Chapter 12Inferences on Categorical Data
Ch 12.1 Goodness-of-Fit Test
Objective A : Goodness-of-Fit Test
Example 1:
Determine the expected counts for each outcome.
Ei   i  npi
E1  700 * 0.15  105
E 2  700 * 0.30  210
E3  700 * 0.35  245
E 4  700 * 0.20  140
Example 2:
How was E1 , E2 ,...... being calculated?
Stat --> Calculator --> Binomial --> Input the following --> Compute!
Since the grand total is (260+ 400 +280 + 50 + 10) = 1000, for X = 0, the
expected count is 1000*0.2401 = 240.1. We use the same procedure to
find the rest of E i .
(a) Determine the  2 test statistic.
2  
Oi  Ei 2
Ei
(260  240.1) 2 (400  411.6) 2 (280  264.6) 2 (50  75.6) 2 (10  8.1) 2




240.1
411.6
264.6
75.6
8.1
= 11.987
=
(b) Determine the degrees of freedom.
DF  k  1  5  1  4
(c) Use StatCrunch to determine the P  value.
Stat --> Calculator --> Chi-Square --> Standard -->
The hypothesis tests of this section are always right-tailed;
Input the following.
P  value = 0.0174
(d) Test the hypothesis at the   0.05 level of significance.
Recall: Ho: The random variable X is binomial with n = 4, p = 0.3.
Since P-value (0.0174) is less than alpha (0.05), reject Ho.
There is sufficient evidence to support the claim that the random variable X is not binomial with
n = 4, p = 0.3.
Example 3:
Use StatCrunch to perform the hypothesis testing of Example 2
at the   0.01 level of significance.
(a) Setup
Ho: The random variable X is binomial with n = 4, p = 0.3.
H1: The random variable X is not binomial with n=4, p=0.3.
(b) P  value
Open StatCrunch --> Input Observed Counts in Var1 and Expected
Counts in Var2 --> Stat --> Goodness-of-fit --> Chi-Square test -->
Select Var1 for Observed and select Var2 for Expected --> Compute!
P  value = 0.0174
(c) Conclusion
Since P-value (0.0174) is less not than alpha (0.01), we do not reject the null
hypothesis.
There is not sufficient evidence to warrant rejection of the random variable
X is binomial with n = 4, p = 0.3.
Example 4:
Total = 53+66+38+96+88+59 = 400
Ho: p1 = 0.12, p2 = 0.15, p3 = 0.12, p4 = 0.23, p5 = 0.23, p6 = 0.15
H1: At least one of the proportion is not equal to the given claim.
E1 = 0.12*400 = 48, E2 = 0.15*400 = 60, E3 = 0.12*400 = 48,
E4 = 0.23*400 = 92, E5 = 0.23*400 = 92, E6 = 0.15*400 = 60.
X
Brown
Observed
53
Expected
48
Yellow
66
60
Red
38
48
Blue
96
92
Orange
88
92
Green
59
60
Open StatCrunch --> Input Observed Counts in Var1 and Expected Counts in Var2 --> Stat -->
Goodness-of-fit --> Chi-Square test -->
Select Var1 for Observed and select Var2 for Expected --> Compute!
Since P-value (0.613) is not less than alpha, do not reject the null hypothesis.
There is not sufficient evidence to warrant rejection of the distribution claimed
by the manufacturer of M&Ms.
Ch 12.2Tests for Independence
Objective A :Tests for Independence
Example 1:
(a) Compute the expected values of each cell under the assumption of
independence.
By assuming an individual opinion and gender are independent,
(row total)(column total)
.
E
( grand total)
row1 total = 196+199 = 395
col1 total = 196+239 = 435
row2 total = 239+249 = 488
col2 total = 199+249 = 448
Grand total = 196 + 199 + 239 + 249 = 883
E11 
(395)( 435)
(395)( 448)
 194.592 E12 
 200.408
883
883
E 21 
(488)( 435)
(488)( 448)
 240.408 E 22 
 247.592
883
883
Summarize the observed counts and expected counts in a table where
the expected counts are expressed in a parenthesis.
Gender
Men
Women
Column Total
Pro Life
196 (194.592)
239 (240.408)
435
Pro Choice
199 (200.408)
249 (247.592)
448
Row Total
395
488
883 (Grand Total)
(b) Verify that the requirements for performing a chi-square test of
independence are satisfied.
(1) All expected frequencies are greater than or equal to 1.
"True"
(2) No more than 20% of the expected frequencies are less than 5. "True"
(c) Determine the  2 test statistic.
2  
Oi  Ei 2
Ei
(196  194.592) 2 (199  200.408) 2 (239  240.408) 2 (249  247.592) 2



194.592
200.408
240.408
247.592
= 0.0374
=
(d) Determine the degrees of freedom.
DF = (r - 1)(c - 1) = (2 - 1)(2 - 1) = 1
(e) Use StatCrunch to determine the P  value.
Stat --> Calculator --> Chi-Square --> Standard -->
The hypothesis tests of this section are always right-tailed;
Input the following.
(f) Test whether an individual's opinion regarding abortion is independent
of gender at the   0.10 level of significance.
Ho: An individual opinion regarding abortion is independent of gender.
H1: An individual opinion regarding abortion is dependent of gender.
Since P-value (0.847) is not less than alpha (0.10), do not reject the null
hypothesis. An individual opinion regarding abortion is independent of
gender.
Example 2:
Use StatCrunch to redo example 1 of testing whether an individual's
opinion regarding abortion is independent of gender at the   0.10
level of significance.
(a) Setup
Ho: An individual opinion regarding abortion is independent of gender.
H1: An individual opinion regarding abortion is dependent of gender.
(b) P  value
Open StatCrunch --> Input the data (see below)
Stat --> Tables --> Contingency --> With Summary --> Under Select Column(s),
click Pro Life and Ctrl click Pro choice --> Under Row Labels, select Gender -->
Under Display, select Expected count --> Compute!
StatCrunch Results:
(c) Conclusion
Since P-value (0.8489) is not less than alpha (0.05), do not reject the null
hypothesis. An individual opinion regarding abortion is independent of
gender.
Example 3:
Test whether prenatal care and the wantedness of pregnancy are independent
at the   0.05 level of significance.
(a) Setup
Ho: Prenatal care and wantedness of pregnancy are independent.
H1: Prenatal care and wantedness of pregnancy are dependent.
(b) P  value
Open StatCrunch --> Input the data (see below)
Stat --> Tables --> Contingency --> With Summary --> Under Select Column(s),
click Less than 3 months, 3 to 5 Months, and More Than 5 Months (use the
Ctrl) --> Under Row Labels, select Wantedness of Pregnancy --> Under
Display, select Expected count --> Compute!
StatCrunch Results:
(c) Conclusion
Since P-value (0.0003) is less than alpha (0.05), reject the null hypothesis.
Prenatal care and wantedness of pregnancy are dependent.
Ch 12.3Comparing Three or More Means
- One-Way Analysis of Variance, ANOVA
Objective A :One-Way ANOVA Test
(Supplemental Materials)
Example 1:
(a) Ho: 1  2  3
H1: At least one of the means is different from the others.
(b)
1.
2.
3.
4.
There are 3 simple random samples.
The 3 samples are independent of each other.
Normal probability plots indicate that the sample data come from a normal population.
If the largest sample standard deviation is no more than twice the smallest sample
standard deviation, we can assume the populations have the same variance.
Stat  Summary Stats  Columns  Input the following, then click Compute!
StatCrunch Results:
0.109 is not larger than twice (0.063)  0.109 is not larger than 0.126. Thus we can assume the
populations have the same variance.
(C) Test 1  2  3
Stat  ANOVA  One Way  Select all three columns (Simple, Go/No Go, Choice)  Compute!
StatCrunch Results:
Note: We could have obtained each sample standard deviation using ANOVA
instead of using Summary Stats.
Let’s try to understand the meaning of each output.
Column statistics
Std. Error for Simple  St. Dev. of Simple divided by the square root of n  sx1 
 sx1 
0.063750817
= 0.026026162
6
s1
n1
Std. Error for Go/No Go  sx 
2
Std. Error for Choice  sx 
3
s2
0.10996803

 0.044894258
n2
6
s3
0.091593486

 0.037392884
n3
6
ANOVA table
SS(Columns) = SS(treatment)
= SS(between samples/groups)
= 0.052888444  Sum of square due to treatment which is a measure of the
variation between the sample means.
MS(Columns) = MS(treatment)
= SS(treatment)/DF of treatment
= 0.052888444/2 = 0.026444222 The mean square due to treatment
SS(Error) = SS(within samples/groups)
= 0.1227325 Sum of square due to error which is a measure of the variation within
each treatment group.
MS(Error) = SS(Error)/DF of error
= 0.1227325/15 = 0.0081821667The mean square due to error
SS(Total) = SS(treatment) + SS(error)
= 0.052888444 + 0.1227325
= 0.17562094
F-Stat = MS(treatment)/MS(error)
= 0.026444222/0.0081821667
= 3.2319339
P-value = a right-tailed area with F-Stat of 3.2319339 (F Distribution)
= 0.0681
Since P-value (0.0681) is not less than alpha (0.05), do not reject Ho: 1  2  3 .
There is not sufficient evidence to support the claim that at least one population mean is
different from the others.
(d) Draw boxplots of the three stimuli.
Stat  Graph  Boxplots  Select all three columns  Check draw boxes horizontally Compute!
The visual display supports 1  2 but  3 = 1  2 is not as obvious.
Ch 12.4 Two - Way Analysis of Variance
(Supplemental Materials)
Objective A : Two - Way ANOVA Test
An example of interaction effect is sleeping pills and alcohol. They are usually
not fatal when taken alone, but can be fatal when combined.
Example 1:
Example 2: Redo Example 1 using StatCrunch.
(a) Verify that the largest sample standard deviation is no more than twice the smallest standard
deviation. If this is true, we can assume the populations have the same variance.
Input the data in StatCrunch.
Stat  Summary Stats  Columns  Input the following, then click Compute!
StatCrunch Results:
The largest Std. dev. is 2.6457513 and the smallest Std. dev. is 1.5275252.
Since 2.6457513 is not larger than twice(1.5275252), we can assume the populations have the same
variance.
(b) Test whether there is an interaction effect between the drug dosage and age.
Inputting the data is a bit tricky for a two-way ANOVA analysis.
Use cut and paste or type the data in to produce three columns.
 Response (HDL), Row Factor (Age), Col Factor (Dosage)  see below.
Stat  ANOVA  Two Way  Input the following  Compute!
StatCrunch Results:
Hit > for the next page of the StatCrunch outputs:
The above graph is called the interaction plots. We look at the level of parallelism among
the lines. Since the lines are roughly parallel, we conclude there is no interaction between
age and drug dosages.
(c) If there is no interaction between age and drug dosages, determine whether there
is sufficent evidence to conclude that the mean increase in HDL cholesterol is different
(i) among each drug dosage group, (ii) for each age group.
(i)
Since the P-value (<0.0001) is less than alpha (0.05), reject Ho. There is sufficient
evidence to support the claim that HDL cholesterol changes with dosage.
Hit > for the next page of the StatCrunch outputs:
The above graph indicates the mean of HDL increases as the drug dosage increases.
(ii)
Since the P-value (0.0838) is not less than alpha (0.05), do not reject Ho. There is
not sufficient evidence to support the claim that HDL cholesterol changes with age.