1) Suppose that your psychology professor, Dr. I. D. Ego, gives a 20 point truefalse quiz to 9 students and wants to know if they were different from groups in
the past who have tended to have an average of 9.0. Their scores from the
current group were: 6, 7, 7, 8, 8, 8, 8, 9, 9, 10. Did the current group perform
differently from those in the past. We'll assume a significance level of a = 0.05.
n=10, “differently from those in the past” suggests two-tailed test
Here we are testing one group (our sample) against typical past performance
(taken as the population), so we should use a one-sample t-test (since we don’t
know the standard deviation of the population)
H0: the current students do not differ from past students ( = 9.0)
HA: the current students do differ from past students ( ≠ 9.0)
mean null distribution = 9.0
df = n – 1 = 10 – 1 = 9
X
X
 X  72 
n
9
6
7
7
8
8
8
8
9
9
10
8.0
X  X  X  X 
2
-2
-1
-1
0
0
0
0
1
1
2
4
1
1
0
0
0
0
1
1
4
SS = 12
s
SS
12

 1.15
n 1
10  1
sX 
s
1.15

 0.37
n
10
t obs 
X   8.0.  9.0

 2.74
sX
0.37
From the t-table, df = 9, two tailed, tcrit = ±2.262
2.74 is in the critical region, so we can reject the null hypothesis and conclude
that the evidence supports the hypothesis that the current students are
performing differently than past students.
2) The effects of fatigue on mental alertness. Group 1 stays awake for 24 hours,
group 2 gets to go to sleep. All groups then get tested to see how well they
detect a light on screen.
two groups: n1 = 5, n2 = 10, mean1 = 35, mean2 = 24, SS1 = 120, SS2 = 270
Here we are testing two groups. There is nothing to suggest that the groups are
related (and there are uneven numbers in the groups too), so we should use a
two-independnet-sample t-test.
n=15, nothing given suggests that the theory predicts how fatigue effects mental
alertness, so assume a two-tailed test
H0: the two groups don’t differ in mental alertness ( – 2= 0.0)
HA: the two groups do differ in mental alertness ( – 2 ≠ 0.0)
df 1= n1 – 1 = 5 – 1 = 4
df 2= n2 – 1 = 10 – 1 = 9
df T= n1 + n2 – 2 = 13
sP2 
SS1  SS2 120  270

 30
df1  df2
49
s X1  X2 
t obs 
X
1
sP2 sP2


n1 n1
30 30

3
5 10
 X2  1  2 
s X1  X2

35  24  (0)
 3.0
3
From the t-table, df = 13, two tailed, tcrit = ±2.162
3.0 is in the critical region, so we can reject the null hypothesis and conclude
that the evidence supports the hypothesis that the groups are different with
respect to mental alertness.
3) Suppose that you want to compare the heights of men and women in your
morning course. Your prediction is that the men are taller than the women in the
course.
 men's heights: 67, 73, 74, 70, 70, 75, 73, 68, 69
 women's heights: 69, 63, 67, 64, 61, 66, 60, 63, 63
Here we are testing two groups. There is nothing to suggest that the groups are
related so we should use a two-independent-sample t-test.
n=15, nothing given suggests that the theory predicts how fatigue effects mental
alertness, so assume a two-tailed test
H0: the two groups don’t differ in height ( – 2= 0.0)
HA: the two groups do differ in height ( – 2 ≠ 0.0)
df 1= n1 – 1 = 9 – 1 = 8
df 2= n2 – 1 = 9 – 1 = 8
df T= n1 + n2 – 2 = 16
mean1 = 71.0, mean2 = 64.0
SS1 = 64.0 , SS2 = 66.0
sP2 
SS1  SS2 64  66

 8.125
df1  df2
88
s X1  X2 
t obs 
X
1
sP2 sP2


n1 n1
8.125 8.125

 1.34
9
9
 X 2  1  2 
s X1  X2

71  64  (0)
 5.22
1.34
From the t-table, df = 16, two tailed, tcrit = ±2.12
5.22 is in the critical region, so we can reject the null hypothesis and conclude
that the evidence supports the hypothesis that the groups are different with
respect to height.
4) Suppose that your physics professor, Dr. M. C. Squared, gives a 20 point
true-false quiz to 10 students and wants to know if they did worse than guessing
(think about what score you would get if just guessing). Their scores were: 6, 7,
7, 8, 8, 8, 8, 9, 9, 10.
This problem is like the one earlier, except that instead of comparing the sample
to a population, you compare it against a theoretical population of guesses. If
you assume that the chance performance on a T/F test is 0.5, then average
performance on a 20 question test would be 10/20. Thus, the mean of the
treatment distribution will be 10.
n=10, “worse than guessing” suggests 1-tailed test
Here we are testing one group (our sample) against typical past performance
(taken as the population), so we should use a one-sample t-test (since we don’t
know the standard deviation of the population)
H0: the current students do not differ from chance (or do better) ( ≥ 10.0)
HA: the current students do worse than chance ( < 10.0)
mean null distribution = 9.0
df = n – 1 = 9 – 1 = 8
X
X
 X  72 
n
9
6
7
7
8
8
8
8
9
9
10
8.0
X  X  X  X 
2
-2
-1
-1
0
0
0
0
1
1
2
4
1
1
0
0
0
0
1
1
4
SS = 12
s
SS
12

 1.15
n 1
10  1
sX 
s
1.15

 0.37
n
10
t obs 
X   8.0.  10.0

 5.40
sX
0.37
From the t-table, df = 9, one tailed, tcrit = ±1.833
5.40 is in the critical region, so we can reject the null hypothesis and conclude
that the evidence supports the hypothesis that the current students are
performing worse than chance.
5) An instructor asks his statistics class, on the first day of classes, to rate how
much they like statistics, on a scale of 1 to 10 (1 hate it, 10 love it). Then, at the
end of the semester, the instructor asks the same students, the same question.
The instructor wants to know if taking the stats course had an impact on the
students feelings about statistics.
The results of the two ratings are presented below. D stands for the difference
between the pre- and post-ratings for each individual.
Student
1
2
3
4
5
6
7
8
9
10
Pre-test
(first day)
1
3
4
7
2
2
4
3
6
8
Post-test
(end of semester)
4
5
6
8
3
2
6
4
6
6
D
DD
D  D
3
2
2
1
1
0
2
1
0
-2
D  1.0
2
1
1
0
0
-1
1
0
-1
-3
4
1
1
0
0
1
1
0
1
9
SSD=18
2
H0: the feelings about do not differ from before and after the class (D = 10.0)
HA: the feelings about do not differ from before and after the class (D ≠ 10.0)
df D= nD – 1 = 10 – 1 = 9
sD 
SSD
18

 1.41
nD  1
10  1
sD 
sD
1.41

 0.45
nD
10
df = 9, two-tailed, tcrit = 2.262.
t obs 
D   1.0  0.0

 2.24
sD
0.45