Chapter 12
Inference for Distributions
and Relationships
Sec$on 12.1
Chi-Square Tests for
Goodness of Fit
Chi-Square Tests for Goodness of Fit
LEARNING TARGETS
By the end of this sec.on, you should be able to:
üSTATE appropriate hypotheses and COMPUTE the expected
counts and chi-square test sta<s<c for a chi-square test for
goodness of fit.
üSTATE and CHECK the Random, 10%, and Large Counts
condi<ons for performing a chi-square test for goodness of fit.
üCALCULATE the degrees of freedom and P-value for a chi-square
test for goodness of fit.
üPERFORM a chi-square test for goodness of fit.
üCONDUCT a follow-up analysis when the results of a chi-square
test are sta<s<cally significant.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Chi-Square Tests for Goodness of Fit
Ramón Rivera Moret
Here’s what the Consumer Affairs Department of
Mars, Inc. says about the distribu>on of color for
M&M’S® Milk Chocolate Candies produced at its
HackeDstown, New Jersey, factory:
Starnes/Tabor, The Prac)ce of Sta)s)cs
Chi-Square Tests for Goodness of Fit
Brown: 12.5%
Red: 12.5%
Yellow: 12.5%
Green: 12.5%
Orange: 25%
Blue: 25%
Ramón Rivera Moret
Here’s what the Consumer Affairs Department of
Mars, Inc. says about the distribu>on of color for
M&M’S® Milk Chocolate Candies produced at its
HackeDstown, New Jersey, factory:
Starnes/Tabor, The Prac)ce of Sta)s)cs
Sta5ng Hypotheses
Ramón Rivera Moret
Jerome’s class wondered if the
distribu>on of color in a large bag of
M&M’S Milk Chocolate Candies differed
from the claimed distribu>on.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Sta5ng Hypotheses
Here are data from a random sample
of 60 candies from that bag:
Ramón Rivera Moret
Jerome’s class wondered if the
distribu>on of color in a large bag of
M&M’S Milk Chocolate Candies differed
from the claimed distribu>on.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Sta5ng Hypotheses
Ramón Rivera Moret
H0: The distribu>on of color in the large bag
of M&M’S Milk Chocolate Candies is the
same as the claimed distribu>on.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Sta5ng Hypotheses
Ha: The distribu>on of color in the large bag of
M&M’S Milk Chocolate Candies is not the same as
the claimed distribu>on.
Ramón Rivera Moret
H0: The distribu>on of color in the large bag
of M&M’S Milk Chocolate Candies is the
same as the claimed distribu>on.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Sta5ng Hypotheses
Ha: The distribu>on of color in the large bag of
M&M’S Milk Chocolate Candies is not the same as
the claimed distribu>on.
Ramón Rivera Moret
H0: The distribu>on of color in the large bag
of M&M’S Milk Chocolate Candies is the
same as the claimed distribu>on.
We can also write the hypotheses in symbols.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Sta5ng Hypotheses
H0: The distribu>on of color in the large bag
of M&M’S Milk Chocolate Candies is the
same as the claimed distribu>on.
Ramón Rivera Moret
Ha: The distribu>on of color in the large bag of
M&M’S Milk Chocolate Candies is not the same as
the claimed distribu>on.
We can also write the hypotheses in symbols.
H0: pbrown = 0.125, pred = 0.125, pyellow = 0.125, pgreen = 0.125, porange = 0.25, pblue = 0.25
Ha: At least two of the pi’s are incorrect
where pcolor = the true propor>on of M&M’S Milk Chocolate Candies in the
large bag of that color
Starnes/Tabor, The Prac)ce of Sta)s)cs
Sta5ng Hypotheses
Ha: The distribu>on of color in the large bag of
ll
a
t
a
M&M’S Milk Chocolate Candies is not the same as
s th
t
s
e
gg
u
s
the claimed distribu>on.
t
tha wrong.
y
a
w
a
are
n
i
n
:
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o
i
N
es
u>
O
h
b
I
t
i
r
T
o
t
U the hypotheses
is
d
hyp in
d
e
We can alsoCA
write
symbols.
e
v
s iz
a>
e
n
r
h
t
e
t
o
al
yp
e
h
h
t
e
th
te pred
a
n
t
i
H0: pbrown
=
0.125,
=
0.125,
pyellow = 0.125, pgreen = 0.125, porange = 0.25, pblue = 0.25
s
s
t
n
’
o
n
>
o
or of the p ’s are incorrect
p
o
Ha: DAt eleast
two
r
i
th p
where pcolor = the true propor>on of M&M’S Milk Chocolate Candies in the
large bag of that color
Ramón Rivera Moret
H0: The distribu>on of color in the large bag
of M&M’S Milk Chocolate Candies is the
same as the claimed distribu>on.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
How many candies of each color should Jerome’s class expect to find in
their sample of 60 candies?
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
How many candies of each color should Jerome’s class expect to find in
their sample of 60 candies?
Brown: (60)(0.125) = 7.5
Red:
(60)(0.125) = 7.5
Yellow: (60)(0.125) = 7.5
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
How many candies of each color should Jerome’s class expect to find in
their sample of 60 candies?
Brown: (60)(0.125) = 7.5
Red:
(60)(0.125) = 7.5
Yellow: (60)(0.125) = 7.5
Green: (60)(0.125) = 7.5
Orange: (60)(0.25) = 15
Blue:
(60)(0.25) = 15
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
How many candies of each color should Jerome’s class expect to find in
their sample of 60 candies?
Brown: (60)(0.125) = 7.5
Red:
(60)(0.125) = 7.5
Yellow: (60)(0.125) = 7.5
Green: (60)(0.125) = 7.5
Orange: (60)(0.25) = 15
Blue:
(60)(0.25) = 15
Calcula(ng Expected Counts in a
Chi-Square Test for Goodness of Fit
The expected count for category i in the distribu5on of a categorical variable is
npi
where pi is the propor5on for category i specified by the null hypothesis.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
How likely is it that differences this large or larger would occur just by
chance in random samples of size 60 from the popula>on distribu>on
claimed by Mars, Inc.?
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
How likely is it that differences this large or larger would occur just by
chance in random samples of size 60 from the popula>on distribu>on
claimed by Mars, Inc.?
The sta>s>c we use to make the comparison is the
chi-square test sta9s9c c2.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
The chi-square test sta9s9c is a measure of how far the observed counts
are from the expected counts. The formula for the sta>s>c is
!
(𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑
𝑐𝑜𝑢𝑛𝑡
−
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑
𝑐𝑜𝑢𝑛𝑡)
𝜒! = #
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑐𝑜𝑢𝑛𝑡
where the sum is over all possible values of the categorical variable.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
The chi-square test sta9s9c is a measure of how far the observed counts
are from the expected counts. The formula for the sta>s>c is
!
(𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑
𝑐𝑜𝑢𝑛𝑡
−
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑
𝑐𝑜𝑢𝑛𝑡)
𝜒! = #
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑐𝑜𝑢𝑛𝑡
where the sum is over all possible values of the categorical variable.
For Jerome’s data, we add six terms—one for each color category:
!
!
!
!
!
!
(12
−
7.5)
(3
−
7.5)
(7
−
7.5)
(9
−
7.5)
(9
−
15)
(20
−
15)
𝜒! =
+
+
+
+
+
7.5
7.5
7.5
7.5
15
15
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
The chi-square test sta9s9c is a measure of how far the observed counts
are from the expected counts. The formula for the sta>s>c is
!
(𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑
𝑐𝑜𝑢𝑛𝑡
−
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑
𝑐𝑜𝑢𝑛𝑡)
𝜒! = #
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑐𝑜𝑢𝑛𝑡
where the sum is over all possible values of the categorical variable.
For Jerome’s data, we add six terms—one for each color category:
!
!
!
!
!
!
(12
−
7.5)
(3
−
7.5)
(7
−
7.5)
(9
−
7.5)
(9
−
15)
(20
−
15)
𝜒! =
+
+
+
+
+
7.5
7.5
7.5
7.5
15
15
= 2.7 + 2.7 + 0.03 + 0.30 + 2.4 + 1.67
= 9.8
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
Problem: Carrie made a 6-sided die in her ceramics class and rolled it 90
>mes to test if each side was equally likely to show up. The table
summarizes the outcomes of her 90 rolls.
(a) State the hypotheses that Carrie should test.
(b) Calculate the expected count for each of the possible outcomes.
(c) Calculate the value of the chi-square test sta>s>c.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
Problem: Carrie made a 6-sided die in her ceramics class and rolled it 90
>mes to test if each side was equally likely to show up. The table
summarizes the outcomes of her 90 rolls.
(a) State the hypotheses that Carrie should test.
(b) Calculate the expected count for each of the possible outcomes.
(c) Calculate the value of the chi-square test sta>s>c.
(a)
H0: The sides of Carrie’s die are equally likely to show up.
Ha: The sides of Carrie’s die are not equally likely to show up.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
Problem: Carrie made a 6-sided die in her ceramics class and rolled it 90
>mes to test if each side was equally likely to show up. The table
summarizes the outcomes of her 90 rolls.
(a) State the hypotheses that Carrie should test.
(b) Calculate the expected count for each of the possible outcomes.
(c) Calculate the value of the chi-square test sta>s>c.
(b) If H0 is true, each of the 6 sides should show up 1/6 of the time. The
expected count is 90(1/6) = 15 for each side.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Comparing Observed and Expected Counts:
The Chi-Square Test Sta<s<c
Problem: Carrie made a 6-sided die in her ceramics class and rolled it 90
>mes to test if each side was equally likely to show up. The table
summarizes the outcomes of her 90 rolls.
(a) State the hypotheses that Carrie should test.
(b) Calculate the expected count for each of the possible outcomes.
(c) Calculate the value of the chi-square test sta>s>c.
(c)
!
!
!
!
!
!
(12
−
15)
(28
−
15)
(12
−
15)
(13
−
15)
(10
−
15)
(15
−
15)
𝜒! =
+
+
+
+
+
15
15
15
15
15
15
= 0.6 + 11.27 + 0.6 + 0.27 + 1.67 +0
= 14.41
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
We used so`ware to simulate taking 1000 random samples of size 60 from
the popula>on distribu>on of M&M’S Milk Chocolate Candies given by
Mars, Inc. Here are the values of the chi-square test sta>s>c for these
1000 samples.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
We used so`ware to simulate taking 1000 random samples of size 60 from
the popula>on distribu>on of M&M’S Milk Chocolate Candies given by
Mars, Inc. Here are the values of the chi-square test sta>s>c for these
1000 samples.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
We used so`ware to simulate taking 1000 random samples of size 60 from
the popula>on distribu>on of M&M’S Milk Chocolate Candies given by
Mars, Inc. Here are the values of the chi-square test sta>s>c for these
1000 samples.
Larger values of c2 give more
convincing evidence against
H0 and in favor of Ha.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
We used so`ware to simulate taking 1000 random samples of size 60 from
the popula>on distribu>on of M&M’S Milk Chocolate Candies given by
Mars, Inc. Here are the values of the chi-square test sta>s>c for these
1000 samples.
Larger values of c2 give more
convincing evidence against
H0 and in favor of Ha.
Our es>mated P-value is
87/1000 = 0.087
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
A chi-square distribu9on is defined by a density curve that takes only
nonnega>ve values and is skewed to the right. A par>cular chi-square
distribu>on is specified by its degrees of freedom.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
A chi-square distribu9on is defined by a density curve that takes only
nonnega>ve values and is skewed to the right. A par>cular chi-square
distribu>on is specified by its degrees of freedom.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
A chi-square distribu9on is defined by a density curve that takes only
nonnega>ve values and is skewed to the right. A par>cular chi-square
distribu>on is specified by its degrees of freedom.
• As the degrees of freedom
(df) increase, the density
curves become less skewed.
• The mean of a par>cular chisquare distribu>on is equal
to its degrees of freedom.
• For df > 2, the mode (peak)
of the chi-square density
curve is at df – 2.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
There are 6 color categories for M&M’s® Milk Chocolate Candies,
so df = 6 – 1 = 5.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
There are 6 color categories for M&M’s® Milk Chocolate Candies,
so df = 6 – 1 = 5.
The P-value is the probability of gegng a c2 value of as large as or larger
than 9.8 when H0 is true.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
There are 6 color categories for M&M’s® Milk Chocolate Candies,
so df = 6 – 1 = 5.
The P-value is the probability of gegng a c2 value of as large as or larger
than 9.8 when H0 is true.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
There are 6 color categories for M&M’s® Milk Chocolate Candies,
so df = 6 – 1 = 5.
The P-value is the probability of gegng a c2 value of as large as or larger
than 9.8 when H0 is true.
The P-value for a test based on Jerome’s data is between 0.05 and 0.10.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
There are 6 color categories for M&M’s® Milk Chocolate Candies,
so df = 6 – 1 = 5.
The P-value is the probability of gegng a c2 value of as large as or larger
than 9.8 when H0 is true.
The P-value for a test based on Jerome’s data is between 0.05 and 0.10.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
There are 6 color categories for M&M’s® Milk Chocolate Candies,
so df = 6 – 1 = 5.
The P-value is the probability of gegng a c2 value of as large as or larger
than 9.8 when H0 is true.
e!
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The P-value for a test based on Jerome’s data is between 0.05 and 0.10.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
Problem: Carrie made a 6-sided die in her ceramics class and rolled it 90
>mes to test if each side was equally likely to show up. The table
summarizes the observed and expected counts. In the preceding example,
we calculated c2 = 14.41. Find the P-value using Table C. Then calculate a
more precise value using technology. Assume the condi>ons for inference
are met.
Starnes/Tabor, The Prac)ce of Sta)s)cs
The Chi-Square Distribu5ons and P-Values
Problem: Carrie made a 6-sided die in her ceramics class and rolled it 90
>mes to test if each side was equally likely to show up. The table
summarizes the observed and expected counts. In the preceding example,
we calculated c2 = 14.41. Find the P-value using Table C. Then calculate a
more precise value using technology. Assume the condi>ons for inference
are met.
df = 6 – 1 = 5
Using Table C:
the P-value is between 0.01 and 0.02
Using technology:
c2 cdf(lower:14.41, upper:10000, df:5)
= 0.0132
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
Condi(ons for Performing a Chi-Square Test
for Goodness of Fit
• Random: The data come from a random sample from the popula>on of
interest.
◦ 10%: When sampling without replacement, n < 0.10N.
• Large Counts: All expected counts are at least 5.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
Condi(ons for Performing a Chi-Square Test
for Goodness of Fit
• Random: The data come from a random sample from the popula>on of
interest.
◦ 10%: When sampling without replacement, n < 0.10N.
• Large Counts: All expected counts are at least 5.
AP® Exam Tip
When checking the Large Counts condi>on, be sure to examine the
expected counts, not the observed counts. And make sure to write and
label the expected counts on your paper or you won’t receive credit.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
The Chi-Square Test for Goodness of Fit
Suppose the condi5ons are met. To perform a test of H0: The stated distribu5on of
a categorical variable in the popula5on of interest is correct compute the chisquare test sta5s5c:
(𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑐𝑜𝑢𝑛𝑡 − 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑐𝑜𝑢𝑛𝑡)!
!
𝜒 =2
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑐𝑜𝑢𝑛𝑡
where the sum is over the k different categories. The P-value is the area to the
right of c2 under the chi-square density curve with degrees of freedom = number
of categories – 1 .
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
a hockey player’s birth month has a big influence on his chance
to make it to the highest levels of the game. Specifically,
because January 1 is the cut-off date for youth leagues in
Canada [where many Na5onal Hockey League (NHL) players
come from], players born in January will be compe5ng against
players up to 12 months younger. The older players tend to be bigger, stronger,
and more coordinated and hence get more playing 5me, more coaching, and have
a beVer chance of being successful. To see if birth date is related to success
(judged by whether a player makes it into the NHL), a random sample of 80 NHL
players from a recent season was selected and their birthdays were recorded. The
one-way table summarizes the data on birthdays for these 80 players.
Canva Pty Ltd/Alamy
Problem: In his book Outliers, Malcolm Gladwell suggests that
Do these data provide convincing evidence that the birthdays of NHL players are
not uniformly distributed across the four quarters of the year?
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
season was selected and their birthdays were recorded. The
one-way table summarizes the data on birthdays for these
80 players.
Canva Pty Ltd/Alamy
Problem: … a random sample of 80 NHL players from a recent
Do these data provide convincing evidence that the birthdays of NHL players are
not uniformly distributed across the four quarters of the year?
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
season was selected and their birthdays were recorded. The
one-way table summarizes the data on birthdays for these
80 players.
Canva Pty Ltd/Alamy
Problem: … a random sample of 80 NHL players from a recent
Do these data provide convincing evidence that the birthdays of NHL players are
not uniformly distributed across the four quarters of the year?
STATE
H0: The birthdays of all NHL players are uniformly distributed across
the four quarters of the year.
Ha: The birthdays of all NHL players are not uniformly distributed
across the four quarters of the year.
We’ll use α = 0.05.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
season was selected and their birthdays were recorded. The
one-way table summarizes the data on birthdays for these
80 players.
Canva Pty Ltd/Alamy
Problem: … a random sample of 80 NHL players from a recent
Do these data provide convincing evidence that the birthdays of NHL players are
not uniformly distributed across the four quarters of the year?
PLAN
Chi-square test for goodness of fit
• Random: The data came from a random sample of NHL players ✓
◦ 10%: We must assume that 80 is less than 10% of all NHL players ✓
• Large Counts: All expected counts = 80(1/4) = 20 ≥ 5 ✓
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
season was selected and their birthdays were recorded. The
one-way table summarizes the data on birthdays for these
80 players.
Canva Pty Ltd/Alamy
Problem: … a random sample of 80 NHL players from a recent
Do these data provide convincing evidence that the birthdays of NHL players are
not uniformly distributed across the four quarters of the year?
DO
(32 − 20)! (20 − 20)! (16 − 20)! (12 − 20)!
𝜒 =
+
+
+
= 7.2 + 0 + 0.8 + 3.2 = 11.2
20
20
20
20
!
df = 4 – 1 = 3
Using Table C: The P-value is between 0.01 and 0.02
Using technology: c2 cdf(lower:11.2, upper:10000, df:3) = 0.011
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
season was selected and their birthdays were recorded. The
one-way table summarizes the data on birthdays for these
80 players.
Canva Pty Ltd/Alamy
Problem: … a random sample of 80 NHL players from a recent
Do these data provide convincing evidence that the birthdays of NHL players are
not uniformly distributed across the four quarters of the year?
CONCLUDE
Because the P-value of 0.011 < α = 0.05, we reject H0. We have
convincing evidence that the birthdays of NHL players are not
uniformly distributed across the four quarters of the year.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
Note: When you run the chi-square test for goodness of fit on the TI-84
calculator, a list of the individual components will be produced and stored in
a list called CNTRB (for contribu>on).
Starnes/Tabor, The Prac)ce of Sta)s)cs
Carrying Out a Test
AP® Exam Tip
You can use your calculator to carry out the mechanics of a significance
test on the AP® Sta>s>cs exam. But there’s a risk involved. If you just
give the calculator answer with no work, and one or more of your values
is incorrect, you will likely get no credit for the “Do” step. We
recommend wri>ng out the first few terms of the chi-square calcula>on
followed by “. . .”. This approach might help you earn par>al credit if you
enter a number incorrectly. Be sure to name the procedure (chi-square
test for goodness of fit) and to report the test sta>s>c (c2 = 11.2),
degrees of freedom (df = 3), and P-value (0.011).
Starnes/Tabor, The Prac)ce of Sta)s)cs
Follow-up Analysis
If the sample data lead to a sta>s>cally significant result,
we can conclude that our variable has a distribu>on
different from the one stated.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Follow-up Analysis
If the sample data lead to a sta>s>cally significant result,
we can conclude that our variable has a distribu>on
different from the one stated.
To inves>gate how the distribu>on is different, start by
iden>fying the categories that contribute the most to the
chi-square sta>s>c.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Follow-up Analysis
If the sample data lead to a sta>s>cally significant result,
we can conclude that our variable has a distribu>on
different from the one stated.
To inves>gate how the distribu>on is different, start by
iden>fying the categories that contribute the most to the
chi-square sta>s>c.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Follow-up Analysis
Starnes/Tabor, The Prac)ce of Sta)s)cs
Follow-up Analysis
The two biggest contribu>ons to the chi-square sta>s>c came
from Jan–Mar and Oct–Dec.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Follow-up Analysis
The two biggest contribu>ons to the chi-square sta>s>c came
from Jan–Mar and Oct–Dec.
In October through December,
8 fewer players were born
than expected.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Follow-up Analysis
The two biggest contribu>ons to the chi-square sta>s>c came
from Jan–Mar and Oct–Dec.
In October through December,
8 fewer players were born
than expected.
In January through March,
12 more players were born
than expected.
Starnes/Tabor, The Prac)ce of Sta)s)cs
Sec5on Summary
LEARNING TARGETS
A6er this sec.on, you should be able to:
üSTATE appropriate hypotheses and COMPUTE the expected
counts and chi-square test sta<s<c for a chi-square test for
goodness of fit.
üSTATE and CHECK the Random, 10%, and Large Counts
condi<ons for performing a chi-square test for goodness of fit.
üCALCULATE the degrees of freedom and P-value for a chi-square
test for goodness of fit.
üPERFORM a chi-square test for goodness of fit.
üCONDUCT a follow-up analysis when the results of a chi-square
test are sta<s<cally significant.
Starnes/Tabor, The Prac)ce of Sta)s)cs