Non-parametric Statistical Tests and Effect Sizes

This assignment focuses on applying non-parametric statistical tests and calculating effect sizes. It includes exercises on rank sum and chi-squared tests, Spearman's correlation coefficient, and various measures of effect size like Cohen's d and omega squared. The assignment also features case studies requiring application of these concepts to analyze real-world data.

Exercise 11:
1. An experiment is conducted to test the claim that James Bond can taste the difference
between a Martini that is shaken and one that is stirred. What is the null hypothesis?
Solution
Null hypothesis (H0) is that James Bond cannot taste the difference between a Martini
that is shaken and one that is stirred.
2. The following explanation is incorrect. What three words should be added to make it
correct?
The probability value is the probability of obtaining a statistic as different (or more
different/extreme) from the parameter specified in the null hypothesis as the statistic
obtained in the experiment. The probability value is computed assuming that the null
hypothesis is true.
3. Why do experimenters test hypotheses they think are false?
Solution
 To establish directionality.
 To place the burden of proof on the alternative
4. State the null hypothesis for:
a. An experiment testing whether echinacea decreases the length of colds.
Solution
H0: Echinacea does not have an effect on the length of colds
b. A correlational study on the relationship between brain size and intelligence.
Solution
H0: Brain size is not correlated with intelligence (H0 :r=0¿

c. An investigation of whether a self-proclaimed psychic can predict the outcome of a
coin flip.
Solution
H0: Self-proclaimed psychic cannot predict the outcome of a coin flip
d. A study comparing a drug with a placebo on the amount of pain relief. (A one tailed
test was used).
Solution
H0: There is no difference in the amount of pain relief by placebo and drug.
5. Assume the null hypothesis is that μ = 50 and that the graph shown below is the sampling
distribution of the mean (M). Would a sample value of M= 60 be significant in a two-
tailed test at the .05 level? Roughly what value of M would be needed to be significant?
Solution
Yes a sample of M = 60 would be significant in a two-tailed test at the 0.05 level.
Roughly the M value should be less than or equal to 40 or greater than or equal to 60.
6. A researcher develops a new theory that predicts that vegetarians will have more of a
particular vitamin in their blood than non-vegetarians. An experiment is conducted and
vegetarians do have more of the vitamin, but the difference is not significant. The
probability value is 0.13. Should the experimenter's confidence in the theory increase,
decrease, or stay the same?
Solution
His confidence should increase that the null hypothesis is false

7. A researcher hypothesizes that the lowering in cholesterol associated with weight loss is
really due to exercise. To test this, the researcher carefully controls for exercise while
comparing the cholesterol levels of a group of subjects who lose weight by dieting with a
control group that does not diet. The difference between groups in cholesterol is not
significant. Can the researcher claim that weight loss has no effect?
Solution
The researcher cannot claim that weight loss has no effect, only that the data do not
support the hypothesis.
8. A significance test is performed and p = .20. Why can't the experimenter claim that the
probability that the null hypothesis is true is .20?
Solution
The p-value just indicates the probability of obtaining a particular statistic only (where in
this case, a proportion) from the sample data under the assumption that the null
hypothesis is true. This is NEVER the same as saying the probability of the null
hypothesis being true is.20.
9. For a drug to be approved by the FDA, the drug must be shown to be safe and effective.
If the drug is significantly more effective than a placebo, then the drug is deemed
effective. What do you know about the effectiveness of a drug once it has been approved
by the FDA (assuming that there has not been a Type I error)?
Solution

10. When is it valid to use a one-tailed test? What is the advantage of a one-tailed test? Give
an example of a null hypothesis that would be tested by a one-tailed test.
Solution
Advantage of a one-tailed test
The advantage of adopting the one-tailed test is an improvement in power to reject the
null hypothesis if the null hypothesis is truly false.
Example;
Null hypothesis (H0): The average weight of students is less than 60 kilograms (i.e.
H0 : μ<60 ¿ .
11. Distinguish between probability value and significance level.
Solution
The probability value (also called the p-value) is the probability of the observed result
found in your research study of occurring (or an even more extreme result occurring),
under the assumption that the null hypothesis is true (i.e., if the null were true). On the
other hand, the significance level (also called the alpha level) is the cutoff value the
researcher selects and then uses to decide when to reject the null hypothesis.
12. Suppose a study was conducted on the effectiveness of a class on "How to take tests."
The SAT scores of an experimental group and a control group were compared. (There
were 100 subjects in each group.) The mean score of the experimental group was 503 and

the mean score of the control group was 499. The difference between means was found to
be significant, p = .037. What do you conclude about the effectiveness of the class?
Solution
If you set alpha =.05: There is sufficient evidence to reject the null hypothesis that there
is no difference between the groups, since p<.05. There is a significant difference
between the mean score of the groups.
13. Is it more conservative to use an alpha level of .01 or an alpha level of .05? Would beta
be higher for an alpha of .05 or for an alpha of .01?
Solution
Choosing a lower significance level, alpha, is more conservative in that you are less
likely to reject the null hypothesis when it is true. Generally, in biostatistics applications,
making a type I error (e.g. declaring a new treatment beneficial when in truth it is not) is
more serious than making a type II error (e.g. failing to show the benefit of a new
treatment, even though it has some benefit). There is a trade-off between error rates--
lower alpha will increase beta.
14. Why is "Ho: "M1 = M2" not a proper null hypothesis?
Solution
The null hypothesis to be formulated should state that there is NO significant difference
between M1 and M2

15. An experimenter expects an effect to come out in a certain direction. Is this sufficient
basis for using a one-tailed test? Why or why not?
Solution
Not really. We need not to care about effects in the opposite direction. Just because we
think something will go one way or another is different.
16. How do the Type I and Type II error rates of one-tailed and two-tailed tests differ?
Solution
One-tailed tests have lower Type II error rates and more power than do two-tailed
tests
17. A two-tailed probability is .03. What is the one-tailed probability if the effect were in the
specified direction? What would it be if the effect were in the other direction?
Solution
The one-tailed probability if the effect were in the specified direction would be 0.015.
The probability would however be 0.985 if the effect were in the other direction.
18. You choose an alpha level of .01 and then analyze your data.
(a) What is the probability that you will make a Type I error given that the
null hypothesis is true?
Solution

The probability of type I error is actually alpha given that the null
hypothesis is true so it is .01
(b) What is the probability that you will make a Type I error given that the
null hypothesis is false?
Solution
When null hypothesis is false, it is impossible to make a type I error. It
means probability that you will make a type I error given the null
hypothesis is false is zero
19. Why doesn't it make sense to test the hypothesis that the sample mean is 42?
Solution
Hypothesis tests must be about parameters, not sample statistics
20. True/false: It is easier to reject the null hypothesis if the researcher uses a smaller
alpha (α) level.
Solution
False
21. True/false: You are more likely to make a Type I error when using a small sample than
when using a large sample.
Solution

False
22. True/false: You accept the alternative hypothesis when you reject the null hypothesis.
Solution
True
23. True/false: You do not accept the null hypothesis when you fail to reject it.
Solution
False
24. True/false: A researcher risks making a Type I error any time the null hypothesis is
rejected.
Solution
True

Exercise 12:
1. The scores of a random sample of 8 students on a physics test are as follows: 60, 62, 67,
69, 70, 72, 75, and 78.
a. Test to see if the sample mean is significantly different from 65 at the .05 level.
Report the t and p values.
Solution
t-value=1.911181
p-value=0.076674
b. The researcher realizes that she accidentally recorded the score that should have been
76 as 67. Are these corrected scores significantly different from 65 at the .05 level?
Solution
Yes the corrected scores makes the scores to be significantly different from 65 at the
0.05 level (p = 0.038)
2. Report the t and p values.
Solution
t-value = 5.0186
p-value = 0.0007

3. The scores on a (hypothetical) vocabulary test of a group of 20 year olds and a group of
60 year olds are shown below.
a. Test the mean difference for significance using the .05 level.
Solution
t-Test: Two-Sample Assuming Equal Variances
20 yr
olds
60 yr
olds
Mean 19.22222 25.375
Variance 30.94444 23.125
Observations 9 8
Pooled Variance 27.29537
Hypothesized Mean
Difference 0
df 15
t Stat -2.42364
P(T<=t) one-tail 0.014238
t Critical one-tail 1.75305
P(T<=t) two-tail 0.028476
t Critical two-tail 2.13145
As can be seen, the p-value is less than 0.05 level, we thus reject the null hypothesis
and conclude that there is significant difference in the scores of the 20 years old and
60 years old.
b. List the assumptions made in computing your answer.
Solution
Normality of the data; we assume that the data are normally distributed
Independence of the variables; the two variables 20 year old and 60 year old are
independent of each other.
The two populations have the same variance

4. The sampling distribution of a statistic is normally distributed with an estimated standard
error of 12 (df = 20).
(a) What is the probability that you would have gotten a mean of 107 (or more extreme)
if the population parameter were 100? Is this probability significant at the .05 level
(two-tailed)?
Solution
z= 107−100
12 = 7
12 =0.5833
Probability= 0.5597
This probability is insignificant at the .05 level (two-tailed)
(b) What is the probability that you would have gotten a mean of 95 or less (one-tailed)?
Is this probability significant at the .05 level? You may want to use the t Distribution
calculator for this problem.
Solution
z= 95−100
12 =−5
12 =−0.41667
Probability= 0.3385
This probability is insignificant at the .05 level (one-tailed)
5. How do you decide whether to use an independent groups t test or a correlated t test (test
of dependent means)?
Solution
Check whether the groups are related or not. If the groups are related then a paired t-test
could be used and if they are unrelated then independent t test is the most ideal.

6. An experiment compared the ability of three groups of subjects to remember briefly-
presented chess positions.
a. Using the Tukey HSD procedure, determine which groups are significantly different
from each other at the .05 level.
Solution
Multiple Comparisons
Dependent Variable: VAR00001
Tukey HSD
(I) VAR00002 (J) VAR00002 Mean Difference
(I-J)
Std. Error Sig. 95% Confidence Interval
Lower Bound Upper Bound
Non-player Beginner -13.75000* 5.09978 .031 -26.3945 -1.1055
Player -30.85000* 5.09978 .000 -43.4945 -18.2055
Beginner Non-player 13.75000* 5.09978 .031 1.1055 26.3945
Player -17.10000* 5.09978 .007 -29.7445 -4.4555
Player Non-player 30.85000* 5.09978 .000 18.2055 43.4945
Beginner 17.10000* 5.09978 .007 4.4555 29.7445
*. The mean difference is significant at the 0.05 level.
All the three groups are significantly different from each other in terms of the average
scores.
b. Now compare each pair of groups using t-tests. Make sure to control for the
familywise error rate (at 0.05) by using the Bonferroni correction. Specify the alpha
level you used.
Solution
t-Test: Two-Sample Assuming Equal Variances
Non-
player
Beginn
ers
Mean 33.04 46.79

Variance
64.533
78
81.552
11
Observations 10 10
Pooled Variance
73.042
94
Hypothesized Mean Difference 0
df 18
t Stat
-
3.5974
8
P(T<=t) one-tail
0.0010
29
t Critical one-tail
1.7340
64
P(T<=t) two-tail
0.0020
59
t Critical two-tail
2.1009
22
t-Test: Two-Sample Assuming Equal Variances
Non-
player
Tournam
ent
players
Mean 33.04 63.89
Variance
64.533
78 244.0299
Observations 10 10
Pooled Variance
154.28
18
Hypothesized Mean
Difference 0
df 18
t Stat -5.5537
P(T<=t) one-tail
1.42E-
05
t Critical one-tail
1.7340
64
P(T<=t) two-tail
2.85E-
05
t Critical two-tail
2.1009
22
t-Test: Two-Sample Assuming Equal

Variances
Begin
ners
Tourna
ment
players
Mean 46.79 63.89
Variance
81.55
211
244.029
9
Observations 10 10
Pooled Variance
162.7
91
Hypothesized Mean
Difference 0
df 18
t Stat
-
2.996
86
P(T<=t) one-tail
0.003
869
t Critical one-tail
1.734
064
P(T<=t) two-tail
0.007
738
t Critical two-tail
2.100
922
As can be seen, the tests further confirms the groups are significantly different from one another.
7. Below are data showing the results of six subjects on a memory test. The three scores per
subject are their scores on three trials (a, b, and c) of a memory task. Are the subjects get-
ting better each trial? Test the linear effect of trial for the data.
a. Compute L for each subject using the contrast weights -1, 0, and 1. That is, compute (-1)
(a) + (0)(b) + (1)(c) for each subject.
Solution
L1 -1*4+0*6+1*7=3
L2 -1*3+0*7+1*8=5
L3 -1*2+0*8+1*5=3
L4 -1*1+0*4+1*7=6
L5 -1*4+0*6+1*9=5
L6 -1*2+0*4+1*2=0

b. Compute a one-sample t-test on this column (with the L values for each subject) you
created.
Solution
M =sample mean= 3+5+3+ 6+5+0
6 =3.667
Standard error of mean= 2.160
√6 =.8819
t= M
Sm = 3.667
0.8819 =4.158
probability of two tailed test is .0088
8. Participants threw darts at a target. In one condition, they used their preferred hand; in the
other condition, they used their other hand. All subjects performed in both conditions (the
order of conditions was counterbalanced).
a. Which kind of t-test should be used?
Solution
Related sample t-test
b. Calculate the two-tailed t and p values using this t test.
Solution
t-Test: Paired Two Sample for Means
prefer
red
non-
prefer
red
Mean 10.6 8.6
Variance 5.3 1.3
Observations 5 5
Pearson Correlation -
0.076

19
Hypothesized Mean
Difference 0
df 4
t Stat
1.690
309
P(T<=t) one-tail
0.083
116
t Critical one-tail
2.131
847
P(T<=t) two-tail
0.166
233
t Critical two-tail
2.776
445
The t-value is 1.690 while the p-value is 0.1662; the p-value is insignificant
c. Calculate the one-tailed t and p values using this t test.
Solution
t-Test: Paired Two Sample for Means
prefer
red
non-
prefer
red
Mean 10.6 8.6
Variance 5.3 1.3
Observations 5 5
Pearson Correlation
-
0.076
19
Hypothesized Mean
Difference 0
df 4
t Stat
1.690
309
P(T<=t) one-tail
0.083
116
t Critical one-tail
2.131
847
P(T<=t) two-tail
0.166
233
t Critical two-tail
2.776
445

The t-value is 1.690 while the p-value is 0.083
9. Assume the data in the previous problem were collected using two different groups of
subjects: One group used their preferred hand and the other group used their non-
preferred hand. Analyze the data and compare the results to those for the previous
problem.
Solution
t-Test: Two-Sample Assuming Equal
Variances
prefer
red
non-
prefer
red
Mean 10.6 8.6
Variance 5.3 1.3
Observations 5 5
Pooled Variance 3.3
Hypothesized Mean
Difference 0
df 8
t Stat
1.740
777
P(T<=t) one-tail
0.059
951
t Critical one-tail
1.859
548
P(T<=t) two-tail
0.119
903
t Critical two-tail
2.306
004
10. You have 4 means, and you want to compare each mean to every other mean.
(a) How many tests total are you going to compute?
Solution
You will have to compute 6 tests

(b) What would be the chance of making at least one Type I error if the Type I error for each
test was .05 and the tests were independent?
Solution
The probability will definitely be larger than 0.05 since as you compute more means then
you increase the risk of committing a type 1 error. The value will be approximately 0.3
(c) Are the tests independent and how does independence/non-independence affect the
probability in (b).
Solution
The tests are not independent. The probability is lower than it would be for the same
number of independent comparisons.
11. In an experiment, participants were divided into 4 groups. There were 20 participants in
each group, so the degrees of freedom (error) for this study was 80 - 4 = 76. Tukey’s
HSD test was performed on the data.
(a) Calculate the p-value for each pair based on the Q value given below. You will want
to use the Studentized Range Calculator.
Solution
Using Studentized range calculator we obtained the p-values as follows;
Comparison group P-value
A-B 0.085
A-C 0.043
A-D 0.017
B-C 0.627
B-D 0.036
C-D 0.051
(b) Which differences are significant at the .05 level?
Solution

The significant differences at the 0.05 level are;
Comparison group P-value Significant or not
A-C 0.043 Significant
A-D 0.017 Significant
B-D 0.036 Significant
12. If you have 5 groups in your study, why shouldn’t you just compute a t test of each group
mean with each other group mean?
Solution
It is time consuming and tedious to compute several tests. Each group would require a
test which is time consuming and tedious
13. You are conducting a study to see if students do better when they study all at once or in
intervals. The other group of 12 participants took a test after studying for three twenty
minute sessions. The first group had a mean score of 75 and a variance of 120. The
second group had a mean score of 86 and a variance of 100.
a. What is the calculated t value? Are the mean test scores of these two groups significantly
different at the .05 level?
Solution
t= X1−X2
√ ( n1 −1 ) s1
2+ ( n2−1 ) s2
2
n1 +n2−2 ( 1
n1
+ 1
n2 ) = 75−86
√ ( 12−1 ) 120+ ( 12−1 ) 100
12+12−2 ( 1
12 + 1
12 ) =−2.569
The t-value computed is greater than the critical t-value thus we reject the null hypothesis
and conclude that the mean test scores of these two groups are significantly different at
the .05 level.

b. What would the t value be if there were only 6 participants in each group? Would the
scores be significant at the .05 level?
Solution
¿ 75−86
√ ( 6−1 ) 120+ ( 6−1 ) 100
6+ 6−2 ( 1
6 + 1
6 ) =−1.817
The t-value computed is less than the critical t-value thus we fail to reject the null
hypothesis and conclude that the mean test scores of these two groups are not
significantly different at the .05 level.
14. A new test was designed to have a mean of 80 and a standard deviation of 10. A random
sample of 20 students at your school take the test, and the mean score turns out to be 85.
Does this score differ significantly from 80?
Solution
t= X−μ
σ / √ n = 85−80
10 / √ 20 =2.236
The t-value computed is greater than the critical t-value thus we reject the null hypothesis
and conclude that the score differ significantly from 80 at the .05 level.
15. You perform a one-sample t test and calculate a t statistic of 3.0. The mean of your
sample was 1.3 and the standard deviation was 2.6. How many participants were used in
this study?
Solution
3=1.3−1.0
2.6 / √ n
√ n= 7.8
0.3
n=26

16. True/false: The contrasts (-3, 1 1 1) and (0, 0 , -1, 1) are orthogonal.
Solution
True
17. True/false: If you are making 4 comparisons between means, then based on the
Bonferroni correction, you should use an alpha level of .01 for each test.
Solution
False
18. True/false: Correlated t tests almost always have greater power than independent t tests.
Solution
True
19. True/false: The graph below represents a violation of the homogeneity of variance
assumption.
Solution
False
20. True/False
Solution
True

Exercise 13: Power
1. Define power in your own words.
Solution
Power is the probability of rejecting a false null hypothesis. The probability of failing to
reject a false null hypothesis is often referred to as β. Therefore power can be defined as:
power = 1 - β.
2. List 3 measures one can take to increase the power of an experiment. Explain why your
measures result in greater power.
Solution
Three measures that can be taken to increase the power of an experiment include;
 Increase alpha
 Conduct a one-tailed test
 Increase the effect size
 Decrease random error
 Increase sample size
3. Population 1 mean = 36
Population 2 mean = 45
Both population variances are 10.
What is the probability that a t test will find a significant difference between means at the
0.05 level? Give results for both one- and two-tailed tests. Hint: the power of a one-tailed
test at 0.05 level is the power of a two-tailed test at 0.10.
Solution
We compute Z as follows;

Z= μ1−μ2
σ
Z= 45−36
10 = 9
10 =0.9
For one-tailed we have
or P(Z > 0.9) = 0.1841
Power=1−0.841=0.8159
For two-tailed we have;
P((|Z| > 0.9) = 0.3681
Power=1−0.3681=0.6319
4. Ranking
Population 1
Mean
N Population 2
Mean
Standard
Deviation
A 29 20 43 12
B 34 15 40 6
C 105 24 50 27
D 170 2 120 10
We compute Z as follows;
For a
Z= μ1−μ2
σ / √n = 43−29
12 =1.16667
Power=1−Φ ( 5.2175 ) =1−0.2433
Power=1−0.2433=0.7567
For b
Z= μ1−μ2
σ / √ n = 40−34
6 =1

Power=1−Φ ( 1.000 )=1−0.3173
Power=1−0.3173=0.6827
For c
Z= μ1−μ2
σ / √n =50−105
27 =−2.03704
Power=1−Φ ( −2.03704 ) =1−0.0416=0.9584
Power=1−0.0416=0.9584
For d
Z= μ1−μ2
σ / √n =120−170
10 =−5
Power=1−Φ (−5 )=1−2
Power=1−0.0001=0.9999
Power Rank Population 1
Mean
N Population 2
Mean
Standard
Deviation
D 0.9999 1 170 2 120 10
C 0.9584 2 105 24 50 27
A 0.7567 3 29 20 43 12
B 0.6827 4 34 15 40 6
5. Tossing a coin
(a) What is the probability that Alan will be able convince Ken that his coin has special
powers by finding a p value below 0.05 (one tailed). Use the Binomial Calculator (and
some trial and error)
Solution

P = 0.05 corresponds to z = 1.65
hypothetical mean of unbiased coin: 100(0.5) = 50
hypothetical variance of unbiased coin: 100(0.5)(0.5) = 25
hypothetical std of unbiased coin: sqrt(25) = 5
Mean of heads appearing 60%: 100(0.6) = 60
Z= X−μ
σ = 60−50
5 =10
5 =2
P ( Z <2 )=0.9772
(b) If Ken told Alan to flip the coin only 20 times, what is the probability that Alan will not
be able to convince Ken (by failing to reject the null hypothesis at the 0.05 level)?
Solution
Hypothetical mean of unbiased coin: 20(0.5) = 10
hypothetical variance of unbiased coin: 20(0.5)(0.5) = 5
hypothetical std of unbiased coin: sqrt(5) = √5
Mean of heads appearing 60%: 20(0.6) = 12
Z= X−μ
σ = 12−10
√5 = 2
√5 =0.894427
P ( Z <0.894427 )=0.8145

Exercise 14:
1. What is the null hypothesis tested by analysis of variance?
Solution
The null hypothesis tested in analysis of variance (ANOVA) is that there is no difference
in means
2. What are the assumptions of between-subjects analysis of variance?
Solution
 One of the assumptions is on the equal variance also known as homogeneity of
variance.
 The data needs to come from a population that is normally distributed.
 The factors need to be independent from each other. This particular assumption
requires that each and every subject provide only one value.
3. What is a between-subjects variable?
Solution
A between-subjects variable is a dependent variable (response variable) in which
different groups of subjects are used for each level of the variable.
4. Why not just compute t-tests among all pairs of means instead computing an analysis of
variance?
Solution
When computing t-tests among all pair of means, it becomes more and more lengthier as
the number of groups increases, as such it requires several steps to compare the two
means; this is tedious and time consuming. ANOVA on the other hand puts all the data
into one number (F) and easily gives one P for testing the null hypothesis.

5. What is the difference between "N" and "n"?
Solution
“N” is the total number observations while “n” is the number of observations in each
group.
6. How is it that estimates of variance can be used to test a hypothesis about means?
Solution
The test of a hypothesis about means is based on two estimates of population variance:
the First is the mean square error that is based on the difference among scores within
groups and estimates the variance; the second is the mean square between that is based on
differences among the sample means
7. Explain why the variance of the sample means has to be multiplied by "n" in the
computation of MSB.
Solution
Since MSB is an estimate of population variance based on the sample means multiplying
by “n” counters the property of the sampling distribution of the mean where its variances
decreases as “n” increases
8. What kind of skew does the F distribution have?
Solution
Positive
9. When do MSB and MSE estimate the same quantity?
Solution
When population means are equal.

10. If an experiment is conducted with 6 conditions and 5 subjects in each condition, what
are dfn and dfe?
Solution
d f n=5−1=4.
d f e=6 x 5−5=25 ,∨5 (6−1)=25
11. How is the shape of the F distribution affected by the degrees of freedom?
Solution
Greater degrees of freedom reduce the skew of the F distribution
12. What are the two components of the total sum of squares in a one-factor between-subjects
design?
Solution
The sum of squares for condition and the sum of squares error
13. How is the mean square computed from the sum of squares?
Solution
The mean square is computed by dividing the sum of squares by the degrees of freedom.
14. An experimenter is interested in the effects of two independent variables on self-esteem.
What is better about conducting a factorial experiment than conducting two separate
experiments, one for each independent variable?
Solution
A factorial experiment can be used to detect interactions between the independent
variables as they relate to self-esteem.

15. An experiment is conducted on the effect of age and treatment condition (experimental
versus control) on reading speed. Which statistical term (main effect, simple effect,
interaction, specific comparison) applies to each of the descriptions of effects.
a. The effect of the treatment was larger for 15-year olds than it was for 5- or 10-year
olds.
Solution
Interaction
b. Overall, subjects in the treatment condition performed faster than subjects in the
control condition.
Solution
Main effects
c. The difference between the 10- and 15-year olds was significant under the treatment
condition.
Solution
Simple effect
d. The difference between the 15- year olds and the average of the 5- and 10-year olds
was significant.
Solution
Specific comparison
e. As they grow older, children read faster.
Solution
Main effect

16. An A(3) x B(4) factorial design with 6 subjects in each group is analyzed. Give the source and
degrees of freedom columns of the analysis of variance summary table.
Solution
Source df
A 3-1=2
B 4-1=3
AB 2*3=6
Error 3*4(6-1)=60
Total (3*4*6)-1=71
17. The following data are from a hypothetical study on the effects of age and time on scores on a
test of reading comprehension. Compute the analysis of variance summary table.
12- year -olds 16 -year-olds
30 minutes 66
68
59
72
46
74
71
67
82
76
60 minutes 69
61
69
73
61
92
95
92
98
94
Solution
Descriptive Statistics
Dependent Variable: Test Scores
Time Age Mean Std. Deviation N
30 minutes
12- year -olds 62.2000 10.20784 5
16- year -olds 74.0000 5.61249 5
Total 68.1000 9.94932 10
60 minutes
12- year -olds 66.6000 5.36656 5
16- year -olds 94.2000 2.48998 5
Total 80.4000 15.07168 10
Total
12- year -olds 64.4000 8.03050 10
16- year -olds 84.1000 11.40614 10
Total 74.2500 13.93925 20

Tests of Between-Subjects Effects
Dependent Variable: Test Scores
Source Type III Sum of
Squares
df Mean Square F Sig.
Corrected Model 3008.950a 3 1002.983 23.503 .000
Intercept 110261.250 1 110261.250 2583.743 .000
Time 756.450 1 756.450 17.726 .001
Age 1940.450 1 1940.450 45.470 .000
Time * Age 312.050 1 312.050 7.312 .016
Error 682.800 16 42.675
Total 113953.000 20
Corrected Total 3691.750 19
a. R Squared = .815 (Adjusted R Squared = .780)
"Time*Age” have a statistically significant effect on the dependent variable, "Test scores" (p =
0.016). We can see from the table above that there was statistically significant difference in mean
test scores between 30 minutes and 60 minutes (p = .001), similarly, there was a statistically
significant differences between those aged 12 years old and those aged 16 years old (p = .000).

18. Define "Three-way interaction"
Solution
A three-way interaction means that one, or more, two-way interactions differ across the
levels of a third variable.
19. Define interaction in terms of simple effects.
Solution
An interaction effect is the simultaneous effect of two or more independent variables on
at least one dependent variable in which their joint effect is significantly greater (or
significantly less) than the sum of the parts.
20. Plot an interaction for an A(2) x B(2) design in which the effect of B is greater at A1 than
it is at A2. The dependent variable is "Number correct." Make sure to label both axes.
Solution
21. Following are two graphs of population means for 2 x 3 designs. For each graph, indicate
which effect(s) (A, B, or A x B) are
nonzero.http://onlinestatbook.com/2/analysis_of_variance/graphics/q8ch13.GIF
Solution
Graph a: A x B is nonzero
Graph b: A and B are nonzero
22. The following data are from an A(2) x B(4) factorial design.
B1 B2 B3 B4
A1 1
3
4
2
2
4
3
4
2
4
5
6

5 5 6 8
A2
1
1
2
2
2
3
2
4
4
6
7
8
8
9
9
8
a. Compute an analysis of variance.
Solution
Tests of Between-Subjects Effects
Dependent Variable: Scores
Source Type III Sum of
Squares
df Mean Square F Sig.
Corrected Model 110.750a 3 36.917 12.801 .000
Intercept 612.500 1 612.500 212.384 .000
A .000 0 . . .
B 110.750 3 36.917 12.801 .000
A * B .000 0 . . .
Error 80.750 28 2.884
Total 804.000 32
Corrected Total 191.500 31
a. R Squared = .578 (Adjusted R Squared = .533)
b. Test differences among the four levels of B using the Bonferroni correction.
Solution
Multiple Comparisons
Dependent Variable: Scores
Bonferroni
(I) B (J) B Mean Difference
(I-J)
Std. Error Sig. 95% Confidence Interval
Lower Bound Upper Bound
B1
B2 -.6250 .84911 1.000 -3.0356 1.7856
B3 -2.6250* .84911 .027 -5.0356 -.2144
B4 -4.7500* .84911 .000 -7.1606 -2.3394
B2 B1 .6250 .84911 1.000 -1.7856 3.0356

B3 -2.0000 .84911 .154 -4.4106 .4106
B4 -4.1250* .84911 .000 -6.5356 -1.7144
B3
B1 2.6250* .84911 .027 .2144 5.0356
B2 2.0000 .84911 .154 -.4106 4.4106
B4 -2.1250 .84911 .111 -4.5356 .2856
B4
B1 4.7500* .84911 .000 2.3394 7.1606
B2 4.1250* .84911 .000 1.7144 6.5356
B3 2.1250 .84911 .111 -.2856 4.5356
Based on observed means.
The error term is Mean Square(Error) = 2.884.
*. The mean difference is significant at the .05 level.
c. Test the linear component of trend for the effect of B.
Solution
t = 7.38, p < .001, significant
d. Plot the interaction.
Solution
e. Describe the interaction in words.
Solution
The dependent variable increases at a more rapid rate across categories of “b” in level 2
of “a.”

Descriptive Statistics
Dependent Variable: Scores
A B Mean Std. Deviation N
A1
B1 2.3750 1.50594 8
B2 3.0000 1.19523 8
B3 5.0000 2.07020 8
B4 7.1250 1.88509 8
Total 4.3750 2.48544 32
Total
B1 2.3750 1.50594 8
B2 3.0000 1.19523 8
B3 5.0000 2.07020 8
B4 7.1250 1.88509 8
Total 4.3750 2.48544 32
23. Why are within-subjects designs usually more powerful than between-subjects design?
Solution
Within-subjects designs control for individual differences between subjects
24. What source of variation is found in an ANOVA summary table for a within-subjects
design that is not in in an ANOVA summary table for a between-subjects design. What
happens to this source of variation in a between-subjects design?
Solution
"Subjects" is not a source of variation in between-subjects designs. It goes into the error
term.
25. The following data contain three scores from each of five subjects. The three scores per
subject are their scores on three trials of a memory task.
4 6 7
3 7 7
2 8 5

1 4 7
4 6 9
a. Compute an ANOVA
Solution
ANOVA
Scores
Sum of Squares df Mean Square F Sig.
Between Groups 49.733 2 24.867 12.644 .001
Within Groups 23.600 12 1.967
Total 73.333 14
b. Test all pairwise differences between means using the Bonferroni test at the .01 level.
Solution
Multiple Comparisons
Dependent Variable: Scores
Bonferroni
(I) Group (J) Group Mean Difference
(I-J)
Std. Error Sig. 95% Confidence Interval
Lower Bound Upper Bound
Group1 Group2 -3.40000* .88694 .007 -5.8652 -.9348
Group3 -4.20000* .88694 .001 -6.6652 -1.7348
Group2 Group1 3.40000* .88694 .007 .9348 5.8652
Group3 -.80000 .88694 1.000 -3.2652 1.6652
Group3 Group1 4.20000* .88694 .001 1.7348 6.6652
Group2 .80000 .88694 1.000 -1.6652 3.2652
*. The mean difference is significant at the 0.05 level.
c. Test the linear and quadratic components of trend for these data.
Solution

Linear: ± = 51.88, p = .002
Quadratic: ± = 2.07, p = .223
26. Give the source and df columns of the ANOVA summary table for the following
experiments:
a. Twenty two subjects are each tested on a simple reaction time task and on a choice
reaction time task.
Solution
b. Twelve male and 12 female subjects are each tested under three levels of drug
dosage: 0 mg, 10 mg, and 20 mg.
Solution
Source df
Gender 2-1=1
Drug Dosage 3-1=2
Gender * Drug Dosage 1*2=2
Error 2*3(12-1)=66
Total (2*3*12)-1=71

Exercise 16:
1. When is a log transformation valuable?
Solution
Log transformations are valuable where the data are highly skewed and the researcher
would be interested in making them to be less skewed.
2. If the arithmetic mean of log10 transformed data were 3, what would be the geometric
mean?
Solution
log10(X)=3, then X=1000. 103 =1000.Thus the anit-log of the arithmetic mean is the
geometric mean=1000
3. Using Tukey's ladder of transformation, transform the following data using a λ of 0.5: 9,
16, 25
Solution
The value of .5 simply suggest that we take the square root of the data. Thus {9,16,25}
becomes {3, 4, 5}
4. What value of λ in Tukey's ladder decreases skew the most?
Solution
-2
5. What value of λ in Tukey's ladder increases skew the most?
Solution
2

6. In the ADHD case study, transform the data in the placebo condition (D0) with λ's of .5,
0, -.5, and -1. How does the skew in each of these compare to the skew in the raw data.
Which transformation leads to the least skew?
Solution
They all reduce skew. Skew is reduced most with λ= -1
Exercise 17:
1. Which of the two chi-square distributions shown below (A or B) has the larger degrees of
freedom? How do you know?
Solution
B; because of the shape of the curve; it is less skewed and has a higher mean
2. Twelve subjects were each given two Favors of ice cream to taste and then were asked
whether they liked them. Two of the subjects liked the first Favor and nine of them liked
the second Favor. Is it valid to use the Chi Square test to determine whether this
difference in proportions is significant? Why or why not?
Solution
Chi square would not be appropriate in this case. The proportions being compared are
within-subject and thus are not independent of each other.
3. A die is suspected of being biased. It is rolled 25 times with the following result:
Conduct a significance test to see if the die is biased.
(a) What Chi Square value do you get and how many degrees of freedom does it have?
Solution

Chi Square = 16.04, df = 5
(b) What is the p value?
Solution
p = .007
4. A recent experiment investigated the relationship between smoking and urinary
incontinence. Of the 322 subjects in the study who were incontinent, 113 were smokers,
51 were former smokers, and 158 had never smoked. Of the 284 control subjects who
were not in- continent, 68 were smokers, 23 were former smokers, and 193 had never
smoked.
(a) Create a table displaying this data.
Solution
Smoking Incontinent Not Incontinent Total
Smokers 113 68 181
Former Smokers 51 23 74
Never Smoked 158 193 351
Total 322 284 606
(b) What is the expected frequency in each cell?
Solution
Smoking Incontinent Not Incontinent Total
Smokers 96.2 84.8 181.0
Former Smokers 39.3 34.7 74.0
Never Smoked 186.5 164.5 351.0

Total 322.0 284.0 606.0
(c) Conduct a significance test to see if there is a relationship between smoking and
incontinence. What Chi Square value do you get? What p value do you get?
Solution
Chi-Square Tests
Value df Asymp. Sig. (2-
sided)
Pearson Chi-Square 22.980a 2 .000
Likelihood Ratio 23.281 2 .000
Linear-by-Linear Association 17.010 1 .000
N of Valid Cases 606
a. 0 cells (0.0%) have expected count less than 5. The minimum
expected count is 34.68.
The Chi-Square value is 22.98 with a p-value of 0.000.
(d) What do you conclude?
Solution
We reject the null hypothesis and conclude that there is strong evidence of
relationship (association) between smoking and urinary incontinence
5. At a school pep rally, a group of sophomore students organized a free raffle for prizes.
They claim that they put the names of all of the students in the school in the basket and
that they randomly drew 36 names out of this basket. Of the prize winners, 6 were
freshmen, 14 were sophomores, 9 were juniors, and 7 were seniors. The results do not
seem that random to you. You think it is a little fishy that sophomores organized the
raffle and also won the most prizes. Your school is composed of 30% freshmen, 25%
sophomores, 25% juniors, and 20% seniors.

(a) What are the expected frequencies of winners from each class?
Solution
Winners
Freshmen 34*0.3 = 10.2
Sophomores 34*0.25 = 8.5
Juniors 34*0.25 = 8.5
Seniors 34*0.20 = 6.8
Total 34*1.00 = 34.0
(b) Conduct a significance test to determine whether the winners of the prizes were
distributed throughout the classes as would be expected based on the percentage of
students in each group. Report your Chi Square and p values.
Solution
χ2=∑ ( O−E )2
E = ( 6−10.2 )2
10.2 + ( 14−8.5 )2
8.5 + ( 9−8.5 )2
8.5 + ( 7−6.8 )2
6.8 =5.3235
The Chi-Square value is 5.3235 with a p-value of 0.149585.
(c) What do you conclude?
Solution
We fail to reject the null hypothesis and conclude that indeed the results do not seem to
be random.
6. Some parents of the West Bay little leaguers think that they are noticing a pattern. There
seems to be a relationship between the number on the kids’ jerseys and their position.
These parents decide to record what they see. The hypothetical data appear below.
Conduct a Chi Square test to determine if the parents’ suspicion that there is a
relationship between jersey number and position is right. Report your Chi Square and p
values.

Solution
Chi-Square Tests
Value df Asymp. Sig. (2-
sided)
Pearson Chi-Square 10.226a 4 .037
Likelihood Ratio 9.750 4 .045
Linear-by-Linear Association 3.561 1 .059
N of Valid Cases 54
a. 2 cells (22.2%) have expected count less than 5. The minimum
expected count is 3.89.
The Chi-Square value is 10.226 with a p-value of 0.037. We reject the null hypothesis
and conclude that there is evidence of a relationship between the number on the kids’
jerseys and their position.
7. True/false: A Chi Square distribution with 2 df has a larger mean than a Chi Square
distribution with 12 df.
Solution
False
8. True/false: A Chi Square test is often used to determine if there is a significant
relationship between two continuous variables.
Solution
False
9. True/false: Imagine that you want to determine if the spinner shown below is biased. You
spin it 50 times and write down how many times the arrow lands in each section. You
will reject the null hypothesis at the .05 level and determine that this spinner is biased if
you calculate a Chi Square value of 7.82 or higher.
Solution
True

Questions from Case Studies
SAT and GPA (SG) case study
10. (SG) Answer these items to determine if the math SAT scores are normally distributed.
You may want to first standardize the scores.
a. If these data were normally distributed, how many scores would you expect there to be in
each of these brackets:
(i) smaller than 1 SD below the mean,
Solution
16.8
(ii) in between the mean and 1 SD below the mean,
Solution
35.7
(iii) in between the mean and 1 SD above the mean,
Solution
35.7
(iv) greater than 1 SD above the mean?
Solution
16.8
b. How many scores are actually in each of these brackets?
Solution
Actual
Smaller than 1 SD below the
mean
24
In between the mean and 1
SD below the mean
36

In between the mean and 1
SD above the mean
19
Greater than 1 SD above the
mean
26
c. Conduct a Chi Square test to determine if the math SAT scores are normally distributed
based on these expected and observed frequencies.
Solution
Chi-square = 15.94, p = .001. Scores are not normally distributed
Diet and Health (DH) case study
11. (DH) Conduct a Pearson Chi Square test to determine if there is any relationship between
diet and outcome. Report the Chi Square and p values and state your conclusions.
Solution
Chi-square = 16.55, p = .001. There is a significant relationship between diet and
outcome
The following questions are from ARTIST (reproduced with permission)
12. A study compared members of a medical clinic who filed complaints with a random
sample of members who did not complain. The study divided the complainers into two
subgroups: those who filed complaints about medical treatment and those who filed
nonmedical complaints. Here are the data on the total number in each group and the
number who voluntarily left the medical clinic. Set up a two-way table. Analyze these
data to see if there is a relationship between complaint (no, yes - medical, yes -
nonmedical) and leaving the clinic (yes or no).

Solution
Medical
complaint
Non-medical
complaint
Total
Yes 26 28 54
No 173 412 585
Total 199 440 639
The chi-square statistic is 7.9547. The p-value is .004796. This result is significant at p
< .05. We conclude that there is a significant relationship between complaint (no, yes -
medical, yes - nonmedical) and leaving the clinic (yes or no).
13. Imagine that you believe there is a relationship between a person’s eye color and where
he or she prefers to sit in a large lecture hall. You decide to collect data from a random
sample of individuals and conduct a chi-square test of independence. What would your
two-way table look like? Use the information to construct such a table, and be sure to
label the different levels of each category.
Solution
Prefers to
sit
White Blue Total
In front 24 56 80
Back 46 34 80
Total 70 90 160
14. A geologist collects hand-specimen sized pieces of limestone from a particular area. A
qualitative assessment of both texture and color is made with the following results. Is
there evidence of association between color and texture for these limestones? Explain
your answer.
Solution
Chi-Square Tests

Value df Asymp. Sig. (2-
sided)
Pearson Chi-Square 17.727a 4 .001
Likelihood Ratio 18.141 4 .001
Linear-by-Linear Association 11.612 1 .001
N of Valid Cases 120
a. 0 cells (0.0%) have expected count less than 5. The minimum
expected count is 6.40.
Yes there is evidence of association between color and texture for these limestones (p-
value < 0.05)
15. Suppose that college students are asked to identify their preferences in political affiliation
(Democrat, Republican, or Independent) and in ice cream (chocolate, vanilla, or
strawberry). Suppose that their responses are represented in the following two-way table
(with some of the totals left for you to calculate):
a. What is the proportion of the respondents who prefer chocolate ice cream?
Solution
80
173 =46.24 %
b. What proportion of respondents are independents?
Solution
26
173 =15.03 %
c. What proportion of Independents prefer chocolate ice cream?
Solution
9
26 =34.62 %

d. What proportion of those who prefer chocolate ice cream are Independents?
Solution
9
80 =11.25 %
e. Analyze the data to determine if there is a relationship between political party
preference and ice cream preference.
Solution
Chi-Square Tests
Value df Asymp. Sig. (2-
sided)
Pearson Chi-Square 23.808a 4 .000
Likelihood Ratio 24.735 4 .000
Linear-by-Linear Association 2.102 1 .147
N of Valid Cases 173
a. 1 cells (11.1%) have expected count less than 5. The minimum
expected count is 3.76.
Based on the above results we conclude that there is evidence of a relationship
between political party preference and ice cream preference.
16. NCAA collected data on graduation rates of athletes in Division I in the mid-1980s.
Among 2,332 men, 1,343 had not graduated from college, and among 959 women, 441
had not graduated.
a. Set up a two-way table to examine the relationship between gender and graduation.
Solution
Men Women Total
Graduated 989 518 1507
Not 1343 441 1784

graduated
Total 2332 959 3291
b. Identify a test procedure that would be appropriate for analyzing the relationship between
gender and graduation. Carry out the procedure and state your conclusion
Solution
The chi-square statistic is 36.8664. The p-value is 0.000. This result is significant at p
< .05 hence we conclude there is significant relationship between gender and graduation.
Exercise 18:
1. For the following data, how many ways could the data be arranged (including the original
arrangement) so that the advantage of the Experimental Group mean over the Control
Group mean is as large or larger then the original arrangement.
Solution
2
2. For the data in Problem 1, how many ways can the data be rearranged?
Solution
252
3. What is the one-tailed probability for a test of the difference.
Solution
0.0079
4. For the following data, how many ways can the data be rearranged?
Solution
1680

5. In general, are rank randomization tests or randomization tests more powerful?
Solution
Randomization tests
6. What is the advantage of rank randomization tests over randomization tests?
Solution
There are tables for rank randomization tests
7. Test whether the differences among conditions for the data in Problem 1 is significant
(one tailed) at the .01 level using a rank randomization test.
Solution
Rank sum = 39 which is significant at .01
Questions from Case Studies
SAT and GPA (SG) case study
8. (SG) Compute Spearman's ρ for the relationship between UGPA and SAT.
Solution
0.716
Stereograms (S) case study
9. (S) Test the difference in central tendency between the two conditions using a rank-
randomization test (with the normal approximation) with a one-tailed test. Give the Z and
the p.
Solution
Z = 2.21, p = 0.0271
Smiles and Leniency (SL) case study

10. (SL) Test the difference in central tendency between the four conditions using a rank-
randomization test (with the normal approximation). Give the Chi Square and the p.
Solution
Chi Square 9.1747, p = .0271
Exercise 19:
1. If the probability of a disease is .34 without treatment and .22 with treatment then what is
the
(a) absolute risk reduction
Solution
ARR=risk ∈control group – risk ∈treatment group
ARR=0.34−0.22=0.12
(b) relative risk reduction
Solution
RRR = (risk in controls – risk in treatment group)
risk in controls
RRR= 0.12
0.34 =35.29 %

(c) Odds ratio
Solution
RR=1-RRR
RR=1-0.3529=0.6471
¿= 1
RR = 1
0.6471 =1.545
(d) Number needed to treat
Solution
NNT = 1
ARR = 1
0.12 =8.33
2. When is it meaningful to compute the proportional difference between means?
Solution
When we want to compare to proportions
3. The mean for an experimental group is 12, the mean for the control group were 8, the
MSE from the ANOVA is 16, and N, the number of observations is 20, compute g and d.
Solution
g = 1, d = 1.054
4. Two experiments investigated the same variables but one of the experiment had subject
who differed greatly from each other whereas the subjects in the other experiment were
relatively homogeneous. Which experiment would likely have the larger value of g?
Solution
The one with the more homogeneous subjects.

5. Why is ω2 preferable to η2?
Solution
η2 has a positive bias.
6. What is the difference between η2 and partial η2?
Solution
η2 is the proportion of variance explained. Partialη2is the proportion of variance
explained not counting the variance explained by other independent variables.
Questions from Case Studies
Teacher Ratings (TR)
7. (TR) What are the values of d and g?
Solution
g = .70; d = .72
8. (TR) What are the values of ω2 and η2?
Smiles and Leniency (SL)
9. (SL)What are the values of ω2 and η2?
Obesity and Bias (OB)
10. For compute ω2 and partial ω2 for the effect of “Weight” in a “Weight x Relatedness”
ANOVA.