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MATH 2441
Probability and Statistics for Biological Sciences
Testing Hypotheses Involving Single Population Means
This section is mainly a recap of the methodology for testing hypotheses of the form:
H0:  = 0
vs
H0:  = 0
or
vs
HA:  > 0
(SPMHT - 1)
HA:  < 0
or of the form
H0:  = 0
vs
(SPMHT - 2)
H0:   0
where
 is the mean value of some target population
and
0 is some specific numerical value relevant to the application at hand.
We've already used hypotheses of the form (SPMHT - 1) and (SPMHT - 2) as illustrations of the basic
hypothesis testing formalism presented in the introductory document on this topic.
Two cases cover most of the situations that arise in applications involving the testing of this type of
hypothesis:
(i) The Large-Sample Case
This case arises whenever the sample size is 30 or larger. Then the sample mean, x , is approximately
normally distributed with
x  
and
x 

n

s
(SPMHT - 3)
n
This means that an appropriate standardized test statistic is
z
x  0
s
n
(SPMHT - 4)
We can summarize the outcome of the hypothesis tests in this case as follows:
Table 1.
Hypotheses:
H0:  = 0
HA:  > 0
reject H0 at a level of
significance  if:
z > z
(single-tailed rejection region)
H0:  = 0
HA:  < 0
z < -z
(single-tailed rejection region)
Pr(z < test statistic value)
H0:  = 0
HA:   0
z > z/2 or z < -z/2
(two-tailed rejection region)
2Pr(z > test statistic value)
© David W. Sabo (1999)
p-value
Pr(z > test statistic value)
Hypothesis Tests for Single Population Means
Page 1 of 15
The vertical bars in the formula for the p-value in the last row of this table stand for taking the absolute
value. (The formulas in Table 1 also apply for samples of any size drawn from a population which is
approximately normally distributed and for which the value of  is known. In such a situation, the quantity s
in formula (SPMHT - 4) for z would be replaced by , of course.)
(ii) The Small Sample Case
For samples of size less than 30 and  unknown, similar formulas arise when the population is
approximately normally distributed. The working formulas for such situations are as in (SPMHT - 4) and
Table 1, except that the symbol z is everywhere replace by the symbol t, since now the standardized test
statistic
t
x  0
s
n
(SPMHT - 5)
has the t-distribution with  = n - 1 degrees of freedom. In this situation, the outcomes of the various
hypotheses tests are summarized by:
Table 2.
Hypotheses:
H0:  = 0
HA:  > 0
reject H0 at a level of
significance  if:
t > t,
(single-tailed rejection region)
H0:  = 0
HA:  < 0
t < -t,
(single-tailed rejection region)
Pr(t < test statistic value)
H0:  = 0
HA:   0
t > t/2, or t < -t/2,
(two-tailed rejection region)
2Pr(t > test statistic value)
p-value
Pr(t > test statistic value)
In calculating the p-values, one would need to use the t-distribution with  = n - 1 degrees of freedom.
(If your situation does not fall into either of these two cases, then you really have only two alternatives:
(i)
look for some method that doesn't have even the relatively minor requirements of the two
cases described above. For instance, one approach that might get at more or less the
same issue that your original hypothesis test was designed to examine is the so-called
"sign test", which focuses on the median, and requires only that the population distribution
be continuous. You need to realize, though, that it really is impossible to get something
from nothing in most of statistics -- a small sample without much definite information about
the population distribution is a very little bit of information in total, and so no method that
has validity can be expected to give very "significant" or "powerful" results. Because of
this, we will not explore such methods further in this course.
(ii)
the best alternative when you are unable to determine that the population is approximately
normally distributed, as required by the small sample case above, is to collect more data
so that you can eventually use the large sample case formulas.)
We'll give a few examples illustrating the application of the hypothesis test procedures outlined above.
Example 1: Is the data given in the standard data set SalmonCa20 evidence to support the claim that the
mean Ca concentration is greater than 65 ppm in salmon fillets sanitized with 20 ppm chlorine dioxide?
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Hypothesis Tests for Single Population Means
© David W. Sabo (1999)
Solution
Since the words "evidence" and "claim" occur in the question, we know that a hypothesis test is being asked
for here. Further, the issue is whether we can conclude that the mean Ca concentration in these fillets is
greater than 65 ppm. This tells us immediately what the hypotheses should be.
let  = the mean Ca concentration in salmon fillets sanitized with 20 ppm chlorine dioxide
Then, to respond to the question asked, we need to test the hypotheses:
H0:  = 65 ppm
HA:  > 65 ppm
The HA we use must be the statement for which the question asks us to seek support  the mean Ca
concentration, , is greater than 65 ppm.
The data includes Ca concentrations for a sample of n = 50 (which is greater than 30) salmon fillets, giving a
mean value, x = 65.96 ppm and a standard deviation, s = 9.889 ppm. Since the sample size is greater than
30, we are able to use the large sample testing procedure summarized in Table 1 above.
No value of  is specified, so we will use  = 0.05. Thus, the rejection criterion is
reject H0 if z > z = z0.05 = 1.645.
So, calculate the value of the standardized test statistic, z:
z
x  0
65.96  65

 0.69
s
9.889
n
50
Since 0.69 is not greater than 1.645, we cannot reject H0 at a level of significance of 0.05. In fact, for these
hypotheses,
p-value = Pr(z > 0.69) = 0.50 - 0.2549 = 0.2451 >> 0.05.
This p-value, the probability that we will be making an error if we decide to reject H0 on the strength of the
data given, is far too high. Thus, the SalmonCa20 data is inconclusive -- we do not have anything near
adequate data to support the claim that the mean Ca content of these fillets is greater than 65 ppm.

Example 2: Since the SalmonCa20 data does not support the claim that the mean Ca content of these
salmon fillets is greater than 65 ppm, can we say that it supports the claim that the mean Ca content is less
than 65 ppm?
Solution
The only way to answer this question properly is to test the hypotheses:
H0:  = 65 ppm
HA:  < 65 ppm
All of the other information in the problem remains as in Example 1, so, we will use the large sample
procedures summarized in Table 1, with  = 0.05. For this pair of hypotheses, the rejection criterion is
Reject H0 if z < -z = -z0.05 = -1.645.
The standardized test statistic, z, has the same value here as in Example 1
© David W. Sabo (1999)
Hypothesis Tests for Single Population Means
Page 3 of 15
z
x  0
65.96  65

 0.69
s
9.889
n
50
since none of the information in the problem has changed. Clearly, since 0.69 is not less than -1.645, we
cannot reject the H0 just above at a level of significance of 0.05. In fact, for these hypotheses, we get
p-value = Pr(z < 0.69) = 0.50 + 0.2549 = 0.7549.
This is even worse than the p-value obtained in Example 1. Of course, we should have expected that it
would make very little sense to reject H0 here in favor of the hypothesis  < 65 ppm, because the data itself
is in direct contradiction to  < 65 ppm. If we had observed a value of x which was less than 65 ppm, that
may or may not have been adequate evidence to conclude that  < 65 ppm. However, it would be a strange
thing indeed to use the observation of a value of x that is larger than 65 ppm to argue that  < 65 ppm.

NOTE: these two examples demonstrate exactly what we mean by "inconclusive" in hypothesis testing.
Just because the data does not support the conclusion that  > 65 ppm does not mean that it will support
the conclusion that  < 65 ppm, or vice versa. In this case, the data is just too close to  = 65 ppm to be
able to tell us anything about either  < 65 ppm or  > 65 ppm at an acceptable level of significance. No
conclusion at all with regard to the value 65 ppm can be supported by this data.
Example 3: A researcher wishes to determine whether the yield of a new variety of sweet potato is different
from the nominal standard yield of 20 t/ha. He sets up fourteen test plots of the potatoes, and for these
obtains a mean yield of 23.26 t/ha with a standard deviation of 1.80 t/ha. Is this data adequate evidence to
conclude that the new variety has a yield different from 20 t/ha?
Solution
The word "different" occurs twice here so it's pretty clear that the hypotheses that need to be tested are
H0:  = 20 t/ha
HA:   20 t/ha
Thus, we require a two-tailed test. Since n = 14, which is smaller than 30, we have a small sample case. It
will be necessary to assume that the population is normally distributed here (that is, the yields of all such
plots of sweet potatoes form a population which is approximately normally distributed). If we were given the
original data, we could attempt to confirm this assumption by constructing a normal probability plot, but since
that data has not been supplied, we are unable to do more than state population normality as an otherwise
unsupported assumption about the population.
Using a level of significance of 0.05, we can state the rejection criteria here as: reject H 0 if
either
t > t0.025, 13 = 2.160
or
t < -t0.025, 13 = -2.160
Using the sample statistics x = 23.26 t/ha and s = 1.80 t/ha, we get for the value of the standardized test
statistic:
t
x  0
23.26  20

 6.78
s
1.80
n
14
But, 6.78 is very much larger than 2.160, so we can reject H0 without hesitation at a level of significance of
0.05. Thus, the data obtained is strong support for the claim that the true mean yield of this new variety of
sweet potato is different from 20 t/ha.
In this case, the p-value is given by
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Hypothesis Tests for Single Population Means
© David W. Sabo (1999)
p=value = 2Pr(t > 6.78,  = 13)
Now, according to our t-tables, the closest we can come to this probability is
Pr(t > 4.221,  = 13) = 0.0005
and so
Pr(t > 6.78,  = 13) < 0.0005
and so
p=value = 2Pr(t > 6.78,  = 13) < 2(0.0005) = 0.001.
That is, using just our printed t-tables, the best we can say is that the p-value here is less than 0.001,
meaning that if we use the data here as support for concluding that the mean yield really is different from 20
t/ha, we have a probability of less than 0.001 of that being an incorrect conclusion. (Using Excel, we can
determine that the p-value here is 0.0000130, a very small value -- hence the rejection of H0 is very sound in
this case.)

Type 2 Error Probabilities
The hypothesis test procedures summarized in Tables 1 and 2 earlier are developed to control the
probability of type 1 errors occurring when the null hypothesis is rejected. However, when we do not reject
the null hypothesis, there is a risk that we are committing a type 2 error. The probability, , of committing a
type 2 error depends on the actual mean value of the population, which is something we don't know or we
wouldn't be testing hypotheses in the first place. While we can't calculate "the value of " for a particular pair
of hypotheses, we can calculate values of  for specific hypothetical values of . This is a useful ability,
because the so-called power of a hypothesis test, 1 - , is an informative measure of how effective a
hypothesis test procedure is in detecting a real effect.
Formulas for  are easily obtained by simply inspecting the setup of the hypothesis test. Consider a righttailed test first:
H0:  = 0
HA:  > 0
We reject H0 in favor of HA if we find that the standardized test statistic, z, (in the large sample case)
satisfies z > z, which is equivalent to the condition
x  0
 z

n

x  0  z 

n
We've used the symbol  here -- if  is unknown, approximate it by the known value of s. Similarly, if the
small sample case is applicable, replace z everywhere by the symbol t.
Thus, with reference to z or equivalently, with reference to x , the situation can be sketched as:
© David W. Sabo (1999)
Hypothesis Tests for Single Population Means
Page 5 of 15

(µ = µ)
area = 
area = 

z
do not reject H0
µ0
reject H0
do not reject H0
z
µ
reject H0
µ0 + z·/n
Now, as indicated in the figure on the right above, we can calculate the probability of not rejecting H0 when
 =  but assuming the same standard deviation, . Then, if  > 0, this non-rejection of H0 will be a type 2
error, and so the calculated probability will be the probability of committing a type 2 error when  =  .
Thus,
(   )  Pr(not reject H0 when    )
 Pr( x   0  z 

when    )
n

0  z 
  

n
 Pr  z 




n


(SPMHT - 6)
Similarly, for a left-tailed test, we reject H0 if
z < -z
or
x  0  z 
n
and so, the probability of not rejecting H0 when  =  is
    Pr( x   0  z 

n
when   )

0  z 
  

n
 Pr  z 




n


(SPMHT - 7)
This will, of course, be a type 2 error probability if  < 0.
Finally, for the two-tailed hypothesis test, H0 is not rejected if
-z/2  z  z/2
or
0  z / 2 
Page 6 of 15
n
 x  0  z / 2 
n
Hypothesis Tests for Single Population Means
© David W. Sabo (1999)
A type 2 error will occur whenever a value of x is observed in this interval, but  =   0. The probability
of such an event is
    Pr(  0  z  / 2

n
 x  0  z / 2

n
when   )
 0  z / 2 
 
0  z / 2 
  

n
n
 Pr 
z





n
n


(SPMHT - 8)
We illustrate the use of formulas (SPMHT - 6) through (SPMHT - 8) with a couple of examples. Although
these formulas look quite complicated at a quick glance, they result from straightforward application of the
familiar method for calculating normal probabilities to the events "H0 is not rejected." You may find it easier
to simply calculate type 2 error probabilities from "scratch" when required rather than trying to memorize
these formulas.
Example 4: For the hypothesis test described above in Exercise 1, determine the probability of making a
type 2 error when the true mean is really 70 ppm.
Solution
Example 1 involves a right-tailed hypothesis test, so we can apply formula (SPMHT - 6) directly here. Since
the hypotheses were
H0:  = 65 ppm
HA:  > 65 ppm
we have 0 = 65 ppm. From the statement of the problem here, we have  = 70 ppm. In Example 1, we
used  = 0.05 and so z = z0.05 = 1.645. Also, use   s = 9.889 ppm, and n = 50. Substituting all of these
numbers into formula (SPMHT - 6) then gives:
  70   Pr( x  65  1.645
9.889
 Pr x  67.30
50
when   70 )
when   70 



67 .30  70 
 Pr  z 
  Pr z   1.93   0.0268
9.889


50 

This is quite an acceptably small number, since it is well below 0.05. Thus, if the hypothesis test is carried
out as set up in Example 1, the probability that H0 will not be rejected if  = 70 (and bigger, of course) is only
0.0268.

Example 5: Compute the probability of making a type 2 error when the true mean yield is 18 t/ha in the
hypothesis test situation described in Example 3 above.
Solution
Since this is a two-tailed test situation, we need to follow the procedure which led to formula (SPMHT - 8)
above. The hypotheses being tested are
© David W. Sabo (1999)
Hypothesis Tests for Single Population Means
Page 7 of 15
H0:  = 20 t/ha
HA:   20 t/ha
Given the sample size of 14 and the level of significance,  = 0.05, the non-rejection region turned out to be
-t0.025, 13 = -2.160  t  2.160 = t0.025, 13
This corresponds to the two conditions
x   0  2.160
s
x   0  2.160
s
 20  2.160
1.80
n
 18.961
14
and
n
 20  2.160
1.80
14
 21 .039
Thus, the required type 2 error probability here is
  18   Pr 18.961  x  21.039
when   18 

 18 .961  18
21 .039  18
 Pr 
t
1.80
 1.80
14
14



  Pr 1.998  t  6.317 


with  = 13. Here we run into the usual problem when calculating t-distribution probabilities: our tables do
not give enough information to calculate this probability precisely. From the row labeled  = 13 in the t-table,
we get the following three relevant pieces of information:
Pr(t > 1.771) = 0.05
Pr(t > 2.160) = 0.025
and
Pr(t > 4.221) = 0.0005
From these, we can say the following
since 1.998 is between 1.771 and 2.160, it must be that
Pr(t > 1.998) is a value between 0.05 and 0.025
and
since Pr(t > 6.317) < Pr(t > 4.221), it must be that
Pr(t > 6.317) is considerably smaller than 0.0005, and so could be considered zero
for practical purposes.
Thus,
( = 18) = Pr(1.998  t  6.317)
= Pr(t > 1.998) - Pr(t > 6.317)
 Pr(t > 1.998)
and so is a value between 0.05 and 0.025.
If you do the exact calculation of this probability using the TDIST() function available with Excel, you get
Pr(t > 1.998) = 0.03354
Pr(t > 6.317) = 0.00001
so that
( = 18) = Pr(t > 1.998) - Pr(t > 6.317) = 0.03353.

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Hypothesis Tests for Single Population Means
© David W. Sabo (1999)
Sample Size Considerations
Consider a right-tailed hypothesis test. You will recall from the previous document introducing the basic
ideas of hypothesis testing (and by looking at the second figure above) that if we simply slide the boundary
between the non-rejection and rejection region leftwards in order to reduce the value of ( = ), we do get
a smaller probability of making a type 2 error but at the cost of increasing the probability of making a type 1
error. For a given set of data, we cannot simultaneously reduce the sizes of both type 1 and type 2 errors.
However, the two error probabilities can be adjusted simultaneously by adjusting the sample size.
Assuming the large sample case applies, it is not too difficult to calculate the sample size required to
achieve desired target values for both  and  with the same random sample.
For a right-tailed test, we need to start with formula (SPMHT - 6):

0  z 
  

n
    Pr  z 




n


and let's say that we would like this to be less than or equal to some value. To make this simple, we can just
as well represent this target value by the symbol  again, understanding that  now stands for a number we
select during the planning stage of the experiment, rather than being a number we calculated later when we
know the data collected did not permit rejection of H0.
Thus, the goal is to satisfy the equation

0  z 
  

n
Pr  z 




n


(*)
But, when we write an equation of the form
Pr(z  c)  
the value c is just what we've been calling z1 - , the value of z which cuts off a right hand tail of area 1 - 
(since c is the boundary of a left-hand tail of area ). But also, since the z-distribution is symmetric about
z = 0, we can write
z1 -  = -z
Thus, we can write
Pr(z  -z )  
(**)
(If you have trouble understanding this last paragraph or so, trying sketching the z-diagram.) Now, if you
compare equations (*) and (**) just above, you'll see that we can write down the equation
0  z  
n

 
  z
(***)
n
Rearrange this equation to solve for n:
0  z

n
© David W. Sabo (1999)
    z 

n
Hypothesis Tests for Single Population Means
Page 9 of 15
So

 0     z   z 


n
and so


 z   z 
n
 0  
The direction of the inequality has changed here because we divided both sides of the previous inequality by
(0 - ), which is a negative number because  is bigger than 0 -- that is, you will have a type 2 error here
only if H0 is not rejected but  > 0 . We can get rid of the minus sign showing explicitly in the numerator by
multiplying numerator and denominator on the right-hand side by -1,
z   z 
n
   0
which just has the effect in the denominator of switching the two terms. Now both sides of this inequality are
positive numbers in the situation of interest, and so we can just square both sides to obtain the desired final
result:


 z   z  
n

    0 
2
(SPMHT - 9)
Example 6: Suppose that for the situation in Example 1, we wish to test the hypotheses
H0:  = 65 ppm
HA:  > 65 ppm
at a level of significance of 0.05, but we wish also to set up the experiment so that
( = 68 ppm) = Pr(make a type 2 error when  is really 68 ppm) = 0.05
as well. (If we calculate ( = 68 ppm) using the method illustrated in Example 4, but assuming a sample
size of 50, we get ( = 68 ppm)  0.3085).
Thus, as far as formula (SPMHT - 9) is concerned, we have
0 = 65 ppm
 = 68 ppm
 = 0.05  z = 1.645
 = 0.05  z = 1.645
and
  s = 9.889 ppm
Substituting these numbers into (SPMHT - 9) then gives


 z   z  
 1.645  1.645  9.889  
n
 
  117 .61

68  65


   0 
2
2
Therefore, we can achieve the two stated goals if we plan to use a sample of 118 or more of the salmon
fillets.

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Hypothesis Tests for Single Population Means
© David W. Sabo (1999)
The sample size formula for a left-tail test looks very similar to (SPMHT-9) above for right-tailed tests. Now,
we start with formula (SPMHT-7), requiring that the type 2 error probability be less than or equal to some
specific target value, again denoted :

0  z 
  

n
Pr  z 




n


The event described in brackets on the left here is a right-hand tail, and since the area under the standard
normal density curve for that tail is to be less than or equal to , we know that the left boundary of that tail
must be greater than or equal to z. This gives the equation
0  z  

n
 
 z
n
which initially can be rewritten as

 0    z   z 

n
Since we are now talking about a left-tailed test, a type 2 error can occur only if H0 is not rejected when
 < 0, or 0 -  is positive. Thus all of the factors in this last relation are positive, and so we can rearrange
at will and square all factors to get what turns out to be


 z   z  
n

  0   
2
(SPMHT - 10)
which is essentially identical to formula (SPMHT - 9) for the case of a right-tailed test.
The formula to be used to estimate sample sizes when an experiment involving a two-tailed test is being
planned is just slightly more problematical, but we get a very similar looking formula after making a
reasonable approximation. As above, we start by specifying a target type 2 error probability, denoted by the
simple symbol , for the situation  =   0. Then, formula (SPMHT - 8) gives
 0  z / 2 
 
0  z / 2 
  

n
n
Pr 
z





n
n


We can rewrite the left-hand side as the difference of two probabilities:


0  z / 2 
  
0  z / 2 
  


n
n
Pr  z 
  Pr  z 







n
n




(#)
The following two figures will help sort things out.
© David W. Sabo (1999)
Hypothesis Tests for Single Population Means
Page 11 of 15


µ
µ
µ0
reject H0
reject H0
reject H0
µ0
reject H0
The figure on the left shows the situation where the true mean,  is greater than the hypothesized mean, 0.
The dashed distribution curve is centered on 0, and determines the location of the rejection region. The
solid line distribution curve is centered on  and represents the "true" distribution of the sample mean. The
probability of making a type 2 error is equal to the area under the solid-line distribution curve for x falling
into the rejection region.
Now, although the rejection region is in two parts, you can see from this figure that by far the greatest part of
the area under the  curve will come from the right-hand half of the rejection region in this case. As a
result, to a good approximation, we can ignore the contribution of the left-hand half of the rejection region to
the probability of making a type 2 error. But this just means that we can ignore the second term on the lefthand side of equation (#) above, leaving

0  z / 2 
  

n
Pr  z 
 



n


But, this is very similar in form to the situation involving just a right-tailed test. We can immediately write
0  z / 2 

n
 
  z
n
from which we get


 z  / 2  z  
n

    0 
2
which looks just like (SPMHT - 9), except that  has been replaced by /2. Since the approximation we
introduced just above amounts to focussing just on the right-hand tail of the two-tailed rejection region, this
is the sort of formula you would expect to get.
The second figure, on the right above, illustrates the situation when  is less than 0. Again you see that
the area under the  curve for one of the two parts -- right-hand part now -- of the rejection region will be
relatively negligible compared to the area for the other part of the rejection region. Going through a
procedure similar to the above, or similar to the left-tailed rejection region case examined earlier, we come
up with the result
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Hypothesis Tests for Single Population Means
© David W. Sabo (1999)


 z  / 2  z  
n

  0   
2
This differs from the formula just above by the reversal of the terms in the denominator. Since in either
situation, reversing the terms in the denominator just causes a sign change, and the sign change disappears
when the indicated squaring is done, the two formulas give identical results in practice. We can write a
single formula which both highlights this sign difference, and yet emphasizes that the final result is the same
in both cases by using the absolute value symbol in the denominator:


 z  / 2  z  
n

  0   
2
(SPMHT - 11)
Example 7: Reconsider the situation described in Example 3 above, where we needed to test the
hypotheses:
H0:  = 20 t/ha
HA:   20 t/ha
at a level of significance of 0.05. To plan the experiment necessary to test these hypotheses so that the
probability of making a type 2 error 0.05 when the true mean is 21 t/ha, we proceed as follows.
Even though Example 3 involves a small sample case, here we are backing up one step to the planning
phase. For planning purposes, we start by assuming a large sample (n > 30 when dealing with population
means) will be required. Thus, we need to use formula (SPMHT - 11) here with the following values:
0 = 20 t/ha
 = 21 t/ha
 = 0.05
 = 0.05


z/2 = z0.025 = 1.96
z = z0.05 = 1.645
and
  s = 1.80 t/ha
Now, substituting these into (SPMHT - 11) gives


 z  / 2  z  
 1.96  1.645 1.80 
n
 
  42.107
20  21
  0   


2
2
Thus, we need to select a random sample of size 43 or bigger to be able to test the hypotheses above at a
level of significance of 0.05, while limiting the probability of making a type 2 error when the true mean is 21
t/ha to 0.05 or less.

NOTE: If we had gotten a value which was less than 30 in the calculation above, it would have indicated
that the small sample case was adequate. However, in that case, the z's in the calculation would have to be
replaced by the corresponding t-values. Since the t-values depend on the value of n, you would have to
carry out a bit of a trial and error process (as was necessary when we considered sample size issues in
connection with confidence interval estimates of the mean using small samples). Essentially, what you
could do is take the value of n given by the large sample formula above, increase it by 2 or 3 to get an
approximately correct value for the small sample case. Then use that value of n to obtain t-values for the
right-hand side of the formula above. If the calculation gives you the same value of n as you assumed, that
would be the correct value to use.
© David W. Sabo (1999)
Hypothesis Tests for Single Population Means
Page 13 of 15
What Influences Sample Size
For one-tailed tests, the required sample size is given by


 z  z   
n

  0    
2
(which is a combination of equations (SPMHT - 9) and (SPMHT - 10) above), and for two-tailed tests, the
sample size is given by


 z / 2  z   
n

  0    
2
Both of these formulas express the required value of n as the square of a quotient that contains three
factors. Each of these factors represents the effect of some aspect of the problem that you should see
intuitively as influencing the quality of the outcome of the hypothesis test.
(z + z) or (z/2 + z) is an expression whose value arises out of the desired probabilities of making
one or the other possible types of errors here. The lower the desired error probability, the
larger the corresponding z value is. The sample size increases as this factor increases 
indicating that to lower the probabilities of a wrong decision, you need to collect more
data.
 is a measure of how variable (or how uniform) the target population is. Generally speaking,
random samples of a particular size will be more representative of the target population
when that population is more uniform than they will be if that population is more variable.
As the degree of variability of the target population increases, sample sizes must increase
to achieve the same degree of representativeness. This shows up here in that n
increases as the value of  increases.
0 -  represents the size of the distinction you are trying to draw about the population. The
hypothesis test is to distinguish the difference between a population with  = 0 and a
population with  = . If these two values are very close together, a very fine distinction is
being attempted, and so more data will be required. You see that this difference, 0 - ,
appears in the denominator of the formula for n. If this difference is smaller, it will result in
the value of n being bigger, and vice versa.
Thus, if you want greater reliability of the conclusion (less chance of a mistaken conclusion), or you are
working with a more diverse population, or you are trying to draw a finer distinction between the target
population and other populations, you will generally need to increase the sample size, which amounts to
increasing the amount of information you have to work with.
If you need to decrease the sample size, you will have to make a decision on which aspect or combination of
aspects of your results you are willing to sacrifice: the probability of making one or both of a type 1 or type 2
error or the degree of distinction you can draw between the target population and some alternative. Since 
is a property of the target population, its value is not adjustable.
The Connection Between Hypothesis Testing and Confidence Interval Estimation
Consider the two-tailed hypothesis test:
H0:  = 0
HA:   0
In the large sample case, we would reject H0 if
Page 14 of 15
Hypothesis Tests for Single Population Means
© David W. Sabo (1999)
z > z/2

x  0  z / 2 
z < -z/2

x  0  z  / 2 
n
or if
n
That is, we do not reject H0 if we observe x to be in the interval
0  z / 2 
n
 x  0  z / 2 
(SPMHT-12)
n
and reject H0 otherwise.
The interval (SPMHT - 12) is sketched in the figure:
do not reject H0

µ0 - z/2 /n
µ0
µ0 + z/2 /n
You notice that it looks a lot like a 100(1 - )% confidence interval estimate centered on 0. Thus, we can
state the criterion for rejecting H0 in a two-tailed hypothesis test involving the population mean as follows:
reject H0:  = 0 vs HA:   0 at a level of significance  if the observed value of x does not fall
within the 100(1 - )% confidence interval centered on 0.
or
do not reject H0 in this case if the observed value of x does fall within this interval.
Thus, satisfying or not satisfying the rejection criteria for the two-tailed hypothesis test is equivalent to
determining where the value of x lies with respect to a confidence interval-type interval about 0.
You could in principle say the same sorts of things about one-tailed hypotheses tests, though the
correspondence is not as natural or insightful because intervals are inherently associated with a two-tailed
situation. Nevertheless, we could, for example, make the statement:
For:
H0:  = 0
HA:  > 0
or for
H0:  = 0
HA:  < 0
do not reject H0 at a level of significance  if the observed value of x lies within the 100(1 - 2)%
confidence interval centered at 0.
© David W. Sabo (1999)
Hypothesis Tests for Single Population Means
Page 15 of 15