Devore J.L., Berk K.N. Modern Mathematical Statistics with Applications
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Thus c is such that 1 F(c), the area under the standard normal curve to the right
of c, is .025 (and not .05!). From Section 4.3 or Appendix Table A.3 , c ¼ 1.96, and
the rejecti on region is z 1.96 or z 1.96. For any a, the two-tailed rejection
region z z
a/2
or z z
a/2
has type I error probability a (since area a/2 is captured
under each of the two tails of the z curve). Again, the key reason for using the
standardized test statistic Z is that because Z has a known distribution when H
0
is
true (standard normal), a rejection region with desired type I error probability is
easily obtained by using an appropriate critical value.
The test procedure for Case I is summarized in the accompanying box, and
the corresponding rejection regions are illustrated in Figure 9.2.
Null hypothesis: H
0
: m ¼ m
0
Test statistic value: z ¼
x m
0
s=
ffiffiffi
n
p
Alternative Hypothesis Rejection Region for Level a Test
H
a
: m > m
0
z z
a
(upper-tailed test)
H
a
: m < m
0
z z
a
(lower-tailed test)
H
a
: m 6¼ m
0
either z z
a/2
or z z
a/2
(two-tailed test)
Use of the following sequence of steps is recommended when testing hypotheses
about a parameter.
1. Identify the parameter of interest and describe it in the context of the problem
situation.
00
z
a
−z
a
−z
a/2
z
a/2
0
Rejection region: either
z curve (probability distribution of test statistic Z when H
0
is true)
abc
Shaded area
= a = P(type I error)
Total shaded area
= a = P(type I error)
Shaded area
= a /2
Shaded
area = a /2
Rejection region: z Ï z
a
Rejection region: z £ −z
a
z Ï z
a/2
or z £ −z
a/2
Figure 9.2 Rejection regions for z tests: (a) upper-tailed test; (b) lower-tailed test; (c) two-tailed test
438
CHAPTER 9 Tests of Hypotheses Based on a Single Sample
2. Determine the null value and state the null hypothesis.
3. State the appropriate alternative hypothesis.
4. Give the formula for the computed value of the test statistic (substituting the null
value and the known values of any other parameters, but not those of any
sample-based quantities).
5. State the rejection region for the selected significance level a.
6. Compute any necessary sample quantities, substitute into the formula for the test
statistic value, and compute that value.
7. Decide whether H
0
should be rejected and state this conclusion in the problem
context.
The formulation of hypotheses (steps 2 and 3) should be done before
examining the data.
Example 9.6 A manufacturer of sprinkler systems used for fire protection in office buildings claims
that the true average system-activation temperature is 130
. A sample of n ¼ 9
systems, when tested, yields a sample average activation temperature of 131.08
F.
If the distribution of activation times is normal with standard deviation 1.5
F, does
the data contradict the manufacturer’s claim at significance level a ¼ .01?
1. Parameter of interest: m ¼ true average activation temperature.
2. Null hypothesis: H
0
: m ¼ 130 (null value ¼ m
0
¼ 130).
3. Alternative hypothesis: H
a
: m 6¼ 130 (a departure from the claimed value in
either direction is of concern).
4. Test statistic value:
z ¼
x m
0
s=
ffiffiffi
n
p
¼
x 130
1:5=
ffiffiffi
n
p
5. Rejection region: The form of H
a
implies use of a two-tailed test with rejection
region either z z
.005
or z z
.005
. From Section 4.3 or Appendix Table A.3,
z
.005
¼ 2.58, so we reject H
0
if either z 2.58 or z 2.58.
6. Substituting n ¼ 9 and
x ¼ 131:08;
z ¼
131:08 130
1:5=
ffiffiffi
9
p
¼
1:08
:5
¼ 2:16
That is, the observed sample mean is a bit more than 2 standard deviations above
what would have been expected were H
0
true.
7. Th e computed value z ¼ 2.16 does not fall in the rejection region
(2.58 < 2.16 < 2.58), so H
0
cannot be rejected at significance level .01. The
data does not give strong support to the claim that the true average differs from
the design value of 130.
■
Another view of the analysis in the previous example involves calculating a 99% CI
for m based on Eq uation 8.5:
x 2:58s=
ffiffiffi
n
p
¼ 131:08 2:58ð1:5=
ffiffiffi
9
p
Þ¼131:08 1:29 ¼ð129:79; 132 :37Þ
9.2 Tests About a Population Mean 439
Notice that the interval includes m
0
¼ 130, and it is not hard to see that the 99% CI
excludes m
0
if and only if the two-tailed hypothesis test rej ects H
0
at level .01.
In general, the 100(1 a)% CI excludes m
0
if and only if the two-tailed hypothesis
test rejects H
0
at level a. Although we will not always call attention to it, this kind
of relationship between hypothesis tests and confidence intervals will occur over
and over in the remainder of the book. It should be intuitively reasonable that the
CI will exclude a value when the corresponding test rejects the value. There is a
similar relationship between lower-tailed tests and upper confidence bounds, and
also between upper-tailed tests and lower confi dence bounds.
b and Sample Size Determination The z tests for Case I are among the few in
statistics for which there are simple formulas available for b, the probability of a
type II error. Consider first the upper-tailed test with rejection region z z
a
. This is
equivalent to
x m
0
þ z
a
s=
ffiffiffi
n
p
,soH
0
will not be rejected if x < m
0
þ z
a
s=
ffiffiffi
n
p
.
Now let m
0
denote a particular value of m that exceeds the null value m
0
. Then,
bðm
0
Þ¼PðH
0
is not rejected when m ¼ m
0
Þ
¼ Pð
X < m
0
þ z
a
s=
ffiffiffi
n
p
when m ¼ m
0
Þ
¼ P
X m
0
s=
ffiffiffi
n
p
< z
a
þ
m
0
m
0
s=
ffiffiffi
n
p
when m ¼ m
0
¼ F z
a
þ
m
0
m
0
s=
ffiffiffi
n
p
As m
0
increases, m
0
m
0
becomes more negative, so b(m
0
) will be small when m
0
greatly exceeds m
0
(because the value at which F is evaluated will then be quite
negative). Error probabilities for the lower-tailed and two-tai led tests are derived in
an analogous manner.
If s is large, the probability of a type II error can be large at an alternative
value m
0
that is of particular concern to an investigator. Suppose we fix a and also
specify b for such an alter native value. In the sprinkler example, company officials
might view m
0
¼ 132 as a very substantial departure from H
0
: m ¼ 130 and
therefore wish b(132) ¼ .10 in addition to a ¼ .01. More generally, consider the
two restrictions P(type I error) ¼ a and b(m
0
) ¼ b for specified a, m
0
, and b. Then
for an upper-tailed test, the sample size n should be chosen to satisfy
F z
a
þ
m
0
m
0
s=
ffiffiffi
n
p
¼ b
This implies that
z
b
¼
z critical value that
captures lower tail area b
¼ z
a
þ
m
0
m
0
s=
ffiffiffi
n
p
It is easy to solve this equation for the desired n. A parallel argument yields the
necessary sample size for lower- and two-tailed tests as summarized in the next
box.
440 CHAPTER 9 Tests of Hypotheses Based on a Single Sample
Alternative Hypothesis Type II Error Probability b(m
0
) for a Level a Test
H
a
: m > m
0
F z
a
þ
m
0
m
0
s=
ffiffiffi
n
p
H
a
: m < m
0
1 F z
a
þ
m
0
m
0
s=
ffiffiffi
n
p
H
a
: m 6¼ m
0
F z
a=2
þ
m
0
m
0
s=
ffiffiffi
n
p
F z
a=2
þ
m
0
m
0
s=
ffiffiffi
n
p
where F(z) ¼ the standard normal cdf.
The sample size n for which a level a test also has b(m
0
) ¼ b at the alternat ive
value m
0
is
n ¼
sðz
a
þ z
b
Þ
m
0
m
0
2
for a one - tailed
(upper or lower) test
sðz
a=2
þ z
b
Þ
m
0
m
0
2
for a two - tailed test
(an approximate solution)
8
>
>
>
<
>
>
>
:
Example 9.7 Let m denote the true average tread life of a type of tire. Consid er testing H
0
:
m ¼ 30,000 versus H
a
: m > 30,000 based on a sample of size n ¼ 16 from a normal
population distribution with s ¼ 1500. A test with a ¼ .01 requires z
a
¼ z
.01
¼ 2.33. The probability of making a type II error when m ¼ 31,000 is
bð31;000Þ¼F 2:33 þ
30;000 31;000
1500=
ffiffiffiffiffi
16
p
¼ F :34ðÞ¼:3669
Since z
.1
¼ 1.28, the requirement that the level .01 test also have b(31,000) ¼ .1
necessitates
n ¼
1500ð2:33 þ 1:28Þ
30;000 31;000
2
¼ð5:42Þ
2
¼ 29:32
The sample size must be an integer, so n ¼ 30 tires should be used.
■
Case II: Large-Sample Tests
When the sample size is large, the z tests for Case I are easily modified to yield
valid test procedures without requiring either a normal population distribution or
known s. The key result was used in Chapter 8 to justify large-sample confidence
intervals: A large n implies that the sample standard deviation s will be close to s
for most samples, so that the standardized variable
Z ¼
X m
S=
ffiffiffi
n
p
9.2 Tests About a Population Mean 441
has approximately a standard normal distribution. Substitution of the null value
m
0
in place of m yields the test statistic
Z ¼
X m
0
S=
ffiffiffi
n
p
which has approximately a standard normal distribution when H
0
is true. The use of
rejection regions given previously for Case I (e.g., z z
a
when the alternative
hypothesis is H
a
: m > m
0
) then results in test procedures for which the significance
level is approximately (rather than exactly) a. The rule of thumb n > 40 will again
be used to characterize a large sample size.
Example 9.8 A sample of bills for meals was obtained at a restaurant (by Erich Brandt). For each
of 70 bills the tip was found as a percentage of the raw bill (before taxes). Does it
appear that the population mean tip percentage for this restaurant exceeds the
standard 15%? Here are the 70 tip percentages:
14.21 20.24 20.10 14.94 15.69 15.04 12.04 20.16 17.85 16.35
19.12 20.37 15.29 18.39 27.55 16.01 10.94 13.52 17.42 14.48
29.87 17.92 19.74 22.73 14.56 15.16 16.09 16.42 19.07 13.74
13.46 16.79 19.03 19.19 19.23 12.39 16.89 18.93 13.56 17.70
11.48 13.96 21.58 11.94 19.02 17.73 20.07 40.09 19.88 22.79
15.23 16.09 19.19 11.91 18.21 15.37 16.31 16.03 48.77 12.31
21.53 12.76 18.07 14.11 15.86 20.67 15.66 18.54 27.88 13.81
Figure 9.3 shows a descriptive summary obtained from MINITAB. The sample mean
tip percentage is >15. Notice that the distribution is positively skewed because there
are some very large tips (and a normal probability plot therefore does not exhibit a
linear pattern), but the large-sample z tests do not require a normal population
distribution.
1. m ¼ true average tip percentage
2. H
0
: m ¼ 15
22.515.0 45.037.530.0
2716 1918
Mean
Median
95% Confidence Intervals
Anderson-Darting Normality Test
A-Squared 4.17
P-Value < 0.005
Mean 17.986
StDev 5.937
Variance 35.247
Skewness 2.9391
Kurtosis 12.0154
70N
Minimum 10.940
1st Quartile
3st
Quartile
14.540
Median 16.840
19.358
Maximum
48.770
95% Confidence Interval for Mean
16.571 19.402
95% Confidence Interval for Median
15.913 18.402
95% Confidence Interval for StDev
5.090 7.124
** * * *
Figure 9.3 MINITAB descriptive summary for the tip data of Example 9.8
442
CHAPTER 9 Tests of Hypotheses Based on a Single Sample
3. H
a
: m > 15
4. z ¼
x 15
s=
ffiffiffi
n
p
5. Using a test with a significance level .05, H
0
will be rejected if z 1.645 (an
upper tailed test).
6. With n ¼ 70,
x ¼ 17:99, and s ¼ 5.937,
z ¼
17:99 15
5:937=
ffiffiffiffiffi
70
p
¼
2:99
:7096
¼ 4:21
7. Since 4.21 > 1.645, H
0
is rejected. There is evidence that the population mean
tip percentage exceeds 15%.
■
Determination of b and the necessary sample size for these large-sample tests
can be based either on specifying a plausible value of s and using the Case I
formulas (even though s is used in the test) or on using the methods to be introduced
shortly in connection with Case III.
Case III: A Normal Population Distribution
with Unknown s
When n is small, the Central Limit Theorem (CLT) can no longer be invoked to
justify the use of a large-sample test. We faced this same difficulty in obtaining a
small-sample confidence interval (CI) for m in Chapter 8. Our approach here will be
the same one used there: We will assume that the population distribution is at least
approximately normal and describe test procedures whose validity rests on this
assumption. If an investigator has good reason to believe that the population
distribution is quite nonnormal, a distribution-free test from Chapter 14 can be
used. Alternatively, a statistician can be consulted regarding procedures valid for
specific families of population distributions other than the normal family. Or a
bootstrap procedure can be developed.
The key result on which tests for a normal population mean are based was
used in Chapter 8 to derive the one-sample t CI: If X
1
, X
2
, ... , X
n
is a random
sample from a normal distribution, the standardize d variable
T ¼
X m
S=
ffiffiffi
n
p
has a t distribution with n 1 degrees of freedom (df). Consider testing H
0
:
m ¼ m
0
against H
a
: m > m
0
by using the test statistic ðX m
0
Þ=ðS=
ffiffiffi
n
p
Þ. That is,
the test statistic results from standardizing
X under the assumption that H
0
is true
(using S=
ffiffiffi
n
p
, the estimated standard deviation of
X, rather than s=
ffiffiffi
n
p
). When H
0
is
true, the test statistic has a t distributio n with n 1 df. Knowledge of the test
statistic’s distribution when H
0
is true (the “null distribution”) allows us to con-
struct a rejection region for which the type I error probability is controlled at the
desired level. In particular, use of the upper-tail t critical value t
a,n1
to specify the
rejection region t t
a,n1
implies that
9.2 Tests About a Population Mean 443
P type I errorðÞ¼PH
0
is rejected when it is trueðÞ
¼ PðT t
a
;n1
when T has a t distribution with n 1dfÞ
¼ a
The test statistic is really the same here as in the large-sample case but is
labeled T to emphasize that its null distribution is a t distribution with n 1df
rather than the standard normal (z) distribution. The rejection region for the t test
differs from that for the z test only in that a t critical value t
a,n1
replaces the
z critical value z
a
. Similar comments apply to alternatives for which a lower-tailed
or two-tailed test is appropriate.
THE
ONE-SAMPLE
t TEST
Null hypothesis: H
0
: m ¼ m
0
Test statistic value: t ¼
x m
0
s=
ffiffiffi
n
p
Alternative Hypothesis Rejection Region for a Level a Test
H
a
: m > m
0
t t
a,n1
(upper-tailed)
H
a
: m < m
0
t t
a,n1
(lower-tailed)
H
a
: m 6¼ m
0
either t t
a/2,n1
or t t
a/2,n1
(two-tailed)
Example 9.9 A well-designed and safe workplace can contribute greatly to increased productiv-
ity. It is especially important that workers not be asked to perform tasks, such as
lifting, that exceed their capabilities. The accompanying data on maximum weight
of lift (MAWL, in kg) for a frequency of four lifts/min was reported in the article
“The Effects of Speed, Frequency, and Load on Measured Hand Forces for a
Floor-to-Knuckle Lifting Task” (Ergonomics, 1992: 833–843); subjects were
randomly selected from the population of healthy males age 18–30. Assuming
that MAWL is normally distributed, does the following data suggest that the
population mean MAWL exceeds 25?
25.8 36.6 26.3 21.8 27.2
Let’s carry out a test using a significance level of .05.
1. m ¼ population mean MAWL
2. H
0
: m ¼ 25
3. H
a
: m > 25
4. t ¼
x 25
s=
ffiffiffi
n
p
5. Reject H
0
if t t
a, n1
¼ t
.05,4
¼ 2.132.
6. Sx
i
¼ 137.7 and Sx
2
i
¼ 3911:97, from which x ¼ 27:54, s ¼ 5.47, and
444 CHAPTER 9 Tests of Hypotheses Based on a Single Sample
t ¼
27:54 25
5:47=
ffiffiffi
5
p
¼
2:54
2:45
¼ 1:04
The accompanying MINITAB output from a request for a one-sample t test has
the same calculated values (the P -value is discussed in Section 9.4).
Test of mu ¼ 25.00 vs mu > 25.00
Variable N Mean StDev SE Mean T P-Value
mawl 5 27.54 5.47 2.45 1.04 0.18
7. Since 1.04 does not fall in the rejection region (1.04 < 2.132), H
0
cannot be
rejected at significance level .05. It is still plausible that m is (at most) 25. ■
b and Sample Size Determination The calculation of b at the alternative value m
0
in Case I was carried out by expressing the rejection region in terms of x (e.g.,
x m
0
þ z
a
s=
ffiffiffi
n
p
) and then subtracting m
0
to standardize correctly. An equivalent
approach involves noting that when m ¼ m
0
, the test statistic Z ¼ðX m
0
Þ=ðs=
ffiffiffi
n
p
Þ
still has a normal distribution with variance 1, but now the mean value of Z is
given by ðm
0
m
0
Þ=ðs=
ffiffiffi
n
p
Þ. That is, when m ¼ m
0
, the test statistic still has a
normal distribution though not the standard normal distribution. Because of this,
b(m
0
) is an area under the normal curve corresponding to mean value
ðm
0
m
0
Þ=ðs=
ffiffiffi
n
p
Þ and variance 1. Both a and b involve working with normally
distributed variables.
The calculation of b(m
0
) for the t test is much less straightforward. This
is because the distribution of the test statistic T ¼ð
X m
0
Þ=ðS=
ffiffiffi
n
p
Þ is quite
complicated when H
0
is false and H
a
is true. Thus, for an upper-tailed test,
determining
bðm
0
Þ¼PðT < t
a;n1
when m ¼ m
0
rather than m
0
Þ
involves integrating a very unpleasant density function. This must be done numeri-
cally, but fortunately it has been done by rese arch statisticians for both one- and
two-tailed t tests. The resu lts are summarized in graphs of b that appear in
Appendix Table A.16. There are four sets of graphs, corresponding to one-tailed
tests at level .05 and level .01 and two-tailed tests at the same levels.
To understand how these graphs are used, note first that both b and the
necessary sample size n in Case I are functions not just of the absolute difference
|m
0
m
0
| but of d ¼ |m
0
m
0
|/s. Suppose, for example, that |m
0
m
0
| ¼ 10.
This departure from H
0
will be much easier to detect (smaller b) when s ¼ 2,
in which case m
0
and m
0
are 5 population standard deviations apart, than whe n
s ¼ 10. The fact that b for the t test depends on d rather than just |m
0
m
0
|is
unfortunate, since to use the graphs one must have some idea of the true value of s.
A conservative (large) guess for s will yield a conse rvative (large) v alue of
b(m
0
) and a conservative estimate of the sample size necessary for prescribed
a and b(m
0
).
Once the alternative m
0
and value of s are selected, d is calculated and its
value located on the horizontal axis of the relevant set of curves. The value of b
is the height of the n 1 df curve above the value of d (visual interpolation is
necessary if n 1 is not a value for which the corresponding curve appears), as
illustrated in Figure 9.4.
9.2 Tests About a Population Mean 445
Rather than fixing n (i.e., n 1, and thus the particular curve from which b is
read), one might prescribe both a (.05 or .01 here) and a value of b for the chosen m
0
and s. After computing d, the point (d, b) is located on the relevant set of graphs.
The curve below and closest to this point gives n 1 and thus n (again, interpola-
tion is often necessary).
Example 9.10 The true average voltage drop from collector to emitter of insulated gate bipolar
transistors of a certain type is supposed to be at most 2.5 V. An investigator selects a
sample of n ¼ 10 such transistors and uses the resulting voltages as a basis for testing
H
0
: m ¼ 2.5 versus H
a
: m > 2.5 using a t test with significance level a ¼ .05. If the
standard deviation of the voltage distribution is s ¼ .100, how likely is it that H
0
will
not be rejected when m ¼ 2.6? With d ¼ |2.5 2.6|/.100 ¼ 1.0, the point on the b
curve at 9 df for a one-tailed test with a ¼ .05 above 1.0 has height approximately .1,
so b .1. The investigator might think that this is too large a value of b for such a
substantial departure from H
0
and may wish to have b ¼ .05 for this alternative value
of m. Since d ¼ 1.0, the point (d, b) ¼ (1.0, .05) must be located. This point is very
close to the 14 df curve, so using n ¼ 15 will give both a ¼ .05 and b ¼ .05 when
the value of m is 2.6 and s ¼ .10. A larger value of s would give a larger b for this
alternative, and an alternative value of m closer to 2.5 would also result in an
increased value of b.
■
Most of the widely used statistical computer packages will also calculate
type II error probabilities and determine necessary sample sizes. As an example, we
asked MINITAB to do the calculations from Example 9.10. Its computations are
based on power, which is simply 1 b. We want b to be small, which is equivalent
to asking that the power of the test be large. For example, b ¼ .05 corresponds to a
value of .95 for power. Here is the resulting MINITAB output.
Power and Sample Size
Testing mean ¼ null (versus > null)
Calculating power for mean
¼ null + 0.1
1
0
d
Value of d corresponding to specified alternative m⬘
b curve for n − 1 df
b when m = m⬘
Figure 9.4 A typical b curve for the t test
446
CHAPTER 9 Tests of Hypotheses Based on a Single Sample
Alpha ¼ 0.05 Sigma ¼ 0.1
Sample
Size Power
10 0.8975
Power and Sample Size
1-Sample t Test
Testing mean
¼ null (versus > null)
Calculating power for mean
¼ null + 0.1
Alpha
¼ 0.05 Sigma ¼ 0.1
Sample
Size
Target
Power
Actual
Power
13 0.9500 0.9597
Notice from the second part of the output that the sample size neces sary to obtain a
power of .95 ( b ¼ .05) for an upper-t ailed test with a ¼ .05 when s ¼ .1 and m
0
is
.1 larger than m
0
is only n ¼ 13, whe reas eyeballing our b curves gave 15. When
available, this type of software is more tru stworthy than the curves.
Exercises Section 9.2 (15–35)
15. Let the test statistic Z have a standard normal
distribution when H
0
is true. Give the significance
level for each of the following situations:
a. H
a
: m > m
0
, rejection region z 1.88
b. H
a
: m < m
0
, rejection region z 2.75
c. H
a
: m 6¼ m
0
, rejection region z 2.88 or z
2.88
16. Let the test statistic T have a t distribution when
H
0
is true. Give the significance level for each of
the following situations:
a. H
a
: m > m
0
,df¼ 15, rejection region
t 3.733
b. H
a
: m < m
0
, n ¼ 24, rejection region
t 2.500
c. H
a
: m 6¼ m
0
, n ¼ 31, rejection region t 1.697
or t 1.697
17. Answer the following questions for the tire prob-
lem in Example 9.7.
a. If
x ¼ 30; 960 and a level a ¼ .01 test is used,
what is the decision?
b. If a level .01 test is used, what is b(30,500)?
c. If a level .01 test is used and it is also required
that b(30,500) ¼ .05, what sample size n is
necessary?
d. If
x ¼ 30; 960, what is the smallest a at which
H
0
can be rejected (based on n ¼ 16)?
18. Reconsider the paint-drying situation of Example
9.2, in which drying time for a test specimen is
normally distributed with s ¼ 9. The hypotheses
H
0
: m ¼ 75 versus H
a
: m < 75 are to be tested
using a random sample of n ¼ 25 observations.
a. How many standard deviations (of
X) below
the null value is
x ¼ 72:3?
b. If
x ¼ 72:3, what is the conclusion using
a ¼ .01?
c. What is a for the test procedure that rejects H
0
when z 2.88?
d. For the test procedure of part (c), what is
b(70)?
e. If the test procedure of part (c) is used, what n
is necessary to ensure that b(70) ¼ .01?
f. If a level .01 test is used with n ¼ 100, what is
the probability of a type I error when m ¼ 76?
19. Themeltingpointofeachof16samplesofabrand
of hydrogenated vegetable oil was determined,
resulting in
x ¼ 94:32. Assume that the distribution
of melting point is normal with s ¼ 1.20.
a. Test H
0
: m ¼ 95 versus H
a
: m 6¼ 95 using a
two-tailed level .01 test.
b. If a level .01 test is used, what is b(94), the
probability of a type II error when m ¼ 94?
c. What value of n is necessary to ensure that
b(94) ¼ .1 when a ¼ .01?
9.2 Tests About a Population Mean 447