Hypothesis test for Population Mean,
based on the sample mean, σ unknown
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The situation is:
 We have a hypothesis about the "true" value of a population mean.
That is, someone (perhaps us) claims that H_{0}: μ = a,
for some value a.
 Not unexpectedly, we do not know the
population standard deviation, σ.
 We will consider an alternative hypothesis which is one of the following
 H_{1}: μ > a,
 H_{1}: μ < a, or
 H_{1}: μ ≠ a.
 We want to test
H_{0} against H_{1}.
 We have already determined
the level of significance that we will use for this test.
The level of significance, α,
is the chance that we are willing to take that
we will make a Type I error,
that is, that we will reject H_{0}
when, in fact, it is true.

Immediately, we recognize that samples
of size n drawn from this population
with have a distribution of the
sample mean that is a
Student's t with n1 degrees of freedom.
At this point we proceed via
the critical value approach
or by the attained significance approach.
These are just different ways to
create a situation where we can finally make a decision.
The critical value approach tended to be used
more often when everyone needed to use the tables.
The attained significance approach is more commonly
used now that we have calculators and computers
to do the computations. Of course either approach can be done
with tables, calculators, or computers. Either approach gives the same final result.


Critical Value Approach
 We determine a sample size n.
This will set the number of degrees of freedom
as we use the Student's t distribution.
 We find the tscore that corresponds to
having the level of significance area more extreme than that tscore,
remembering that if we are looking at being either too low or too high
then we need half the area in both extremes.
 Then, we take a random sample of size n from the population.
 We compute the sample mean,
and the sample standard deviation, s
 Compute
s_{x} = s / sqrt(n)
and use that value to
compute (t)(s_{x}).
 Set the critical value (or values in the case of a twosided test)
such that it (they) mark the value(s) that is (are) that distance,
(t)(s_{x}),
away from the mean given
by H_{0}: μ = a.
 If the sample mean,
, is more extreme than the critical value(s)
then we say that
"we reject H_{0} in favor of
the alternate H_{1}". If the
sample mean
is not more extreme than the critical value(s)
then we say "we have insufficient evidence to reject H_{0}".
 
Attained significance Approach
 Then, we take a random sample of size n from the population.
 We compute the sample mean,
and the samle standard deviation, s.
 Assuming that H_{0} is true,
we compute the probability of getting the value
or a value more extreme than the sample mean.
We can do this using the fact that the distribution of
sample means for samples of size n is a Student's t distribution
with n1 degrees of freedom
with mean=μ and standard deviation=s/sqrt(n).
 If the resulting probability is smaller than or equal to the
predetermined level of significance then we say that
"we reject H_{0} in favor of
the alternate H_{1}". If the
resulting
probability is not less than the predetermined level of significance
then we say "we have insufficient evidence to reject H_{0}".

We will work our way through an example to see this.
Assume that we have a population of values
and that we do not know the standard deviation of those values in the population.
I claim that the mean of the population, μ is equal to 134.
You do not believe me.
You think the mean of the population is not equal to 134.
Notice that you are not saying what the true mean is, just that
it is not 134.
It is hard for you to tell me that I am wrong, so you decide that
you will take a sample of the population, compute the sample mean
and the sample standard deviation.
Then you will see
just how strange that sample mean is, especially knowing the sample standard deviation.
You happen to know that sample means from samples of size 36 from
a population will follow a Student's t distribution
with 35 degrees of freedom and will have
the standard deviation of the sample mean, s_{x},
be equal to s / √36.
Until we take a sample we will not know the value s
and, therefore, we will not be able to
compute s_{x}.
However, if were to turn out that s_{x} is around 2.5
then clearly
if you get a sample mean of 200, you are going to
tell me that I am completely wrong.
The value 200 is over 25 standard deviations above the
value 134, the value we would expect to get if the true mean is 134.
Random sampling is just not going to produce that kind of rare event.
Well, it could happen, but the odds against it are so great you just say it is
too unlikely to happen and reject my hypothesis.
On the other hand, if the sample mean turns out to be 134.7 then
you would never think that getting such a sample mean
would justify your claim that I am wrong.
You have seen many examples where the sample mean
changes with each sample and where the value of the sample mean is often
more than one standard deviation
away from the true population mean.
In fact, you recall that for 35 degrees of freedom about
67.6% of the values
in a Student's t distributed
population are within 1 standard deviation of the mean.
That means that about 32.4%
of the values are more than 1 standard deviation away from the mean.
We expect that nearly 1/3 of the
time a sample of 36 items taken from
our population (where for this part of the discussion we are assuming
that sample yields
s_{x}=2.5) will have values
for the sample maean that are
more than 2.5 above or below the true mean.
If the true mean is 134 then getting
a 36item sample mean of that population that
is 0.7 or more above or below 134 should
happen really often (in fact, nearly 49% of the time).
If we are using the critical value approach,
the question becomes, at what point is the sample mean far enough away
from the supposed 134 for you to say that getting such a sample mean is just too
unlikely; for you to risk claiming that my 134 value
cannot be right because if it were then it is just too unlikely
that you would get a 36item random sample with such an extremely
different sample mean.
You answer that question by using the level of significance
that you specify. If you are only willing to make a Type I error
1% of the time, then you want to find a value for the sample mean
that is so extreme that only 1% of the Student's t distribution
of 36item sample means will be that or more extremely different from
my hypothesis value of 134.
For this example, an extreme value can be too high or too low.
You need to split the 1% error that you are willing to make
between these two options.
You know (from the tables, a calculator, or qt() )
that less than 0.5% (0.005 square units) of the area under
the Student's t curve for 35 degrees of freedom
is less than the tscore = 2.724 [and, correspondingly, less than 0.5%
of the area under the standard normal curve is greater than 2.724].
Therefore, you know that being more than 2.724 standard deviations above or below
the true population mean
will happen less than 1% of the time.
We can do all of that computation, but we still do not
know the value of s_{x} because
we have yet to actually take a sample.
Now you take your sample. The values of that sample are shown in
Table 1.
The computations that we did above are pretty straight forward.
We could do them based on the values stored in L1,
in R
using:
n<length(L1)
n
t < qt(0.005,n1,lower.tail=FALSE)
t
samp_sd < sd(L1)
samp_sd
s_x < samp_sd / sqrt(n)
s_x
to_be_extreme < t*s_x
to_be_extreme
crit_low < 134  to_be_extreme
crit_high < 134 + to_be_extreme
crit_low
crit_high
samp_mean < mean(L1)
samp_mean
Unfortunately, I cannot show the R computation
for the mean of the values in Table 1 above
because the reproductions of the R screens are static
whereas the values in the table are dynamic, they change
whenever the page is refreshed.
It should be enough, at this point in the course, to say
that if you use the gnrnd4() command as given above, then the data will be in
L1 and the commands will produce the values we just computed above.
I did, however, capture two earlier random tables, and I can reporduce them here.
The first is shown in Table 1a.
Figure 1a shows generating the table and making those computations in R.
Figure 1a
You should be able to reproduce these values in your R session.
The result in this case is that the sample mean is not extreme enough to
reject H_{0}.
The second static example is shown in Table 1b.
Figure 1b shows generating the table and making those computations in R.
Figure 1b
In this second static example the sample mean is greater than the upper critical value.
Therefore, in this case we would reject H_{0}.
The presentation above walked us
through the critical value approach for
one dynamic and two static examples.
Alternatively, you could use the attained significance approach.
To do that, after determining that you are willing to
make a Type I error 1% of the time,
you take your 36item sample.
Starting with the dynamic example above,
we already have that sample back in Table 1.
You compute both the sample mean
and the sample standard deviation for that data. We have that above,
namely,
, respecively. Following that you look to see just
how likely it is to get a sample mean that extreme or more extreme from the 134 value
assuming that H_{0} is true, i.e.,
that the true population mean is 134.
You know that the distribution of 36item sample means will
be a Student's t with 35 degrees of freedom and will have
a mean of 134 and a standard deviation of
The computations that we did above are pretty straight forward.
We could do them based on the values stored in L1, in R using:
n<length(L1)
n
samp_sd < sd(L1)
samp_sd
s_x < samp_sd / sqrt(n)
s_x
samp_mean < mean(L1)
samp_mean
diff < samp_mean  134
diff
t<diff/s_x
t
if( samp_mean < 134 ){
p < pt(t,35)
pkind < "lower"} else
{ p<pt(t,35,lower.tail=FALSE)
pkind < "upper"}
pkind
p
p*2
You should note that the script included some "logic" to determine which of the two
tails we should be using. The logic shown, the "if" construction,
is sufficient in the case where we are using a "twotailed" test.
It will require some refinement later. Also, note that the
script gives both a value for p and p*2.
We still need to know which to use. In these "twotailed" tests
we will use the p*2 value.
After the dynamic example above, I included two static examples
so that you could see the critical value script in action.
We return to those same static examples to see
the attained significance script in action.
The first static example was in Table 1a.
Figure 1aattained shows generating that data and then making the
R
computations to use the attained significance approach.
Figure 1aattained
You should be able to reproduce these values in your R session.
The result in this case is that the
attained significance is not below the 0.01 level of the test
so we cannot
reject H_{0}.
The second static example is repeated in Table 1b.
Figure 1battained shows generating that data and then making the
R
computations to use the attained significance approach.
Figure 1battained
You should be able to reproduce these values in your R session.
The result in this case is that the
attained significance is below the 0.01 level of the test
so we
reject H_{0}.
Another dynamic example follows, this time with a bit less dialogue.
We have a population which we
know to be approximately normal.
We do not know that the population standard deviation.
There is a claim, a hypothesis, that the population mean is
μ = 14.2, but we believe, for whatever reason, that
the true population mean is higher than that.
We want to test H_{0}: μ = 14.2
against the alternative
H_{1}: μ > 14.2 at
the 0.05 level of significance,
meaning we are willing to make a Type I
error one out of twenty times.
This is a "onesided test" because the only time we will
reject H_{0} is if we get a sample mean significantly greater
than the H_{0} value of 14.2.
Because the population is normal we can get
away with a smaller sample size. In fact,
we decide to use a sample of size n=17.
The distribution of 17item sample means will be
approximately Student's t with
with 16 degrees of freedom and with
standard deviation equal to approximately s/sqrt(17),
where s is the sample standard deviation.
First we will do this with the critical value approach.
The critical value we want needs to 5% of the area under the curve
for the Student's t above it.
On a standard student's t with 16 degrees of freedom,
the tvalue=1.746 has about 5% of the area above it.
Our critical value will be 1.746*s above the null hypothesis
mean of 14.2, but
we cannot find our critical values until we know the value of
s.
We take a sample, the values of which are shown in Table 2.
The computations that we did above are pretty straight forward.
We could do them, excluding finding the sample mean,
with a bit more accuracy, in R
using:
n<length(L1)
n
t < qt(0.05,n1,lower.tail=FALSE)
t
samp_sd < sd(L1)
samp_sd
s_x < samp_sd / sqrt(n)
s_x
to_be_extreme < t*s_x
to_be_extreme
crit_high < 14.2 + to_be_extreme
crit_high
samp_mean < mean(L1)
samp_mean
We cannot demo those commands for the values in Table 2
because those values are dynamic, changing with each reload of the page.
However, you should be able to use the gnrnd4() command above to create the
same table and then run those commands to produce the same results that
you see here.
To do this same problem using the attained significance approach
we would ask "What is the probability that
if H_{0} is true
we would find a random sample of 17 items with a sample mean that is
Here is a third example, one given with even less discussion.
We have a population and a null hypothesis
H_{0}: μ = 239.7
with the alternative hypothesis
H_{1}: μ < 239.7.
We want to test H_{0} at the
0.035 level of significance.
We do not know the population standard deviation.
There is a real concern that the population is not normally distributed.
To address that concern we choose a sample size greater than 30.
Our sample size will be n=68.
For the critical value approach, we note that
the tscore for 67 degrees of freedom with the area 0.035 to its left is
about 1.841.
We take a random sample of 68 items.
The values are reported in Table 3.
The computations that we did above are pretty straight forward.
We could do them, excluding finding the sample mean,
with a bit more accuracy, in R
using:
n<length(L1)
n
t < qt(0.035,n1)
t
samp_sd < sd(L1)
samp_sd
s_x < samp_sd / sqrt(n)
s_x
to_be_extreme < t*s_x
to_be_extreme
crit_low < 239.7 + to_be_extreme
crit_low
samp_mean < mean(L1)
samp_mean
We cannot demo those commands for the values in Table 3
because those values are dynamic, changing with each reload of the page.
Howver, you should be able to use the gnrnd4() command above to create the
same table and then run those commands to produce the same results that
you see here.
To do this same problem using the attained significance approach
we would ask "What is the probability that
if H_{0} is true
we would find a random sample of 68 items with a sample mean that is
The three dynamic (the samples change each time the page is reloaded or refreshed)
examples above walk through problems of testing
the null hypothesis for the population,
where we do not know the population standard deviation,
by drawing a sample and using either the critical value
or the attained significance approach.
In that "walking through" the problems we begin to recognize that
the steps we will take are almost identical for each problem.
We should be able to capture those steps in some R function.
Consider the following function:
hypoth_test_unknown < function(
H0_mu, H1_type=0, sig_level=0.05,
samp_size, samp_mean, samp_sd)
{ # perform a hypothesis test for the mean=H0_mu
# when we do not know sigma and the alternative hypothesis is
# != if H1_type==0
# < if H1_type < 0
# > if H1_type > 0
# Do the test at sig_level significance, for a
# sample of size samp_size that yields a sample
# mean = samp_mean and a sample standard deviation = samp_sd
s_x < samp_sd/sqrt(samp_size)
if( H1_type == 0)
{ t < abs( qt(sig_level/2,samp_size1))}
else
{ t < abs( qt(sig_level,samp_size1))}
to_be_extreme < t*s_x
decision < "Reject"
if( H1_type < 0 )
{ crit_low < H0_mu  to_be_extreme
crit_high = "n.a."
if( samp_mean > crit_low)
{ decision < "do not reject"}
attained < pt( (samp_meanH0_mu)/s_x, samp_size1)
alt < paste("mu < ", H0_mu)
}
else if ( H1_type == 0)
{ crit_low < H0_mu  to_be_extreme
crit_high < H0_mu + to_be_extreme
if( (crit_low < samp_mean) & (samp_mean < crit_high) )
{ decision < "do not reject"}
if( samp_mean < H0_mu )
{ attained < 2*pt((samp_meanH0_mu)/s_x,samp_size1)}
else
{ attained < 2*pt((samp_meanH0_mu)/s_x, samp_size1,
lower.tail=FALSE)
}
alt < paste( "mu != ", H0_mu)
}
else
{ crit_low < "n.a."
crit_high < H0_mu + to_be_extreme
if( samp_mean < crit_high)
{ decision < "do not reject"}
attained < pt((samp_meanH0_mu)/s_x, samp_size1,
lower.tail=FALSE)
alt < paste("mu > ",H0_mu)
}
ts < (samp_mean  H0_mu)/(samp_sd/sqrt(samp_size))
result < c(H0_mu, alt, s_x, samp_size, sig_level, t,
samp_mean, samp_sd, ts, to_be_extreme,
crit_low, crit_high, attained, decision)
names(result) < c("H0_mu", "H1:", "std. error", "n",
"sig level", "t" ,"samp mean",
"samp stdev", "test stat", "how far",
"critical low", "critical high",
"attained", "decision")
return( result)
}
This is a long function, but it is really just the steps that we
have taken in the previous examples. One of the things that make the
function so long is that we have to build into it all of
the "logic" that we apply to the problems as we do them. Thus, we had to build in
the idea of splitting the significance level
between both high and low values for the alternative hypothesis of being
"not equal to". We also wanted to build in the process of deciding if we want to reject or not reject
the null hypothesis.
As much as I would like to demonstrate how this function works for the dynamic
examples above, I cannot do that here since those examples
each contained random samples that change every time you load the page.
However, I can repeat each of the static cases given above for the first
dynamic situation, and follow that with
two more static examples, one for each of the final two dynamic situations
given above.
Revisit Example 1a:
We decided to take a 36item sample.
Then we found that the two critical values were
127.55 and 140.45. Then we took a sample
shown in Table 1a above. From that sample
we determined, in Figure 1a,
that the sample mean is 132.6111
and the sample standard deviation is 3.841709.
Recall that H_{0} had μ=134,
that this is a twosided test, that the level of significance of the test is
1%, and that the sample size is 36.
With all of that we can use our function via the
command hypoth_test_unknown(134,0,0.01,36, 132.6111, 3.841709)
as shown in Figure 2.
Figure 2
Figure 2 contains all of the results that we generated above
for example 1a. It just does that as the result of one command.
You might note that in order to construt that command
we did need to knwo the sample size, the sample mean, and the sample standard deviation.
Revisit Example 1b:
We can do the same thing for example 1b. Here the command is just
hypoth_test_unknown(134, 0, 0.01, 36, 136.0556, 3.051411 )
and the resulting output is shown in Figure 3.
Figure 3
Again, the output includes all the values we had to construct above
for example 1b.
A static example along the lines of example 2:
We have that same population, known to be normal, and we want to test
H_{0}: μ = 14.2
against the alternative
H_{1}: μ > 14.2
at the 0.05 level of significance. We
take a sample of size 17 and get the values shown in Table 4.
In order to perform our test in R we need to generate the sample there,
find the sample sizze, mean , and standard deviation, and then formulate the
command to run the hypoth_test_inknown() finction.
The first part of that task can be done via
gnrnd4( key1=1567321604, key2=0001000149 )
samp_mean < mean(L1)
samp_sd = sd(L1)
n < length(L1)
c(n,samp_mean,samp_sd)
The results of that are shown in Figure 4.
Figure 4
Once we know the values we can formulate the commad
hypoth_test_unknown(14.2, 1, 0.05, 17, 14.77, 0.75974 )
to produce the results shown in Figure 5.
Figure 5
The nice part of using the above approach is tht we get to
see what is happening.
However, because hypoth_test_unknown()
gives us so much information,
we could actually get this down to a more concise pair of statements,
anmely,
gnrnd4( key1=1567321604, key2=0001000149 )
hypoth_test_unknown(14.2, 1, 0.05,
length(L1), mean(L1), sd(L1) )
Which, together, produce the result shown in Figure 6.
Figure 6
The result is the same, but the big advantage is that we did
not have to type in the intermediate values.
Furthermore, the statement of the function can be used again with
a new problem where we need only change at most the first three parameters.
For example, Table 5 holds a new sample for example 2.
We can perform the test on this sample via the commands
gnrnd4( key1=1321651604, key2=0001000145 )
hypoth_test_unknown(14.2, 1, 0.05,
length(L1), mean(L1), sd(L1) )
which produce the result shown in Figure 7.
Figure 7
A different result, but there was very little change in the commands we used.
A static example along the lines of example 3:
We have that same population and we want to test
H_{0}: μ = 239.7
against the alternative
H_{1}: μ < 239.7
at the 0.035 level of significance. We
take a sample of size 68 and get the values shown in Table 6.
Using the strategy that we just developed, we can
generate the data and run the test via
gnrnd4( key1=1750376704, key2=0008302375 )
hypoth_test_unknown(239.7, 1, 0.035,
length(L1), mean(L1), sd(L1) )
To generate the result shown in Figure 8.
Figure 8
This seems to be a much more efficient solution to the
problem than going through it as we did above for the
dynamic version of the test.
Below is a listing of the R commands used in generating this page.
#critical value approach
gnrnd4( key1=726463504, key2=0000400133 )
L1
n<length(L1)
n
t < qt(0.005,n1,lower.tail=FALSE)
t
samp_sd < sd(L1)
samp_sd
s_x < samp_sd / sqrt(n)
s_x
to_be_extreme < t*s_x
to_be_extreme
crit_low < 134  to_be_extreme
crit_high < 134 + to_be_extreme
crit_low
crit_high
samp_mean < mean(L1)
samp_mean
gnrnd4( key1=144963504, key2=0000400137 )
L1
n<length(L1)
n
t < qt(0.005,n1,lower.tail=FALSE)
t
samp_sd < sd(L1)
samp_sd
s_x < samp_sd / sqrt(n)
s_x
to_be_extreme < t*s_x
to_be_extreme
crit_low < 134  to_be_extreme
crit_high < 134 + to_be_extreme
crit_low
crit_high
samp_mean < mean(L1)
samp_mean
gnrnd4( key1=302583504, key2=0000400135 )
L1
n<length(L1)
n
t < qt(0.005,n1,lower.tail=FALSE)
t
samp_sd < sd(L1)
samp_sd
s_x < samp_sd / sqrt(n)
s_x
to_be_extreme < t*s_x
to_be_extreme
crit_low < 134  to_be_extreme
crit_high < 134 + to_be_extreme
crit_low
crit_high
samp_mean < mean(L1)
samp_mean
# attained approach
gnrnd4( key1=726463504, key2=0000400133 )
L1
n<length(L1)
n
samp_sd < sd(L1)
samp_sd
s_x < samp_sd / sqrt(n)
s_x
samp_mean < mean(L1)
samp_mean
diff < samp_mean  134
diff
t<diff/s_x
t
if( samp_mean < 134 ){
p < pt(t,35)
pkind < "lower"} else
{ p<pt(t,35,lower.tail=FALSE)
pkind < "upper"}
pkind
p
p*2
gnrnd4( key1=1833071604, key2=0001000150 )
L1
n<length(L1)
n
samp_sd < sd(L1)
samp_sd
s_x < samp_sd / sqrt(n)
s_x
samp_mean < mean(L1)
samp_mean
diff < samp_mean  134
diff
t<diff/s_x
t
if( samp_mean < 134 ){
p < pt(t,35)
pkind < "lower"} else
{ p<pt(t,35,lower.tail=FALSE)
pkind < "upper"}
pkind
p
p*2
# second dynamic example
gnrnd4( key1=1833071604, key2=0001000150 )
L1
n<length(L1)
n
t < qt(0.05,n1,lower.tail=FALSE)
t
samp_sd < sd(L1)
samp_sd
s_x < samp_sd / sqrt(n)
s_x
to_be_extreme < t*s_x
to_be_extreme
crit_high < 14.2 + to_be_extreme
crit_high
samp_mean < mean(L1)
samp_mean
diff < samp_mean  14.2
diff
t<diff/s_x
t
if( samp_mean < 14.2 ){
p < pt(t,16)
pkind < "lower"} else
{ p<pt(t,16,lower.tail=FALSE)
pkind < "upper"}
pkind
p
p*2
gnrnd4( key1=1277746704, key2=0005002380 )
n<length(L1)
n
t < qt(0.035,n1)
t
samp_sd < sd(L1)
samp_sd
s_x < samp_sd / sqrt(n)
s_x
to_be_extreme < t*s_x
to_be_extreme
crit_low < 239.7 + to_be_extreme
crit_low
samp_mean < mean(L1)
samp_mean
diff < samp_mean  239.7
diff
t<diff/s_x
t
if( samp_mean < 239.7 ){
p < pt(t,67)
pkind < "lower"} else
{ p<pt(t,67,lower.tail=FALSE)
pkind < "upper"}
pkind
p
p*2
source("../hypo_unknown.R")
# revisit 1a
hypoth_test_unknown(134,0,0.01,36, 132.6111, 3.841709 )
# revisit 1b
hypoth_test_unknown(134, 0, 0.01, 36, 136.0556, 3.051411 )
#static sample for example 2
gnrnd4( key1=1567321604, key2=0001000149 )
samp_mean < mean(L1)
samp_sd = sd(L1)
n < length(L1)
c(n,samp_mean,samp_sd)
hypoth_test_unknown(14.2, 1, 0.05, 17, 14.77, 0.75974 )
# or done in just two statements
gnrnd4( key1=1567321604, key2=0001000149 )
hypoth_test_unknown(14.2, 1, 0.05,
length(L1), mean(L1), sd(L1) )
gnrnd4( key1=1321651604, key2=0001000145 )
hypoth_test_unknown(14.2, 1, 0.05,
length(L1), mean(L1), sd(L1) )
gnrnd4( key1=1750376704, key2=0008302375 )
hypoth_test_unknown(239.7, 1, 0.035,
length(L1), mean(L1), sd(L1) )
hypoth_test_known(239.7, 43.6, 1, 0.035, 68, 230.028 )
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©Roger M. Palay
Saline, MI 48176 January, 2016