Transcript 13.1 Goodness of Fit Test

13.1 Goodness of Fit Test
AP Statistics
Chi-Square Distributions
The chi-square distributions are a family of
distributions that take on only positive values
and are skewed to the right. A specific chisquare distribution is determined by its
degrees of freedom.
Properties:
1. The total area under a chi-square curve is equal
to 1.
2. Each chi-square curve (except when df = 1)
begins at 0 on the horizontal axis, increases to a
peak, and then approaches the horizontal axis
asymptotically from above.
3. Each chi-square curve is skewed to the right. As
the number of degrees of freedom increase, the
curve becomes more and more symmetrical and
looks more like a normal curve (see Figure 13.2
page 732 ).
According to the m&m/Mars company, in 1995
“…the new mix of colors of m&m’s plain
chocoloate candies will contain 30 percent
browns, 20 percent yellows and reds, and 10
percent each of oranges, greens, and blues.”
However, the mix of colors has been known to
change every few years.
Your task today is to determine whether or not the
current mix of colors matches that of 1995.
We want to see if there is sufficient evidence to
reject the company’s 1995 claim. To do this, we’ll
be introduced to a new type of test—the Chisquare Goodness of Fit Test.
A Goodness of Fit Test is used to determine
whether a population has a certain
hypothesized distribution. The null hypothesis
is that the population proportions are equal to
the hypothesized proportions. The alternative
is that at least one of the proportions differ
from the hypothesized proportions. If all
expected counts are at least 1 and 80% of
them are greater than 5, then
2

O  E
2
X 
E
has an approximately Chi-Square Distribution
with df = (k – 1).
• Open a bag of milk chocolate m&m’s and
carefully count how many of each color are in
the sample. Record the observed data in the
“observed” row of the table below.
• Using the statement from the m&m/Mars
company, determine how many of each color
you expected to see. Note, you’ll have to
figure this out using the total number of
m&m’s in your sample bag. Enter these
counts in the “expected” row below.
• If your bag reflects the distribution advertised in 1995,
there should be little difference between the observed and
expected counts. To quantify the difference we’ll calculate
a total which we’ll call “Chi-Square” or X2.
• For each color, perform this calculation:
Observed  Expected
2
Expected
Enter each value in the last row of the table. Add up all of
these “component” values to find X2.
• If this total value is small, we have little evidence to suggest
a difference in distributions. However, the larger X2 gets,
the more evidence we have to suggest the company’s claim
may no longer be applicable to bags of milk chocolate
m&m’s.
Brown
Yellow
Red
Orange
Green
Blue
Total
Observed
5
5
2
2
3
4
21
Expected
6.3
4.2
4.2
2.1
2.1
2.1
21
.2683
.1524
1.1524
.0048
.3857
1.719
3.6825
O  E 
2
E
To determine the likelihood of observing a
difference between observed and expected as
extreme as the one we observed, we must
look up the p-value on a Chi-Square table.
Chi-square distributions are skewed right and
specified by degrees of freedom. In a
Goodness of Fit test, the degrees of freedom
equal one less than the number of categories.
• Find the p-value for our test by looking up X2
for 5 degrees of freedom. Sketch the curve
and observed X2 below. Interpret the result in
the context of the problem. X2cdf(X2, 1E99, df)
Since p is large (> α), there is not significant
evidence to reject the 1995 claim.
Steps:
1. Identify the population of interest and the
parameter(s) that you want to draw conclusions
about. State hypotheses in words and symbols.
2. Choose the appropriate inference procedure
and verify the conditions for using it.
Chi-Square Conditions:
1.
2.
All individual expected counts are at least 1
No more than 20% of the expected counts are less than 5
3. Carry out the inference procedure (calculate the
T.S., df, and p-value).
4. Interpret your results in the context of the
problem.
Example 1: (13.13 p. 744)
A “wheel of fortune” at a carnival is divided into four equal parts:
Part I:
Part II:
Part III:
Part IV:
Win a doll
Win a candy bar
Win a free ride
Win nothing
You suspect that the wheel is unbalanced (i.e., not all parts of the
wheel are equally likely to be landed upon when the wheel is spun).
The results of 500 spins of the wheel are as follows:
Part:
Frequency:
I
95
II
105
III
135
IV
165
Perform a goodness of fit test. Is there evidence that the wheel is not
in balance?
Since the wheel is divided into four equal parts, if it is in balance, then the four outcomes
should occur with approximately equal frequency. Here are the observed and expected
values:
Part:
I
II
III
IV
Observed: 95
105
135
165
Expected: 125
125
125
125
Ho: The wheel is balanced (the four outcomes are uniformly distributed)
Ha: The wheel is not balanced
We will use a chi-square goodness of fit test to measure the strength of the evidence against
the hypothesis that the wheel is balanced. Since all expected counts are greater than 5, we
can proceed with the test.
df = 3
X
2

2
2
2
2

95  125 105 125 135 125 165 125




125
125
 7.2  3.2  0.8  12.8
 24
125
125

P X 32  24  2.4980105
X2cdf(24, 1E99, 3)
p<α
Reject Ho
We have significant evidence to conclude that the wheel is not balanced. Since “Part IV: Win
nothing” shows the greatest deviation from the expected result, there may be reason to
suspect that the carnival game operator may have tampered with the wheel to make it
harder to win.
Example 2:
A statistics student suspected that his 1982 penny was not a fair coin, so he held it upright on a
table top with a finger of one hand and spun the penny repeatedly by flicking it with the index
finger of his other hand. In 200 spins of the coin, it landed with tails side up 122 times.
(a) Perform a goodness of fit test to see if there is sufficient evidence to conclude that spinning
the coin does not produce an equal proportion of heads and tails.
Ho : The distribution of heads and tails from spinning a 1982 penny shows equally
likely outcomes.
Ha : Heads and tails are not equally likely.
We will use a chi-square goodness of fit test to measure the strength of the evidence
against the hypothesis that the penny is a fair coin. Since all expected counts are
greater than 5, we can proceed with the test.
2
2

78  100 122 100
2
df = 1
X 

100
100
 4.84  4.84

 9.68

P X12  9.68  .00186
p < α Reject Ho
We have significant evidence to conclude that spinning a 1982 penny does not
produce equally likely results.
(b) Use a one-proportion inference procedure to determine whether spinning
the coin is equally likely to result in heads or tails.
p = probability of getting tails when the coin is spun
Ho : p = 0.5
Ha : p ≠ 0.5
Assume SRS.
Assume population > 10(200)
np = n(1-p) = 100 > 10
1-prop-z-test
.61 .50
z
 3.1113
.501  .50
200
p < α Reject Ho.
Pz  3.1113and z  3.1113  .00186
There is significant evidence to conclude that heads and tails are clearly not
equally likely (α = 0.05)
(c) Compare your results for parts (a) and (b).
The p-values are identical.
Same conclusion.