Practice Problems
Summation Notation
2-8+9+0 = -15
[(14*-2)+(2*0)+(-8*0)+(9*-1)+(0*10)+(5*6)+(-3*7)]= -28
Then a new vector is created:
= [(19-2),(19+0),(19+0),(19-1),(19+10),(19+6),(19+7)]
Then a new vector is created:
= [(17-2),(17+0),(17+0),(17-1),(17+10),(17+6),(17+7)]
-2+0+0-1+10+6+7 = 20
[(-2+14*-2)+(0+2*0)+(0+-8*0)+(-1+9*-1)+(10+0*10)+(6+5*6)+(73*7)] = -8
= [(-2+14*-2),(0+2*0),(0+-8*0),(-1+9*-1),(10+0*10),(6+5*6),(73*7)]
I1
I2
I3
J1
30
17
-2
J2
10
-1
0
J3
12
10
27
J4
47
9
5
30+10+12+47+17-1+10+9-2+0+27+5 = 164
17-1-2+0 = 14
Z-Scores
P(-1≤z≤1) = .68
This is the probability lies in the middle 2 standard deviations
of the distribution (-1,1)
P(z≥4) = ~0
Look at the “smaller portion” column in z-table.
P(-3.5≤z) = .0002
Look at the “smaller portion” column in z-table.
P(-3.5≤z≥2.5) = 0.9997674
This can be obtained by taking the larger portion from the z
table.
P(-1≤z or z≥2) = .1587 + .0228 = .1815
This can be done by taking the right 50% of the distribution
plus the “mean to z” column value.
Or, you can just take the “larger value,” which is .9772
P(-1≥z or z≥-2)
= P(-1≥z≥-2)
The probability of z being between -1 and -2
This can be computed by taking the “mean to z” for each
For -2 the mean to z is .4778, for -1 the mean to z is .3413 
we can get the p-value by taking the difference between the two:
.4778-.3414 = .1364
See the graph below:
Report the p-value for the following:
P(z≥1)= .1587
Look at the smaller portion.
P(-1≤z≤2)
Since this crosses the middle of the distribution we can take
two mean to z values and add them:
.34+.4772 = .8172
The average systolic blood pressure in the United States is 124
with a standard deviation of 16. My blood pressure is 155.
Should I be concerned that my blood pressure is excessively
high? Why or why not?
z = (155-124)/16 = 1.9375
p = .0262
I am definitely concerned! (Not real data so I am really OK)
At what point would my blood pressure be so I that I should
begin worrying? Provide a real-world explanation.
Using an alpha of .05, our critical value would be 1.65 sd’s and
X=1.65*16+124 (x=z*sd+mu; reverse our standardization)
X=150.4
Chi-Squares
Psychopathology is reported to have the highest probability of
manifestation between 15 and 25. I have collected a random
sample of 100 undergraduates at Loyola. 7 persons meet criteria
major depressive disorder, 2 persons meet criteria for
schizophrenia, and 15 persons meet criteria for alcohol abuse.
I want to know if my frequencies (or proportions) match those of
the population with regard to these specific disorders: .04,
.01, .05 respectively.
What test should I use and why?
Chi-square goodness of fit test should be used. We are only
interested in one variable and want to see if our observed
frequencies match with those theoretical expected values.
Analyze these data and report the chi-square statistic and
associated p-value.
Major Dep.
Schizophrenia
Alcohol Abuse
Observed
7
2
15
Expected
.04*100 = 4
.01*100 = 1
.05*100 = 5
[(7-4)^2/4] + [(2-1)^2/1] + [(15-5)^2/5] = 23.25
df = k-1 = 3-1 = 2
p < .000001
(χ2(2) = 23.25, p < .05)
What assumptions are we making with this test? Are you concerned
any assumptions may be violated? Why or why not?
We are assuming these data are independent, normally
distributed.
I get concerned about these data because I don’t know if some of
these students met criteria for multiple disorders (e.g. alcohol
abuse and depression). If that was the case and some of these
cases were counted multiple times, we have violated the
assumption of independence. I would also feel a lot better
about these data if there were more observations in each of the
cells and the expected values were higher for this sample; a
bigger sample would help out here. For example, there were only
2 persons found who met criteria for schizophrenia.
What if I know that there are expected differences between
genders and the conditional distribution of each of these
disorders? How would I test this – what test should be used?
We would need to have a chi-square test of association to
examine the relationship between these two variables.
Analyze these data and report the expected values, chi-square
statistic and p-value. Give a real world interpretation of the
results.
EXPECTED VALUES COMPUTED AND PLACED IN PARANTHESES
Males
Females
Total
MDD
Schizophrenia
Substance
Abuse
11
[(14*15)/24]
= (8.75)
Total
2
[(14*7)/24]
=
(4.083333)
5
[(10*7)/24]
=
(2.916667)
7
1 [(14*2)/24]
= (4.083333)
1 [(10*2)/24]
= (0.8333333)
4
[(10*15)/24]
= (6.25)
10
2
15
24
14
(OBSERVED-EXPECTED)^2/EXPECTED
Males
Females
Total
MDD
Schizophrenia
1.062925
2.694403
7
2.328231
0.03333335
2
Substance
Abuse
0.5785714
.81
15
Total
14
10
24
CHI-SQUARE =
[(2-4.083333)^2/4.083333] + [(1-4.083333)^2/4.083333] + [(118.75)^2/8.75] + [(5-2.196667)^2/2.916667] + [(10.8333333)^2/0.8333333]+[(4-6.25)^2/6.25]
= 7.507464
df = (3-1)*(2-1) = 2
p = 0.02
(χ2(2) = 7.507464, p < .05)
Comment on the model assumptions and whether or not you think it
is plausible that they were violated.
We are assuming these data are independent, normally
distributed.
I get concerned about these data because I don’t know if some of
these students met criteria for multiple disorders (e.g. alcohol
abuse and depression). Adding the dimension of gender does not
exacerbate this concern but it does thin out the cell
frequencies even more. Notice that only 4/6 of the expected
values are greater than 5.
Compute an effect size of your choice for these data (the chisquare test of association).
I am computing two different ones (note, you cannot compute phi
for these data).
C = sqrt(7.507464/(7.507464+24)) = .49
Phi_C = sqrt(7.507464/(24*(2-1))) = .55
This is a moderately large effect size!
[New Wentworth Example]
Wentworth has a propensity for jumping up on the kitchen table.
While I was brainstorming solutions to this problem, I wondered
if he is more likely to jump on the table after he was given
catnip (maybe we should cut down on the catnip). I randomly
chose 15 days in the month when I would give him some catnip and
record whether or not he jumped on the table afterwards. On the
other days, I observed his regular activity without giving him
any catnip and recorded whether or not he jumped on the table.
Here is what the data look like:
Catnip
Jump 11
Not Jump 4
No Catnip
8
7
Compute the odds ratio for this data and interpret.
OR = (11*7)/(8*4) = 2.40625
The odds of Wentworth jumping on the table given he had catnip
are 2.40625 (141%) times the odds of Wentworth jumping on the
table given he did not have catnip. There is strong evidence of
an association between catnip and Wentworth jumping on the
table.
Or…
OR = (8*4)/(11*7) = 0.4155844
The odds of Wentworth jumping on the table given that he did not
have any catnip are 0.4155844 (42%) times the odds of Wentworth
jumping on the table given that Wentworth did have catnip.
Compute the risk ratio for this data, interpret.
Catnip
Jump 11
Not Jump 4
Total 15
No Catnip
Total
8
19
7
11
15
30
RiskCatnip = (11/15) = 0.7333333
RiskNoCatnip = (8/15) = 0.5333333
RR = .7333/.5333 = 1.375023
The risk of Wentworth jumping on the table given he had catnip
are 1.37 (37%) times the risk of him jumping on the table given
he did not have catnip.
Two psychiatrists are conducting an epidemiological study of
bipolar disorders. They are required to diagnose persons in the
study has having bipolar I, bipolar II, or other. Both doctors
rate all 37 individuals in the study. The following table
provides their ratings.
Compute Cohen’s Kappa and interpret its meaning for these data.
Doctor 2
Doctor 1
Bipolar I
Bipolar II
Other
Total
Bipolar I
5
9
2
16
Bipolar II
8
7
1
16
Other
4
1
0
5
Total
17
17
3
37
Doctor 2
Bipolar I
Doctor 1
Bipolar II
Other
Total
Bipolar I
5
9
(7.351351)
2
16
Bipolar
II
8
7
(7.351351)
1
16
Other
4
1
0
5
(0.4054054)
Total
17
17
3
37
First compute the expected values (we are only interested in the
diagonals)
BP I = 16*17/37 = 7.351351
BP II = 16*17/37 = 7.351351
Other = 5*3/37 = 0.4054054
Compute K
K = ((5+7+0)-(7.351351+7.351351+0.4054054))/(37(7.351351+7.351351+0.4054054)) = -0.1419753
A negative K indicates that the two doctors are agreeing less
than chance would have predicted. Or, we could say these two
doctors are disagreeing more than chance would have predicted.
Substantively this means these two doctors are likely using
different criteria in making these diagnoses.
School district X collects test anxiety data on all if its 5000
students. A teacher in that school is interested in knowing how
his students compare to the whole district in terms of test
anxiety. The mean and sd of the district are 25 and 15
respectively. The teacher’s classroom mean test anxiety score
is 19 for his 33 students.
What are the null and alternative hypotheses?
H0: Mean(classroom) = Mean(District)
(The teacher’s classroom is no different than the district)
H1: Mean(Classroom) ≠ Mean(District)
(The teacher’s classroom is different than the district)
Given the information provided, what test should you use, why?
Since the population parameters are known, we can simply use a
Z-test for this problem.
Conduct the analysis and interpret the results.
We want the difference between the population and sample means.
Z = (19-25)/(15/sqrt(33)) = -2.297825
P = .011
Based on this test statistic I would reject the null hypothesis
that these two groups (classroom and district) are equal and say
that the mean test anxiety for this teacher’s classroom is
significantly less than the mean of the district.
Suppose this teacher was still not satisfied and decided he
wanted to lower his student’s test anxiety even further. He
heard about applied relaxation training at a conference he
recently attended. His student’s test anxiety was measured at
the beginning of the school year (where the mean was 19). He
will implement his intervention for 2 months and measure his
students’ anxiety again in order to see if their anxiety lowered
even further. The mean difference (time1-time2) and sd of the
mean difference are 1.25 and 1.07 respectively.
What are the null and alternative hypotheses?
H0: Mean(time1) = Mean(time2)
(There is no difference between time 1 (pretest) and time 2
(posttest) – no evidence of a treatment effect.)
H1: Mean(time1) ≠ Mean(time2)
(The mean level of test anxiety at time 2 is less than time 1
(before the treatment was implemented))
NOTE: I have specified a non-directional two-tailed hypothesis
test but you could make an argument here for a one-tailed
(directional) test if you are only interested in know if there
is improvement in test scores. Remember that the two-tailed
test is more conservative.
What test would be most appropriate to answer this teacher’s
question?
This is a dependent sample since the same students are being
measured multiple times and being compared to themselves. We
would want to use a dependent samples t-test.
Conduct the analysis and interpret the results.
t = (1.25-0)/(1.07/sqrt(33)) = 6.710938
df = 32
p < .0001
Based on this test statistic I would reject the null hypothesis
that the two testing times are equal. There is some evidence of
a treatment effect based on these data.
Suppose this same teacher had an end-year goal of getting his
students mean test anxiety score down to 16.5 and their actual
end-of-year score was 17.93 (sd=12). How could this teacher
examine whether or not he may have met his goal? How can he
test the difference between his observed mean and his goal mean
for statistical significance?
The teacher could use a one-sample t-test comparing the observed
mean of 17.93 with the theoretical (goal) mean of 16.5.
What are the null and alternative hypotheses?
H0: mean(observed) = mean(theoretical)
H1: mean(observed) ≠ mean(theoretical)
Conduct the analysis and interpret the results.
t = (17.93-16.5)/(12/sqrt(33))
df = 32
p = .25
Based on these data I would retain the null hypothesis that the
theoretical mean and the observed mean are equal. This teacher
should be reasonably pleased with his results – 16.5 is just as
plausible as 17.93 to be the true end-year value of test
anxiety.
Some practice from the book
# 7.6
x<-c(40,58,72,73,76,78,52,72,84,70,72)
t.test(x,mu=50)
#
One Sample t-test
#data: x
#t = 4.6426, df = 10, p-value = 0.0009184
#alternative hypothesis: true mean is not equal to 50
#95 percent confidence interval:
# 59.31385 76.50433
#sample estimates:
#mean of x
# 67.90909
#Or...
t<-(mean(x)-50)/(sd(x)/sqrt(length(x)))
t
# [1] 4.642561
1-pt(t,length(x)-1)
# [1] 0.0004592003
# 7.7
# 7.7 One could argue a one-tailed test is appropriate because
#the
# researcher may only be interested in whether the students are
# "greater than" the average and does not care whether or not
they
# are less than average.
# 7.10 (applied to 7.6)
CI.U<-mean(x)+(sd(x)/sqrt(length(x)))*2.228
CI.L<-mean(x)-(sd(x)/sqrt(length(x)))*2.228
# 7.16
# (tx1 = 12 hours before surgery)
# (tx2 = 10 minutes before surgery)
tx1<c(10,6.5,8,12,5,11.5,5,3.5,7.5,5.8,4.7,8,7,17,8.8,17,15,4.4,2)
tx2<c(6.5,14,13.5,18,14.5,9,18,42,7.5,6,25,12,52,20,16,15,11.5,2.5,2
)
t.test(tx1,tx2,paired=T)
# Paired t-test
# data:
tx1 and tx2
# t = -2.4827, df = 18, p-value = 0.02313
# alternative hypothesis: true difference in means is not equal
to 0
# 95 percent confidence interval:
#
-14.216028
-1.183972
# sample estimates:
# mean of the differences
-7.7
# Endorphin levels are significantly higher at 10 minutes prior
# to surgery.
# 7.17 We use a dependent samples test because we are comparing
two
# observations (different times) from the same set of
individuals.
# 7.18
SE<-sqrt(var(tx1-tx2))
CI.L = -7.7 - 2.110*13.51916/sqrt(19)
CI.U = -7.7 + 2.110*13.51916/sqrt(19)
The distributions above were drawn from the same population. Use
what you know about the central limit theorem to explain what
you see.
The four distributions above share the same mean (50) but their
variance are clearly different. The first graph is the widest,
indicating that the distribution of sample means must be
repeated draws of small sample sizes. The last one is very thin
(low variance) indicating that it must be a sampling
distribution of large sample sizes (relative to the other
three).
True/False – Explain your answers
As alpha increase so does statistical power____T_____
As we increase alpha our type II error rate will fall. That is
to say that we will be less strict about determining statistical
significance and will be rejecting the null more often (but
making more type I errors).
If a research finding shows a strong effect (association or
difference) one can also expect to have a large type II error
probability ____F______
The larger the effect, the more likely one will correctly reject
the null hypothesis (i.e. the conclusion is more clear the
larger the effect size).
An effect size of .8 (from a two-independent samples t-test)
will always mean the test has strong statistical power____F_____
All things being equal, yes, a .8 effect size will have more
power than a smaller effect size. However, it is possible that
statistical assumptions were violated in the process thus
reducing its power. It is also possible that the sample is
small, making it less likely to reject the null hypothesis than
perhaps smaller effect sizes.
One way for me to boost power is to collect more data_____T_____
Increasing the sample size is the best way for a researcher to
boost power.
I carried out a program evaluation of a school district’s new
after school program to prevent truancy. There were 500
students who participated in the program and 488 who were
evaluated but did not participate in the program. The mean
levels of truancy for the experimental group and control group
were 17.58 (sd=5.1) and 21.2 (sd=4.3) respectively.
What test should be used?
Two-independent samples t-test.
Compute the test statistic and effect size for this data.
Sp2=((500-1)*5.1^2+(488-1)*4.3^2)/(500+488-2)
= 22.29576
t=( 22.29576-17.58)/sqrt((22.29576/500)+( 22.29576/488))
= 15.69484
df=500+488-2
= 986
tc = 1.96
I would reject the null hypothesis that the experimental and
control group are equal in their levels of truancy after the
program implementation.
Conduct a power analysis and interpret the results.
nh = 2*500*488/(500+488)
= 494
d = (21.2-17.58)/ 22.81703
= 0.1586534
delta = 0.1586534*sqrt(494/2)
= 2.49
power ≈ .71 using a two-tailed alpha of .05
The following is output from a regression analysis. I am
predicting cultural trauma with age. Interpret the results.
################################################################
Call:
lm(formula = trauma ~ age)
Residuals:
Min
1Q
-2.65410 -0.62412
Median
0.01143
3Q
0.63711
Max
2.97065
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 18.53400
0.61876
29.95
<2e-16 ***
age
0.32573
0.03083
10.57
<2e-16 ***
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.9681 on 998 degrees of freedom
Multiple R-squared: 0.1006,
Adjusted R-squared: 0.09972
F-statistic: 111.7 on 1 and 998 DF, p-value: < 2.2e-16
################################################################
This analysis shows a significantly strong relationship between
age and cultural trauma (b=.32, p<.05). For each year increase
in age there is a .32 unit increase in cultural trauma. R2adj =
.09972 indicating roughly 9% of the variance in cultural trauma
is accounted for by age. The overall model is significant
(F=111.7, p<.05) and the RMSE=.9681 indicating adequate model
quality.
Write the regression equation.
Predict someone’s level of cultural trauma when they are 15
years old.
Say their actual score was 20, what is the error (or residual)?
For the same output above, manually reproduce the intercept and
slope parameter estimates. r = .32
Mean
SD
Age
20.04793 0.993616
Trauma 25.06418 1.020315
r = .32
N = 1000
sxy = .32*0.993616*1.020315
= 0.3244164
b = 0.3244164/ 0.993616^2
= 0.3285925
a = 25.06418 - 0.3285925*20.04793
= 18.47658
Manually test the statistical significance of the slope estimate
you reproduced.
sb = 0.9681/(0.993616*sqrt(1000-1))
= 0.03082612
t = (0.32441640)/0.03082612
= 10.52
tc = 1.96
We would reject the null hypothesis that b=0.
Perhaps I am interested in knowing whether or not males and
females significantly differ in their levels of cultural trauma.
I conduct a t-test in R. Show me how to manually produce the ttest results below. The sd for males (0) is 1.03 and the sd for
females (1) is 1.02.
Two Sample t-test
data: cultural trauma by gender
t = 1.0073, df = 998, p-value = 0.314
alternative hypothesis: true difference in means is not equal to
0
95 percent confidence interval:
-0.06132259 0.19068238
sample estimates:
mean in males (group 0) mean in females (group 1)
25.03901
24.97433
t = (24.97433-25.03901)/((1.03^2+1.02^2)/2)
= -0.06156189
df = 998
tc = 1.96
My recalculation of the output confirms that we should retain
the null hypothesis that males and females are equal in their
levels of cultural trauma.
Based on the following information about cultural trauma and
age, compute the covariance estimate.
Mean
SD
Age
20.04793 0.993616
Trauma 25.06418 1.020315
r = .32
N = 1000
sxy = .32*0.993616*1.020315
= 0.3244164
Compute the adjusted correlation (radj) for these data and
compare it to r.
radj = sqrt(1-(((1-.32^2)*(1000-1))/998))
= 0.3185916 [NOT MUCH DIFFERENT THAN r]