>>We just discussed hypothesis testing for a population mean when sigma, the standard
deviation of the population, is known. But remember what we concluded about sigma being
known: when would that ever happen? When would we ever NOT know the mean of the population
– when would be forced to take a sample to draw conclusions about a population mean
– when the standard deviation of the population was known? It’s hard enough to wrap our
heads around the standard deviation measure in the first place, let alone knowing what
it is for a population when we don’t even know the population mean! We said it was unrealistic
to ever think we’d know the population standard deviation or variance.
Instead, odds are we’re going to have to estimate the population standard deviation
from a sample. Our sample of size n will give us both y-bar and s. When that situation arises
we have three cases to consider. The first case is when the population is normally distributed,
and we use the t distribution no matter the size of the sample. The second case is when
the sample size is greater than the magic number of thirty such that we can assume normality,
and we still use the t distribution. In the third case, a non-normal population and a
small sample size keep us from assuming normality, therefore we use nonparametric techniques
that we’ll discuss later in the course.
Recall from the previous unit on confidence intervals that a t distribution is a bell-shaped
distribution with many properties of the normal distribution, but it’s tailored to the situation
where we estimate sigma using s.
We use a t value in the calculation of the test statistic, t-naught. The interpretations
of the critical value and p-value approaches are the same for the t distribution as they
were for the z distribution. For the critical value approach, the critical value is found
from looking up the t value associated with alpha for a one tailed test or alpha over
two for a two-tailed test. For the p-value approach the probability of the tail, or tails,
associated with the test statistic are found with the t tables.
And the value of the test statistic, t-naught, should look familiar. It’s the difference
between y-bar and mu divided by s over root n. The same kind of interpretation holds:
how different is the sample evidence from the hypothesized value in standard deviation
units, where the standard deviation is measured with s, not sigma.
Let’s take a look at an example.
Glengarry Valves, a manufacturer of valves in many industrial applications, produces
a butterfly valve composed of two semicircular plates on a common spindle that is used to
permit flow in one direction only. The semicircular plates are supplied by a vendor with a specification
that the plates have a tensile strength of five pounds per millimeter. A universal testing
machine was used to test the tensile strength for a random sample of 20 such plates. As
a quality control manager at Glengarry Valves, say you’re going to change vendors if the
tensile strength for the valve plates is less than four point eight pounds per millimeter.
Let's try to determine if a vendor change needs to be made with 95% confidence. And
let’s use the critical value approach to start with.
So the statement we’re keying in on is “less than four point eight pounds per millimeter.”
The words “less than” imply strictly less than. No equality. So “less than four point
pounds per millimeter” must be the in the alternative hypothesis. The statement that
contains equality, “greater than or equal to four point eight pounds per millimeter”
must be in the null hypothesis.
As we’re testing a population mean when sigma is unknown, we’ll use a t distribution.
The degrees of freedom for the t distribution are twenty minus one or nineteen. The less
than sign in the alternative hypothesis suggests that the test is a lower tail test, so the
rejection region consisting entirely of alpha will be in the lower tail. Recall that the
t distribution is symmetric, so looking up a value of point oh five in the upper tail
gives us the positive form of the critical value, which is one point seven two nine.
Naturally, we’re dealing with a lower tail test; the critical value will be negative
one point seven two nine.
The mean tensile strength of the twenty sample plates is four point six nine pounds per millimeter,
with a standard deviation of point two five six pounds per millimeter. Therefore, the
test statistic t-naught is four point six nine minus four point eight divided by point
two five six over root twenty. That is, the sample mean four point six nine is one point
nine two standard deviations to the left of four point eight.
When we compare the test statistic to the critical value, we see that negative one point
nine two lies in the rejection region, beyond negative one point seven two nine. As such,
we reject the null hypothesis and conclude that the mean tensile strength of the butterfly
valve plates provided by our supplier is less than four point eight pounds per millimeter.
We draw this conclusion with ninety-five percent confidence. As such, we should find a new
supplier to more effectively meet specifications.
Let’s take a look at another example and use the p-value approach.
Say you’re a production engineer at Glen Ross Roasters, and suppose you have purchased
a filling machine for bags of pumpkin spice coffee that is supposed to fill each bag with
sixteen ounces of coffee. Assume that the weights of bag fills are approximately normally
distributed. A random sample of twelve bags was collected and weighed, and we want to
know if our filling machine works as advertised.
What would the hypotheses be for this problem? We just want to know if the machine works
at filling a coffee bag at sixteen ounces. If the fill is too much, we lose money by
providing too much coffee for the marketed size. If the fill is too little, customers
may balk at not getting their money’s worth. So in this case, we want to know if average
fill is equal to sixteen ounces or not equal to sixteen ounces. Naturally, equality must
be in the null statement. And the alternative would be strict inequality.
Since we’re dealing with a population mean when the population standard deviation is
unknown, we use a t distribution for our test statistic. The mean fill of the twelve sample
coffee bags is fifteen point nine three ounces, and the standard deviation is point one seven
six ounce. Test statistic t-naught is fifteen point nine three minus sixteen divided by
point one seven six over root twelve. That’s negative one point three eight.
Now, to calculate the p-value. Let’s look at the tail associated with negative one point
three eight. Since we have to multiply the tail area by two to get the p-value, we find
the area below negative one point three eight. Note that if we find the area above negative
one point three eight, when we multiply by two, we’d get a p-value greater than one,
which won’t work. So we find the area below negative one point three eight. The degrees
of freedom for this t distribution is twelve minus one or eleven. So we look at the row
associated with eleven degrees of freedom. The t table is different from the z table,
as we have a different distribution for each degree of freedom. The columns in the t table
are associated with a tail area. So we’re probably not going to find an exact value
in the tail associated with our test statistic. We’ll have to find a set of bounds for the
area. So let’s look across the row for eleven degrees of freedom. The first t value is one
point three six three, and it’s associated with point one in the tail. The next t value
over is one point seven nine six is associated with point oh five. The magnitude of our test
statistic is one point three eight, which lies between one point three six three and
one point seven nine six. That means the tail area must be between point oh five and point
one. We don’t know exactly what it is, according to the table, but we know the tail area is
between point oh five and point one. But the p-value is found by multiplying the tail area
by two since it’s a two-tailed test. That means the p-value is bounded between point
one and point two.
We compare this p-value to a value of alpha. We don’t know the p-value exactly, but the
bounds of the p-value are pretty large. At its smallest, the p-value is bigger than point
one. Generally, we’d never reject the null with a p-value that large. As we don’t reject
the null, we conclude that the mean fill of the coffee bags is no different than sixteen
ounces. The machine is doing a good job of filling coffee bags based on our sample of
twelve bags.
Note that the process doesn’t change. The idea behind a test statistic doesn’t change
– it’s a measure of how far apart the sample evidence, y-bar, and the hypothesized
population mean, mu-naught, are in standard deviation units. It’s a measure of the correctness
of the null hypothesis based on sample evidence. And the critical value approach doesn’t
change – we always find a critical value and compare the test statistic to it to determine
if it lies in the rejection region. And the p-value approach doesn’t change – we always
find a tail area and compare it to our level of alpha. And the process is the same for
hypotheses dealing with population proportions or population standard deviations, which we’ll
discuss later on.