COGS 14B / Introduction to Statistical Analysis

COGS 14B / Introduction to Statistical Analysis
HOMEWORK 4 – Spring 2013
1. Describe what statistical test you should use, including the
direction of the test.
(a) You want to know whether brain area F responds more to pictures of
feet than anything else. You have 20 participants, and you measure the
activity in each person’s area F, when each person sees feet vs. when
they see non-foot objects.
Upper-tailed one sample t test
(b) UCSD students take a standardized test and the mean of their score is
60. The national test norm has a mean of 50 and a standard deviation of
10. Are UCSD students smarter than 95% of the test-takers?
Upper-tailed z test
(c) You want to know who has the higher geek quotient (GQ): Cognitive
Science professors at UCSD, or Psychology professors at UCSD. You
measure the GQ of each faculty member in each department
Two-tailed two-independent t-test
2. Answer questions for the following t test.
(a) If a sample consists of 29 individuals, then what is the df value for the
t statistic? 28
(b) For a sample of n = 15, what t values determine the two-tailed critical
region for α = .05 ?
+2.145, -2.145
(c) Two samples, each with n = 10 subjects, produce a pooled variance of
20. Based on this information, the estimated standard error for the
sample mean difference would be _2_.
(d) A hypothesis test produces a t statistic of t = 2.20. If the researcher
is using a two-tailed test with α = .05, how large does the sample have to
be in order to reject the null hypothesis? n =13
(e) A t distribution with __infinite___ degrees of freedom is just the z
distribution
3. Which set of sample characteristics is most likely to produce a
significant t statistic? (a)
(a) large sample size and a small sample variance
(b) large sample size and a large sample variance
(c) small sample size and a small sample variance
(d) small sample size and a large sample variance
4. Give the sample size (i.e., number of experimental participants)
for the following statistics and tests:
(a) Degree of freedom = 22, independent samples t-test
24 participants
(b) Degree of freedom = 19, one sample t-test
20 participants
5. A library system lends books for periods of 2 days. This policy is
being reevaluated in view of a possible new loan period that could
be either longer or shorter than 21 days. To aid in making this
decision, book-lending records were consulted to determine the
loan periods that patrons actually used. A random sample of eight
records revealed the following loan period in days: 21, 15, 12, 24,
20, 21, 13 and 16. Test the null hypothesis with a 0.05 level of
significance.
Statistical Hypotheses:
Ho: µ = 21
H1: µ 21
Decision Rule: Reject Ho at 0.05 level of significance if
given df=7
or
Calculations
Decision
Retain Ho at the 0.05 level of significance because t=-2.12 is less
negative than -2.365 (more generally, it lies between the two critical
values ±2.365).
Interpretation
No evidence that, on average, library patrons borrow books for longer or
shorter periods than 21 days.
6. A research team wishes to determine whether alcohol
consumption causes a deterioration in driving performance. A
total of one hundred and twenty volunteer subjects are randomly
assigned, in equal numbers, to two groups, one in which before
the driving test subjects drink a glass of orange juice, and one in
which, they drink a glass of orange juice containing two ounces of
vodka (treatment group). Performance was measured by the
number of errors made on a driving simulator. For subjects in the
treatment group, the mean number of errors equals 26.4, and for
subjects in the control group, the mean number of errors equals
18.6. The estimate standard error equals 2.4.
a) What test should the research team use?
t-test for two independent samples
b) Test the team’s hypothesis at the 0.05 level of significance.
Population 1: drivers with alcohol
Population 2: drivers without alcohol
Dependent variable: number of errors
Statistical Hypotheses :
Ho: (µ1-µ2)≤0
H1: (µ1-µ2)>0
Decision rule:
Reject Ho at the 0.05 level of significance if t≥1.671, given df=60+602=118
Calculations:
t = [(26.4-18.6) –(0-0)] / 2.4 = 7.8/2.4 = 3.25
Decision:
Reject Ho at the 0.05 level of significance because t =3.25 exceeds
1.658
Interpretation:
Alcohol consumption causes an increase in mean performance errors on
a driving simulator
7. What is the critical F ratio for a one factor ANOVA at the given
alpha level and with the following degrees of freedom (df)?
alpha=.05, df between=3, df within=13
3.41
alpha=.01, df between=3, df within=13
5.74
alpha=.05, df between=6, df within=22
2.55
8. A researcher is interested in whether or not people's gas usage
varies with political affiliation. To find out, she surveys people of
five different political affiliations (Democrat, Green, Libertarian,
Republican, and Independent) and asks them how many gallons
of gas they used in the past week. She surveys five people of
each political affiliation and starts to analyze her data with a one
factor ANOVA (see table). Please complete the table for her and
evaluate the p-value.
Source
Between
Within
Total
SS
523
388
911
df
a)4
b)20
c)24
MS
d)131
e)19
F
f)6.89
a) Is the test statistic (F), significant?
Yes, it is significant at .01, because 6.89 > 4.43 (CV with 4 and 20 df)
b) Given this result, would it be appropriate to calculate the effect size
(η2)? What would that be?
Yes, since F is significant it makes sense to calculate the effect value:
η2 = SSbetween / SStotal = 523/911 = 0.57
According to Cohen’s guidelines this is a large effect (0.25 or greater)
c) With this information, can you calculate Tukey’s HSD for multiple
comparisons? If yes, go ahead and do it. If not, why not?
Yes, you can. Let’s do it with a 0.05 level of significance.
HSD = q √(MSwithin / n)
MSwithin = 19 (form the ANOVA summary table above)
n = 5 (five people per political affiliation)
q (from table) = 4.23 (with α= .05, k=5 (political affiliations), dfwithin=20
HSD = q √(MSerror / n) = 4.23 √(19/5) = 4.23 √(3.8)
= 4.23 * 1.95 = 8.25
Differences between pairs of means across the four political groups would
have to be equal or greater than 8.25 to be considered statistically
significant at the .05 level of significance.
d) What information is missing in this problem for the data in (c) to be
useful?
What is missing is the information regarding the values of the observed
means (gallons of gas used) for each political affiliation. Without that
information the differences between means (for the comparison) cannot
be computed, let alone compared against the value of the Tukey’s HSD
test.
9. ANOVA tests have many steps, so the following question is
designed to walk you through the process of computing an Fscore for an ANOVA test.
A researcher wants to know if red bull or coffee has an effect on alertness
in participants who have been awake for 24 straight hours. They perform
a test of alertness that is operationalized to measure alertness on a scale
of 0-9, with 9 being very alert and 0 being not at all alert. Group 1 is a
control group, who drinks 8 oz. of water, Group 2 drinks 8 oz. of red bull
and Group 3 drinks 8 oz of coffee.
Group 1
1
2
1
Group 2
3
5
4
Group 3
8
4
2
The hypotheses are as follows:
H0: µ1=µ2=µ3 (all three means are the same)
H1: H0 is not true (at least one of the means is different)
a) Compute the mean for each group. Compute the grand mean (the
mean of all the data). Hint: Since in this case each group has the same
number of members, you can calculate the grand mean by finding the
mean of the group means.
Mean1= 1.3333
Mean2= 4
Mean3= 4.6667
Grand mean = 3.3333
b) Compute the Sums of Squares between (follow example shown in
class. See slides).
Compute the Sums of Squares within (follow example shown in class. See
slides).
SSbetween: (T12 /n1) + (T22 /n2) + (T32 /n3) – Grand T2/N
(42/3) + (122/3) + (142/3) - 302/9= 18.6637
SSwithin: (x112 + x122 + x132 + x212 + … + x332) – [(T12 /n1) + (T22
/n2) + (T32 /n3)]
(12 + 22 + 12 + 32 + 52 + 42 + 82 + 42 + 22) – [(42/3) + (122/3) +
(142/3)] = 21.33333
c) Compute the total sums of squares directly using what you computed
in (b)
SStotal=21.3333 + 18.6667 = 40
d) Calculate total degrees of freedom, between degrees of freedom, and
within degrees of freedom:
dftotal = total number of observations-1, that is (n-1)
dfbetween = number of groups -1, that is (k-1)
dfwithin = total number of observations- number of groups, that is (n-k)
dfbetween = 3 -1= 2
dfwithin = 9- 3 = 6
dftotal = 9-1 = 8
e) calculate the mean square within and mean square between groups
MSwithin = SSwithin / dfwithin
MSbetween = SSbetween / dfbetween
MSbetween = 18.6667 / 2 = 9.3333
MSwithin = 21.3333 / 6 = 3.555
f) Given that F= MSbetween/MSwithin, if we suspect a difference between
groups would we expect a large number or a number close to one?
If we expect a difference between groups we would expect a large F value
because that would indicate that the variability between groups is
exceedingly larger than the variability within groups (random error).
g) What does an F-ratio substantially large than 1 mean?
An F-ratio of approximately 1 means that the two variabilities are very
close and so the between group variability could possibly be explained by
the general variation found from the population variation. A larger F
reflects a between group variability being larger than the within group
variability.
h) Calculate the F-ratio using the above equation.
F = 9.3333/3.555 = 2.625
i) Look at the F-table (from our class' website) and given the degrees of
freedom involved in this question No 1, determine the Critical Values at
0.01 and 0.05 level of significance. Given our hypotheses, what can we
conclude about our data and at what level of significance?
With 2 degrees of freedom in the numerator and 6 degrees of freedom in
the denominator, the critical value for F is 5.14 at a .05 significance level
and 10.92 at a .01 significance level.
Since F = 2.625<5.14, we fail to reject the hypothesis. It is possible that
the variation between groups could be explained by the general variation
in the population. There is no evidence that the variability could be
explained by the treatment received by each of the groups.
j) Are we just as well off doing a series of t-tests comparing the means
of: group 1 to group 2; group 1 to group 3; and group 2 to group 3. Why
or why not? Explain.
While using a series of t-tests would be possible here, there is one main
advantage of using an ANOVA test to compare three or more groups: by
using an ANOVA we minimize the probability of a Type 1 error.
10) Cognitive scientists investigating the development of social
behavior studied twelve toddlers in four different conditions. As a
result of these observations, each toddler got an aggression score
for each of the conditions. An ANOVA table led to the following
partial results:
Source
Between
Within
Subject
Error
Total
SS
1200
8800
5500
3300
10000
df
MS
3
44
11
33
47
F
400
4.00*
100
a) Complete the summary table.
b) Would these results lead to reject the null hypothesis that the mean
population aggression scores is the same for all four conditions?
Yes. F is significant at the .05 level.
b) Would it be appropriate to estimate the effect size? Would it be a
small, medium, or large effect according to Cohen’s guidelines?
Yes, it would be appropriate because the null hypothesis was rejected.
The effect size would be:
ηp 2
= SSbetween / (SStotal – SSsubject)
= 1200 / (10000 – 5500)
= 1200 / 4500 = 0.267
or
= SSbetween / (SSbetween + SSerror)
= 1200 / (1200 + 3300)
= 1200 / 4500 = 0.267
According to Cohen’s guidelines this is a large effect (>0.25).
d) If the researchers were to look for differences between specific pairs of
population means across the four conditions (multiple comparisons), what
would the statistic for the Tukey’s HSD be, if tested at a .05 level of
significance?
HSD = q √(MSerror / n)
MSerror = 100
n = 12 (twelve toddlers)
q (from table) = 3.85 (with α= .05, k=4 (conditions), dferror=33 (~30)
HSD = q √(MSerror / n) = 3.85 √(100/12) = 3.85 √8.33
= 3.85 * 2.89 = 11.13
Differences between pairs of means across the four conditions would have
to be equal or greater than 11.13 to be considered statistically significant
at the .05 level of significance.