Transcript One Tailed Tests - Wayne State College

One Tailed Tests
Here we study the hypothesis test for
the mean of a population when the
alternative hypothesis is an
inequality.
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Let’s follow an example and then highlight some main points
about hypothesis testing.
At its drive-up windows McDonald’s has had a mean service
time of 163.9 seconds. McDonald’s would like to improve on
this.
The null hypothesis in this case would be that the mean of the
population of all service times would be greater than or equal
to 163.9. We write the null as Ho: μ ≥ 163.9
The alternative, then, would be that the mean of the population
of all service times are less than 163.9. We write H1: μ <
163.9. In this case the alternative hypothesis contains the
improvement the company is attempting to make. The null
hypothesis is a maintaining of the status quo.
In a hypothesis test we will either 1) accept (fail to reject) the 2
null hypothesis or 2) reject the null and go with the alternative.
163.9
Here I show the distribution of sample means
of service times. The center is at 163.9. Note
some samples would have means less than
163.9 even when in reality the population has
mean of 163.9.
Sample
means
service
time
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Remember a Type I Error would have us reject the null when it is
true. In this case a Type I Error would suggest the company is
improving service time when it is in fact not doing so. We want to be
careful about this, so, we will only reject the null when the
probability of an observed event is really low.
What do I mean by the phrase, “probability of an observed event?”
In a study like this we do not look at every drive-up case. This would
be expensive. So a sample is taken. From the sample we calculate a
statistic and figure a probability of this value or a more extreme
value. This is the event to which I refer.
On the next slide I show the distribution of sample means, assuming
the mean is exactly 163.9 seconds. To have the standard error of this
distribution we need to know the sample size and have either the
standard deviation in the population or the sample standard deviation
as an estimate of it. Here we say we do not know the population
standard deviation, so we use the sample value and know we have to
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use the t distribution.
Level of
significance
Reject
null
Accept null
Critical
value
163.9
In terms of a t = 0
Sample
means in
seconds
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Now, in the graph on the previous slide I put the mean of the
distribution at 163.9.
You will notice on the left side of the distribution I put another
vertical line and wrote next to it the phrase level of
significance. I also labeled the value on the horizontal axis as
“critical value.” This is the lowest level the sample mean could
be and still have the claim true.
Here is the basic logic of the hypothesis test. Say in our one
sample we get a value farther away from the center than the
critical value. Although it could happen, the probability is
small. This will lead us to reject the null.
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Now, let’s refer back to the graph on slide 5. The area I called
the level of significance is the “low” probability I referred to
before. If our sample value falls in this area past the critical
value we will reject the null. But we pick the level of
significance to be low, like .05 or .01, and thus we make our
chance of making a Type I Error low.
Next I want to develop some equivalent ways of conducting the
hypothesis test. They are all similar and related.
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Critical t method
Level of
significance =
Reject
null
Accept null
Critical
value of t
=
t
distribution
In terms of a t = 0
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Say that alpha, or the level of significance we want is .05 and also
say we will use a sample of 25 stores to observe service time.
Thus with df = 25 – 1 = 24 and all .05 in the lower tail we have a
critical value of -1.7109. We put all of the alpha =.05 in the lower
tail because our alternative is an inequality.
Will you please put the alpha and critical t values in the graph on
the previous slide?
Up until this point I have not used a real sample mean value. I
have talked in the hypothetical. The hypothetical allowed me to
think of critical values. (In practical terms you would probably
have a sample mean from data at this time, but we have not used it
yet.)
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To complete the test we have to take the sample mean and
actually calculate the t value for it. This will be called the test
statistic tstat.
Say in our sample we get an X bar = 152.7 and a sample
standard deviation of 20 seconds. Thus,
tstat = (sample mean minus hypothesis value)/(sample standard
dev/sqrt(n)) = (152.7 – 163.9)/(20/sqrt(25)) = -2.80.
Since the test statistic is farther from a t of 0 than the critical
value is we reject the null and go with the alternative in this
example. McDonald’s has done something to speed up its
service time (reduce the time to serve customers).
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p value method
Level of
significance = .05
Reject
null
Accept null
Critical
value of t
= -1.7109
0
t
distribution
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p-value
What is a p-value? Let’s develop a story.
When we said the level of significance was picked to be .05 we
could find the critical value of -1.7109 in terms of a t value.
Then we had a sample mean of 152.7 and a corresponding tstat
of minus 2.80. On the number line this tstat is farther from the
center than the critical value. And for this reason we rejected the
null.
Now, if you go to the graph on the previous slide, and if you put
your right pinky, palm up on the critical value, the area to the left
of your pinky was chosen to be .05.
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Now, move your pinky to the left to the value of the sample
mean, or its corresponding t. The area under the curve to the left
of your pinky is less than .05. This actual area is called the pvalue. It is just the level of significance of the sample statistic
we obtained.
An equivalent way to test an hypothesis then is to reject the null
if the p value is less than the level of significance established for
the test.
In the t table we see in row df = 24 the value 2.7969 under .005.
With a symmetric distribution -2.7969 has a lower tail area of
.005. Our sample value is even less than that so the p-value is
less than .005. We reject the null here.
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Remember, all this is a one tailed test on the low side of the
mean. What if we had a one tailed test on the high side of the
mean? Same logic. Generic example:
Sample size is 50 so df = 49. Alpha is picked to be .01. The
critical t is found in the table as 2.4049.
Do not reject null
Reject null
t
2.4049
Say in a sample the sample mean is 6.034 and sample standard
deviation is .02. SO our tstat = (6.034 – 6.03)/(.02/sqrt(50)) =
1.41 and since this is not as big as 2.4049 we do not reject the
null.
The p-value is somewhere between .05 and .10 because 1.41 is
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between 1.6766 and 1.2991.
Let’s summarize some things.
When an alternative hypothesis H1 has a not equal sign we
have a two tailed test. We take alpha and divide by 2 to get
critical values. Reject the null if our tstat is more extreme than
the critical values. If we go to the p-value approach we have to
take our tail area and multiply by 2. Reject the null if the pvalue for the tstat is less than or equal to alpha.
When an alternative hypothesis H1 has an inequality (either a
less than or a greater than sign) sign we have a one tailed test.
Alpha is all in one tail to get the critical value. Reject the null
if our tstat is more extreme than the critical value. If we go
with the p-value approach the tail area with the tstat is the pvalue. Reject the null if the p-value for the tstat is less than or
equal to alpha.
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