1. No category

SOCI 102 - Exam 3 Study Guide Using Sample Exam
Multiple Choice
Identify the choice that best completes the statement or answers the question.
1.In what circumstances is the t statistic used instead of a z-score for a hypothesis test?
a. The t statistic is used when the sample size is n = 30 or larger.
b. The t statistic is used when the population mean is unknown.
c. The t statistic is used when the population variance (or standard deviation) is unknown.
d. The t statistic is used if you are not sure that the population distribution is normal.
ANSWER: C.
2.To calculate a t statistic for a specific sample, what information is needed about the population from which the sample
was obtained?
a. the value for N
b. the value for
c. the value for or 2
d. All of the other options are needed to compute t.
ANSWER: B.
3.The estimated standard error, sM, provides a measure of the "average" or standard discrepancy between __________.
a. a score and the population mean (X and )
b. a sample mean and the population mean (M and )
c. a score and the sample mean (X and M)
d. None of the other options is correct.
ANSWER: B.
4.A sample of n = 25 individuals is selected from a population with = 80, and a treatment is administered to the sample.
Which set of sample characteristics is most likely to lead to a decision that there is a significant treatment effect?
a. M = 85 and small sample variance
b. M = 85 and large sample variance
c. M = 90 and small sample variance
d. M = 90 and large sample variance
ANSWER: C. To put the question another way, which set will produce a more extreme value of t?
If you try some test values, the answer is clear. Lets use 4 as the “small” sample variance, and “100” as the large sample
variance. Any two values would do, as long as the “small” value is actually smaller than the “large” value.
Choice A.
M = 85 and small sample variance
s2  4
s2
4

 .16  .40
n
25
85  80 5
t
  12.50
.40
.4
sM 
Choice B.
M = 85 and large sample variance
s 2  100
s2

n
85  80
t

2
sM 
100
 42
25
5
 2.50
2
Choice C
M = 90 and small sample variance
s2  4
s2
4

 .16  .40
n
25
90  80 10
t

 25
.40
.4
sM 
Choice D.
M = 90 and large sample variance
s 2  100
s2
100

 42
n
25
90  80 10
t

5
2
2
sM 
5.A sample of n = 25 scores is obtained from a population with
will produce the most extreme value for the t statistic?
a. M = 85 with s2 = 10
b. M = 90 with s2 = 10
c. M = 85 with s2 = 100
d. M = 90 with s2 = 100
= 80. Which of the following sets of sample statistics
ANSWER: B. You can use a similar approach as above to see why.
Choice A.
M = 85 with s2 = 10
s2
10

 .40  .63
n
25
85  80
5
t

 7.94
.63
.63
sM 
Choice B.
M = 90 with s2 = 10
s2
10
sM 

 .40  .63
n
25
90  80 10
t

 15.87
.63
.63
Choice C.
M = 85 with s2 = 100
s2
sM 

n
85  80
t

2
Choice D.
100
 42
25
5
 2.50
2
M = 90 with s2 = 100
s2
100

 42
n
25
90  80 10
t

5
2
2
sM 
6.If a sample consists of 30 individuals, then what is the df value for the t statistic?
a. 29
b. 30
c. 31
d. cannot be determined from the information given
ANSWER: A. N-1 = 30-1 = 29.
7.With
a. t =
b. t =
c. t =
d. t =
= .05 the two-tailed critical region for a sample of n = 10 participants would have boundaries of __________.
1.96
1.833
2.262
2.228
ANSWER: C. For n = 10 we would have 9 degrees of freedom, so looking at the table of critical t values we find:
8.A sample of n = 25 scores produces a t statistic of t = –2.05. If the researcher is using a two-tailed test, the correct
statistical decision is __________.
a. The researcher can reject the null hypothesis with = .05 but not with = .01.
b. The researcher can reject the null hypothesis with either = .05 or = .01.
c. The researcher must fail to reject the null hypothesis with either = .05 or = .01.
d. It is impossible to make a decision about H0 without more information.
ANSWER: C. Here the degrees of freedom are 25-1 = 24, so the critical value for t would be:
Since the obtained t (-2.05) does not lie beyond ± 2.064, we can’t reject H0.
9.A research study uses a single sample of participants to evaluate the effect of a treatment. The results of the hypothesis
test are reported as follows: “t(14) = 2.73, p < .05.” Based on this report, how many individuals were in the sample?
a. 13
b. 14
c. 15
d. cannot be determined from the information provided
ANSWER: C. Since df = n – 1, then n = df + 1. So n = 14+1 = 15.
10.Even a very small treatment effect can be statistically significant if __________.
a. the sample size is big and the sample variance is small
b. the sample size and the sample variance are both big
c. the sample size is small and the sample variance is big
d. the sample size and the sample variance are both small
ANSWER: A. As sample sizes get larger, the value of degrees of freedom also gets larger, and the related critical values
of t get smaller. Look at how the values change as you move down each column of the critical values of t
table.
In addition, as sample variance gets smaller, the value of t gets larger. Look at question 4 to see how this is.
11.Which of the following research situations is most likely to use an independent-measures design?
a. Evaluate the effectiveness of a diet program by measuring how much weight is lost during
4 weeks of dieting.
b. Evaluate the effectiveness of a cholesterol medication by comparing cholesterol levels
before and after the medication.
c. Evaluate the difference in verbal skills between 3-year-old girls and 3-year-old boys.
d. Evaluate the development of verbal skills between age 2 and age 3 for a sample of girls.
ANSWER: C. Hard to be both a 3-year-old girl and a 3-year-old boy.
12.What is the pooled variance for the following two samples?
a.
b.
c.
d.
Sample 1: n = 6 and SS = 200.
Sample 2: n = 10 and SS = 300
250
500
500/14
500/16
ANSWER: C.
s 2p 
SS1  SS 2
200  300
500 500



(6  1)  (10  1) 5  9 14
df 1  df 2
13.As the sample variance increases, the value of the t statistic __________.
a. increases (moves away from zero)
b. decreases (moves toward zero)
c. approximates the value of M1 – M2
d. the value of t is not influenced by sample variance
ANSWER: B. The larger the sample variance, the closer to a t of 0 you get. See question 4.
14.Which of the following sets of data would produce the largest value for an independent-measures t statistic?
a. The two sample means are 10 and 12 with sample variances of 20 and 25.
b. The two sample means are 10 and 12 with variances of 120 and 125.
c. The two sample means are 10 and 20 with variances of 20 and 25.
d. The two sample means are 10 and 20 with variances of 120 and 125.
ANSWER: C. Basically, the bigger the difference between the sample means, the bigger the value of t, and n addition, the
smaller the sample variance, the larger the value of t. You put these both together, and you would expect that the larger
difference between the means and smaller sample variance would give you the largest t of all. For simplicity let’s assume
equal sample sizes, let’s say n = 10, so the pooled variance formula can be used as: s M 1  M 2 
Choice A.
The two sample means are 10 and 12 with sample variances of 20 and 25.
s M1  M 2 
t
s12 s 22
20 25



 2  2.5  4.5  2.12
n1 n2
10 10
M 1  M 2 10  12  2


 .94
2.12
2.12
s M1 M 2
Choice B.
The two sample means are 10 and 12 with variances of 120 and 125.
s12 s 22

(pg. 262)
n1 n2
s M1  M 2 
t
s12 s 22
120 125



 12  12.5  24.5  4.95
10 10
n1 n2
M 1  M 2 10  12  2


 .40
4.95
4.95
s M1 M 2
Choice C.
The two sample means are 10 and 20 with variances of 20 and 25.
s M1  M 2 
t
s12 s 22
20 25



 2  2.5  4.5  2.12
10 10
n1 n2
M 1  M 2 10  20  10


 4.72
2.12
2.12
s M1 M 2
Choice D.
The two sample means are 10 and 20 with variances of 120 and 125.
s M1  M 2 
t
s12 s 22
120 125



 12  12.5  24.5  4.95
10 10
n1 n2
M 1  M 2 10  20  10


 2.02
4.95
4.95
s M1 M 2
15.What is the pooled variance for the following two samples?
Sample 1: n = 6 and SS = 56
Sample 2: n = 4 and SS = 40
a. 5
b.
c. 12
d. 9.6
ANSWER: C
s 2p 
SS1  SS 2
56  40
96
96


 12

(6  1)  (4  1) 5  3 8
df 1  df 2
16.Two samples, each with n = 6 subjects, produce a pooled variance of 20. Based on this information, the estimated
standard error for the sample mean difference would be __________.
a. 20/6
b. 20/12
c. the square root of (20/6 + 20/6)
d. the square root of (20/5 + 20/5)
ANSWER: C
s M1  M 2 
s 2p
n1

s 2p
n2

20 20

6
6
17. An independent-measures research study uses two samples, each with n = 8 participants. If the data produce a t
statistic of t = 2.10, then which of the following is the correct decision for a two-tailed hypothesis test?
a. Reject the null hypothesis with = .05 but fail to reject with = .01.
b. Reject the null hypothesis with either = .05 or = .01.
c. Fail to reject the null hypothesis with either = .05 or
d. cannot answer without additional information
= .01.
ANSWER: C. With n = 8 in each group, each group contributes 8-1=7 degrees of freedom.
df = (n1-1)+(n2-1) = (8-1)+(8-1) = 7+7 = 14
Our obtained t (2.10) is not beyond the critical value from the table (±2.145).
18.The alternative hypothesis for an independent-measures t test states __________.
a. 1 – 2 = 0
b. M1 – M2 0
c. 1 – 2 0
d. M1 – M2 = 0
ANSWER: C. That there is a difference between the means.
19.The results of an independent-measures research study are reported as "t(5) = –2.12, p > .05, two tails." How many
subjects participated in the study?
a. 6
b. 4
c. 7
d. insufficient information given
ANSWER: C. You can think of the formula for the degrees of freedom:
df = (n1-1)+(n2-1)
as
df = n1 + n2 - 2
To recalculate the original value of n (total) you need to add 2 to the df value. So 5+2 = 7
20.For an independent-measures research study, the value of Cohen's d or r2 helps to describe __________.
a. the risk of a Type I error
b. the risk of a Type II error
c. how much difference there is between the two treatments
d. whether the difference between the two treatments is likely to have occurred by chance
ANSWER: C. Both Cohen’s d and r2 are measures of effect size.
21.For which of the following situations would a repeated-measures study be appropriate?
a. Compare attitude scores for males versus females.
b. Compare personality scores for individuals diagnosed with an eating disorder and those
who are not diagnosed.
c. Compare salary levels for college graduates and those who did not graduate from college.
d. Compare reaction times before and after taking a pain medication.
ANSWER: D. In all other 3 choices, membership in one group keeps you from being part in the other.
22.What is the value of MD for the following set of difference scores?
a.
b.
c.
d.
Scores: 3, –8, 6, –4, –2
–1
–5
23
23/5 = 4.6
ANSWER: A
MD 
 D  3 - 8  6 - 4 - 2   5  1
n
5
5
23.A research report describing the results from a repeated-measures study states, "The data showed a significant
difference between treatments, t(22) = 4.71, p < .01." From this report you can conclude that __________.
a. a total of 22 individuals participated in the study
b. a total of 23 individuals participated in the study
c. a total of 24 individuals participated in the study
d. It is impossible to determine the number of participants from the information given.
ANSWER: B. For a repeated-measures test the degrees of freedom are based on the number of pairs of scores, since each
pair yields only one difference, so if df = n – 1, then n = df + 1 = 22 + 1 = 23.
24.What is the value of the estimated standard error for the following set of D-scores?
a.
b.
c.
d.
Scores: 4, 8, 4, 4
12
2
4
1
ANSWER: D. First calculate ΣD and ΣD2 so we can calculate SS for the D scores.
ΣD = 4 + 8 + 4 + 4 = 20
ΣD2 = 16 + 64 + 16 + 16 = 112
SS   D 2 
( D ) 2
(20) 2
400
 112 
 112 
 112  100  12
4
4
n
SS
12
12
s2 


4
n 1 4 1 3
sM D 
s2

n
4
1
4
25.For the repeated-measures t statistic, df = __________.
a. n1 + n2 – 2
b. (n1 – 1) + (n2 – 1)
c. n – 1
d. n1 + n2 – 1
ANSWER: C. For a repeated-measures test the degrees of freedom are based on the number of pairs of scores, since each
pair yields only one difference.
26.With = .05 and a sample of n = 12 subjects in a repeated-measures experiment, the two-tailed critical region for the t
statistic has boundaries of __________.
a. t = ±2.228
b. t = ±1.812
c. t = ±1.796
d. t = ±2.201
ANSWER: D. With n = 12 in a repeated-measures test df = n – 1 = 12 – 1 = 11, so looking at the table we get:
27.Which of the following sets of data is most likely to produce a significant t statistic?
a. MD = 2 and SS = 10
b. MD = 2 and SS = 100
c. MD = 10 and SS = 10
d. MD = 10 and SS = 100
ANSWER: C. Another way to ask the question, is which will produce a more extreme value for t. Lets use some value for
n (say n = 10), plug them into equations and see:
Choice A.
MD = 2 and SS = 10
s2 
SS
10
10


 1.11
n  1 10  1 9
sM D 
t
MD
sM D
s2
1.11

 .11  .33
n
10
2

 6.06
.33
Choice B.
MD = 2 and SS = 100
s2 
SS
100 100


 11.11
n  1 10  1
9
sM D 
t
MD
sM D
s2
11.11

 1.11  1.05
n
10
2

 1.90
1.05
Choice C.
MD = 10 and SS = 10
s2 
SS
10
10


 1.11
n  1 10  1 9
sM D 
t
MD
sM D
s2
1.11

 .11  .33
n
10
10

 30.30
.33
Choice D.
MD = 10 and SS = 100
s2 
SS
100 100


 11.11
n  1 10  1
9
sM D 
t
MD
sM D
s2
11.11

 1.11  1.05
n
10
10

 9.52
1.05
Once again we see that the larger difference and the smallest sample variance gives us the largest value for t.
28.An advantage of a repeated-measured design (compared to an independent-measures design) is that it reduces the
contribution of error variability due to __________.
a. MD
b. degrees of freedom
c. the effect of the treatment
d. individual differences
ANSWER: D. Since you are comparing people to themselves, there are no differences between individuals.
Short Answer
You are in charge of recycling at a large company with many branches. A state law mandates that companies such as
yours must recycle at least 25% of their waste. To determine if your company is in compliance with this law, you collect
information from a sample of n = 4 of your branches. Their rates of recycling are: 20, 28, 20, 20. Use an alpha level of
.05.
X
X2
20
400
28
784
20
400
20
400
ΣX = 88
ΣX2 = 1984
H0: μ ≤ 25
H1: μ > 25
M 
X
n

88
 22
4
( X ) 2
SS   X 2 
 1984 
n
SS
48
48
s2 


 16
n 1 4 1 3
sM 
s2
16

 42
n
4
t obt 
22  25  3

 1.50
2
2
(88) 2
7744
 1984 
 1984  1936  48
4
4
df = 4 – 1 = 3
Is there evidence that your company is in compliance? No
How would you report your results? t(3) = -1.50, p = n.s.
The following data are from an independent-measures experiment comparing two treatment conditions. Do these data
indicate a significant difference between the treatments at the .05 level of significance?
Treatment 1
13
9
7
11
ΣX1 = 40
Treatment 2
9
5
5
9
ΣX2 = 28
Treatment 12
169
81
49
121
ΣX12 = 420
H0: μ1-μ2 = 0
H1: μ1-μ2 ≠ 0
X
40
 10
n
4
 X 2  28  7
M2 
n
4
( X 1 ) 2
(40) 2
1600
2
 420 
 420 
 420  400  20
SS1   X 1 
4
4
n1
M1 
1

SS 2   X 2 
2
s 2p 
n2
(28) 2
784
 212 
 212 
 212  196  16
4
4
SS1  SS 2
20  16
36
36



6
df 1  df 2
(4  1)  (4  1) 3  3 6
s M1  M 2 
t obt 
( X 2 ) 2
s 2p
n1

s 2p
n2

6 6
  3  1.73
4 4
M 1  M 2 10  7
3


 1.73
1.73 1.73
s M1  M 2
df = (4-1) + (4-1) = 6
Treatment 22
81
25
25
81
ΣX22 = 212
Is there a difference between the conditions? No. Our critical value for t would be 2.447, and our obtained t was only 1.73.
How would you report your results? t(6) = 1.73, p = n.s.
The following data are from a dependent-measures experiment comparing scores before and after a treatment. Do these
data indicate a significant increase in scores at the .05 level of significance?
Subject
A
B
C
D
Pre-Treatment
34
41
38
29
Post-Treatment
39
48
41
36
D
-5
-7
-3
-7
ΣD = -22
D2
25
49
9
49
ΣD2 = 132
H0: μD ≥ 0
H1: μD < 0
MD 
 D   22  5.5
n
SS   D 2 
4
( D ) 2
 132 
n
SS
11
11
s2 

  3.67
n 1 4 1 3
sM D 
t obt 
(22) 2
484
 132 
 132  121  11
4
4
s2
3.67

 .92  .96
n
4
M D  5.5

 5.74
.96
sM D
df = n – 1 = 4 – 1 = 3
Did the treatment increase the scores? Yes, our obtained value of t (-5.74) is well beyond the critical value of – 2.353, so
we can reject the null hypothesis. In addition, our obtained value of t is also larger than the critical values for p = .025 and
.01, but not larger than the critical value for p = .005, so we can report the p value only as < .01.
How would you report your results? t(3) = -5.74, p < .01
SOCI 102 - E3 - FA09
Answer Section
MULTIPLE CHOICE
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29.
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C
B
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A
C
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D
A
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SHORT ANSWER
30.
ANS:
H0: = 25. For these data, M = 22, s2 = 16, and t(3) = –1.50. Fail to reject H0.
31.
ANS:
For treatment 1, M = 10 and SS = 20. For treatment 2, M = 7 and SS = 16. For these data the pooled variance is 6, the
= 1.73, and the t statistic is t(6) = 1.73. Fail to reject H0.
standard error is
32.
ANS:
The difference scores are 5, 7, –3, 7. For these data, D = 16 and MD = +4.