IE241 Final Exam
1. What is a test of a statistical
hypothesis?
Decision rule to either reject or not
reject the null hypothesis.
2.
In a poll taken among college students, 50 of 200 women
favored a certain proposition, whereas 100 of 300 men favored
it. Is there a real difference of opinion on this proposition?
Ho: pf = pm
Ha: pf ≠ pm
Critical value of t = 1.96 for this 2-tailed test at α = .05 because
t = z because of large sample sizes
50/200 = 0.25 female proportion
100/300 = 0.33 male proportion
150/500 = 0.30 overall proportion
t
p f  pm
pq pq

n f nm

.25  .33
 .08

 1.912
(.3)(. 7) (.3)(. 7) .041833

200
300
Because the observed t > -1.96, we cannot reject the null
hypothesis.
3. What are the components of the
decision rule?
test statistic and critical region
4. Two sets of 12 children each were taught speed swimming by
two different methods. After the instruction was over, each class
swam ten laps in the pool. For group A, the mean time = 8
minutes and the variance = 9. For group B, the mean = 6 minutes
and the variance = 16. Is there a difference between the two
teaching methods?
Note: Hypotheses are always stated in
Ho: μA = μB
terms of parameters, so you can’t have
Ha: μA ≠ μB
Ho: X A  X B
The critical value of t =2.074 for this 2-tailed test with (12-1)+(12-1)
df at α = .05
t
XA  XB
s A2 sB2

nA nB

86
2

 1.39
9 16 1.44

12 12
Because the observed t < 2.074, we cannot reject the null
hypothesis.
5. What is the logic behind the use of the
critical region?
Under the null hypothesis, the test statistic
can fall anywhere in its distribution. But the
probability α that the test statistic will fall in
the critical region (one of its extreme tails) is
very small so if it does fall there, we choose
not to believe in the null hypothesis.
Note: Many of you answered this by
describing the use of the critical region
(rejection of Ho), instead of the logic behind
its use. The question doesn’t ask how it’s
used, but why it is used that way.
6.
Of 64 offspring of a certain cross between guinea pigs, 34 were red,
10 were black, and 20 were white. According to the genetic model,
these colors should be in the ratio 9:3:4. Are the data consistent with
the genetic model?
red
black
white
total
Obs freq
34
10
20
64
Exp freq
(9/16)64 =36
(3/16)64 =12
(4/16)64 =16
64
Ho: data are consistent with model 9:3:4
Ha: data are not consistent with model 9:3:4
2
2
2
(
34

36
)
(
10

12
)
(
20

16
)
2 


 1.44
36
12
16
For this goodness-of-fit test with (3-1) df at α=.05, the critical value
of χ2 = 5.99. Since the observed value of χ2 < 5.99, we cannot
reject Ho. These data do not support the conclusion that the data
are inconsistent with the model.
7. Explain the two types of error when we
do a hypothesis test.
If we reject the null hypothesis when it
is fact true, we make a type 1 error
whose probability α we can control.
If we do not reject the null hypothesis
when it is in fact false, we make a
type 2 error, whose probability β is
minimized for all test with the same α.
8. Given that X is normally distributed, assume that the results of a
sample of size 16 had a mean = 62 and s = 10. (a) Test the
hypothesis that σ = 8. (b) Test the hypothesis that μ = 50.
(a) Ho: σ = 8
(n  1) s 2 (16  1)102
2
 

 23.438
2
2
Ha: σ ≠ 8

8
Note: This is a 2-tailed test because it asks “is σ = 8 or not.”
Critical value of χ2 = 25.0 with (16-1) df at α= .05. Since the
observed χ2 < 25.0, we cannot reject Ho.
t
X   62  50 12


 4.8
s
10
2.5
n
16
(b) Ho: μ = 50
Ha: μ ≠ 50
Note: This is a 2-tailed test because it asks “is μ = 50 or not.”
Critical value of t = 2.131 for this 2-tailed test at α = .05 with 15
df. Since the observed value of t > 2.131, we reject the null
hypothesis and declare that the mean ≠ 50.
9. Why can we not accept the null
hypothesis if the test statistic does not
fall in the critical region?
We never accept the null hypothesis,
but only fail to reject it with the data at
hand because a different, more
powerful test may yield data which will
enable us to reject the null hypothesis.
10. A cigarette manufacturer sent each of two laboratories
samples of tobacco. Each lab made 5 determinations of the
nicotine content in mg. If the results are as shown below, (a) are
the two labs measuring the same thing? (b) Justify the
assumptions, if any, of your analytic method.
Lab 1
Lab 2
mean
24.4
27
variance
5.3
8.5
(a) Ho: μ1= μ2
Ha: μ1≠ μ2
t
X1  X 2
s12 s 22

n1 n2

24.4  27
5.3 8.5

5
5

 2.6
 1.57
1.66
Critical value of t = 2.306 for (5-1)+ (5-1) df at α=.05 for this
2-tailed test. Since the observed t > -2.306, we cannot reject
the null hypothesis.
10(b). The assumption of the t test for
small samples is that the variances are
equal. We can test this by the F test.
2
2
Ho: s1  s 2
s
8.5
F

 1.604
2
2
5.3
s
Ha: s1  s 2
The critical value of F = 6.39 with (51), (5-1) df at α= .05.
Since the observed F < 6.39, we
cannot reject the null hypothesis. Thus
the assumption is justified.
2
2
2
1
11. What is power?
Power is the complement of β, which
means it is the probability of
rejecting the null hypothesis when the
null hypothesis is in fact false.
12.
A metallurgist made 5 determinations of the melting point of
manganese and found an average of 1268˚ and s = 2.92˚. (a) Test
that the mean of this population is no more than 5˚ more than
published value of 1260˚. After thinking about his measurements, the
metallurgist decided that they should have a standard deviation of 2˚
or less. (b) Are his data consistent with this?
t
X   1268  1265
3


 2.297
s
2.92
1.306
n
5
(a) Ho: μ ≤ 1265
Ha: μ > 1265
The critical value of t = 2.132 for this 1-tailed test with (5-1) df at
α= .05. Since the observed t > 2.132, we reject the null hypothesis.
(n  1) s 2
(5  1)2.922

 8.526
2
2
 
(b) Ho: σ2 ≤ 22
2

Ha: σ2 > 22
The critical value of χ2 = 9.49 for this 1-tailed test with (5-1) df at
α= .05. Since the observed χ2 < 9.49, we cannot reject the null
hypothesis.
2
13. If alpha is .05 and we reject the null,
what is the probability that we are right?
There is no probability that we are right.
After we make the decision, we are
either right or wrong and we do not
know which. But the data support the
decision we made.
14. A researcher classified 172 children according to intelligence
and apparent family income, with results as shown below. Are
these variables related?
Dull
Average
Very smart
8 (13)
32 (34)
23 (16)
63
Well clothed
14 (15)
46 (41)
15 (19)
75
Poorly clothed
13 ( 7)
16 (19)
5 ( 8)
34
Very well clothed
Total
35
94
43
Total
172
The expected values for cellij are obtained by (ni.)(n.j)/n ; the
expected values for the cells are shown in parentheses. The
critical value of χ2 = 9.49 with (3-1)(3-1) df at α=.05.
(8  13) 2 (32  34) 2 (23  16) 2 (14  15) 2 (46  41) 2
 




13
34
16
15
41
(15  19) 2 (13  7) 2 (16  19) 2 (5  8) 2




 13.363
19
7
19
8
2
Since the observed χ2 > 9.49, we reject the null hypothesis
and declare the variables related.
15. Explain how are hypotheses are classified.
Hypotheses are either simple or composite.
A simple hypothesis is a test of a particular
value of the alternative hypothesis against the
value of the null hypothesis.
A composite hypothesis is a test of more
than one value of the alternative hypothesis
against the value of the null hypothesis.
Note: Many of you put null and alternative,
so I gave you half credit.
16. Twelve pairs of students have been matched with respect to intelligence.
One member of each pair is given a practice test before a college
admissions test, and the other is not. The test scores are shown below.
Is the practice test effective? If so, compute a 95% confidence interval
for the difference between the means. Explain why the interval did or
did not cover 0.
Pair
1
2
3
4
5
6
7
8
9
10
11
12
Stimulant
75
73
83
83
92
97
92
85
94
91
87
93
No Stim
74
72
85
81
91
96
92
83
95
89
86
93
Diff
1
1
-2
2
1
1
0
2
-1
2
1
0
Ho: μdiff = 0
Ha: μdiff > 0
X diff  0.67
sdiff  1.23
t
X diff 0.67 0.67


 1.876
s diff
1.23 0.355
12
n
Note: This is a 1-tailed test because the question asks if the practice
test is effective, that is, is it better than no practice test.
The critical value of t = 1.796 with (12-1) df for this 1-tailed test at α
= .05. Since the observed t > 1.796, we reject the null hypothesis.
16 (b)
The 95% confidence interval for the mean
difference is
1.23
0.67  (2.201)
 0.67  0.782
12
The confidence interval is
-0.115 < μ < 1.45
This confidence interval covers 0 even though the
null hypothesis was rejected because the test was
1-tailed and confidence intervals are always
2-sided. This is why the t value has changed from
1.796 to 2.201.
17. Explain how are statistical tests are
classified.
Statistical tests may be either 1-tailed or 2tailed. In a 1-tailed test, the critical region of
probability α is in either the right or left tail of
the test statistic distribution. In a 2-tailed
test, the critical region consists of both tails
of the test statistic distribution, each with
probability ½ α.
Note: t, χ2, and F are types of test statistics,
not types of statistical tests, but I gave you
half credit.
18. Given the two sets of sample results below, (a) test that the
means differ by no more than 1 unit. (b) Then test the hypothesis
that both samples came from populations with the same variance.
Sample 1
mean
4.13
1.96
variance
2.25
4
15
15
n
(a) Ho: μ1- μ2 ≤ 1
Ha: μ1- μ2 > 1
Sample 2
t
X1  X 2
s12 s 22

n1 n2

4.13  1.96  1
2.25 4

15 15

1.17
 1.813
.6455
The critical value of t = 1.701 for (15-1) + (15-1) df in this 1tailed test at α = .05. Since the observed value of t >1.701, we
reject the null hypothesis.
18 (continued)
(b) Ho: σ12 = σ22
Ha: σ12 ≠ σ22
4
F
 1.78
2.25
The critical value of F = 2.44 for (15-1),(15-1) df at
α= .05. Since the observed F < 2.44, we cannot
reject the null hypothesis.
19. (a) What assumption is required for a
t test? (b) How can you check this
assumption?
The t test requires homogeneity of
variance (equal variances) when df is
small. The test is robust to this
assumption for large samples.
The two variances involved in the t test
can be checked for equality by using an
F test.
20. The progeny of a certain mating were classified by an attribute
into three groups, as shown below. According to the genetic
model, the frequencies should be in the ratio p2 : 2pq : q2. Are the
data consistent with the model if p = .3?
Group 1
Group 2
Group 3
Total
Observed
10
53
46
109
Expected
.32(109) =10
.42(109)=46
.49(109)=53 109
Ho: p=.3
Ha: p ≠.3
(10  10) 2 (53  46) 2 (46  53) 2
 


 1.989
10
46
53
2
The critical value of χ2 for (3-1) df with α=.05 is 5.99. Since the
observed χ2 < 5.99, we cannot reject the null hypothesis.
21. Why do we always put the larger variance
over the smaller variance for the F test?. What
about the left tail?
The two tails of the F distribution are
reciprocals of one another. So if we test by
putting the larger variance over the smaller
variance, we use the right tail and use F as
the critical value of the test statistic.
This is equivalent to putting the smaller
variance over the larger and using 1/F as the
critical value of the test statistic for the left
tail.
22. What is significance testing and how
does it differ from hypothesis testing?
Significance testing is equivalent to
hypothesis testing in all respects except
that the size α of the critical region is
not chosen beforehand, as it is in
hypothesis testing.
Instead it is chosen by the probability
of a value as large as or larger than the
value of the observed test statistic.