uniform.
• α = 0.10
•
Test Statistic:
• d.f. = 6 – 1 = 5
•
Rejection Region •
Decision:
0 9.236
0.10
χ 2
•
Conclusion:

Solution: Performing a Goodness of Fit Test
Color
Brown
Yello w
Red
Blue
Orang e
Observed frequency
80
95
88
83
76
Expected frequency
83.33
83.33
83.33
83.33
83.33
O – E
-3.33
11.67
4.67
-.33
-7.33
(O-E) 2 (𝑶 − 𝑬) 𝟐
11.0889
𝑬
.133
136.189
1.634
21.809
.1089
53.729
.262
.001
.645
78 83.33
-5.33
28.409
.341

2
Green
  (

)
E
2
= .133 + 1.634 + .262 + .001 + .645 + .341
= 3.016

Solution: Performing a Goodness of Fit Test
•
H
0
: Distribution of different-colored candies in bags of dark chocolate M&M’s is uniform. (Claim)
•
H a
: Distribution of different-colored candies in bags of dark chocolate M&M’s is not uniform.
• α = 0.01
•
Test Statistic:
• d.f. = 6 – 1 = 5
•
Rejection Region
χ 2 ≈ 3.016
0.10
•
Decision: Fail to Reject H
0
There is not enough evidence at the 10% level of significance to
0
3.016
9.236
χ 2 reject the claim that the distribution is uniform.

Section 10.1 Summary
•
Used the chi-square distribution to test whether a frequency distribution fits a claimed distribution