Welcome to an example of a hypothesis test
of two population means
with known population standard deviations.
Because we have known population standard deviations,
the test statistic is a z-score and we perform
a two sample z test, not a two sample T test.
And now let's take a look at our example.
We assume independent groups
with known population standard deviations.
The mean lasting time of two competing floor waxes
is to be compared.
20 floors are randomly assigned to test each wax.
Both populations have a normal distribution.
The data are recorded in table 10.8 below.
We have the wax, the sample mean number of months
floor wax lasts,
and we have the population standard deviation.
Does the data indicate that wax one is more effective
than wax two?
Test at a 5% level of significance.
Let's begin by listing all of the given information.
20 floors are randomly assigned to test each wax
and therefore the sample size for both waxes is 20,
which means n sub one is equal to 20
and so is n sub two,
where the subscripts indicate which wax.
The sample mean for wax one is three
x bar sub one equals three.
The population standard deviation is 0.33 sigma sub one
is equal to 0.33.
Now looking at wax two, and notice how x bar sub two is 2.9
and sigma sub two is 0.36.
Because we are testing at a 5% level of significance,
alpha is equal to 0.05 and now let's determine the null
and alternative hypothesis.
The question is, does the data indicate that wax one
is more effective than wax two?
If it's more effective then the sample mean will be greater
than the sample mean for wax two
and therefore the alternative hypothesis is mu sub one
is greater than mu sub two.
And if mu sub one is not greater than mu sub two,
we have mu sub one is less than or equal to mu sub two,
which is the null hypothesis.
Before we determine the z-score and the p value
using the TI 84, let's take a look at this
graph shown here above.
This is a distribution for X bar sub one
and minus X bar sub two.
So if there's no difference between the means,
the difference is zero, which is why we have zero
at the center of the distribution.
And now if we find the difference of the sample means,
we have three minus 2.9, which is equal to 0.1,
which is why we have 0.1 marked off here
on the horizontal axis.
And therefore the p value is equal to the area
under the curve, shaded to the right of 0.1,
which also indicates the p value
is equal to the probability.
The difference of the means is greater than
or equal to 0.1.
And now let's find the z-score and p value on the TI 84.
To do this, we press stat,
right arrow to tests, we're performing a two sample z test,
not a T test because we have known
populations' standard deviations.
We want option three.
I've already entered the key information,
but let's review it.
The input is stats, down.
Sigma sub one is 0.33, down.
Sigma sub two is 0.36, down,
x bar sub one is three, down,
n sub one is 20, down.
x bar sub two is 2.9, down,
n sub two is 20, down.
We're testing to see whether mu sub one
is greater than mu sub two.
And therefore we select the third option
greater than mu sub two, which is already highlighted.
If it wasn't, we would arrow over, press enter,
down, down to calculate and press enter.
The z-score's approximately 0.9157.
The p value is approximately 0.1799.
Let's record this information.
Let's take a closer look at this z-score.
If we are performing this hypothesis test by hand,
we would take the alpha of 0.05 and determine
the corresponding z-score, which I've already done here.
Notice the z-score is 1.645
where the projection region is shaded here to the right.
Then by hand, we would compute the z-score
using the form they're shown here at the bottom
of the screen, where again, mu sub one minus mu sub two
is assumed to be zero.
Notice we get the same z-score
we had on the TI 84 of 0.9157.
We'll notice the z-score of 0.9157
would be approximately here, outside the rejection region,
which means we would not reject the null hypothesis,
but let's draw the same conclusion,
comparing the p value to alpha.
Notice how the p value is high or greater than alpha
and when the p value is high, the null must fly
or more formally because the p value is greater than alpha
or because alpha is less than the p value,
we do not reject the null hypothesis,
which means at the 5% level of significance
from the sample data, there is not sufficient evidence
to conclude that the meantime wax one last is longer
than the meantime wax two last.
I hope you found this helpful.