Welcome to an example of a hypothesis test
for two population means with unknown standard deviations.
A study is done by a community group
in two neighboring colleges
to determine which one graduates students
with more math classes.
College A samples 11 graduates.
Their averages is four math classes
with a standard deviation of 1.5 math classes.
College B samples nine graduates.
Their average is 3.5 math classes
with a standard deviation of one math class.
The community group believes
that a student who graduates from college A
has taken more math classes on average.
Both populations have a normal distribution.
Perform a hypothesis test at a 1% significance level.
Let's begin by listing all the given information.
For college A, we will use subscript A.
For college B, we will use subscript B.
College A samples of 11 graduates; n sub A is equal to 11.
The average is four math classes
with a standard deviation of 1.5 math classes,
and therefore x-bar sub A is equal to four,
s sub A is equal to 1.5.
College B samples nine graduates,
and therefore n sub B is nine.
Their average is 3.5 math classes
with a standard deviation of one math class.
And therefore x-bar sub B is equal to 3.5,
s sub B is equal to one.
And now lets set up the alternative and null hypotheses.
The community group believes that a student
who graduates from college A
has taken more math classes on average.
And therefore, the alternative hypothesis
is Mu sub A is greater than Mu sub B.
And if this isn't true,
Mu sub A must be less than or equal to Mu sub B,
which is the null hypothesis.
And we can also give these hypotheses
as shown here on the right by subtracting Mu sub B
in both inequalities,
where the alternative hypothesis
is Mu sub A minus Mu sub B is greater than zero,
and the null hypothesis is Mu sub A minus Mu sub B
is less than or equal to zero.
Now before we determine the t-score and p-value
on the TI-84,
let's take a look at this graph here shown on the left.
Notice using the sample data,
x-bar sub A minus x-bar sub B
is equal to four minus 3.5, which is 0.5.
So the difference of the sample means is 0.5.
If they weren't different, the difference would be zero,
which is at the center of the distribution shown here.
And then we have 0.5 on the right.
So the p-value is equal to the probability
that x-bar sub A minus x-bar sub B
is greater than 0.5.
And the p-value is also the area
of the shaded region here on the right.
Notice how we have a right tail test.
And now let's go to the calculator
and determine the t-score and p-value.
We press Stat, right arrow to test,
and then select option four, two sample t-test.
I've already entered the information here,
but let's review it.
The input is Stats, down.
X-bar sub one is x-bar sub A which is four.
So for college A, the sample mean is four, down.
The sample standard deviation is 1.5, down.
N is 11, down.
X-bar sub two is x-bar sub B which is 3.5, down.
Sample standard deviation is one, down.
Sample size n is nine, down.
We're testing to see whether Mu sub A
is greater than Mu sub B.
And therefore, we select greater than,
which is the third option on the right and press Enter.
And then down, pooled,
we select no, down and down to calculate and press Enter.
We have a t-score of approximately 0.8899
and a p-value of approximately 0.1928.
Let's go ahead and record this information.
Before we compare the p-value to alpha,
let's take a closer look at the t-score
of approximately 0.8899.
If we were doing this by hand,
we would take the alpha of 0.01
and find the corresponding z-score
which I've already done here.
Notice the z is equal to 2.326 when alpha is 0.01
for a right tail test.
And we have the rejection region on the right.
Again if we were doing this by hand,
we would use the t-score formula shown here
and calculate the same z-score that 84 gave us,
which is 0.8899.
But notice how this z-score is not in the rejection region,
it would be approximately here,
and therefore we do not reject the null hypothesis.
But let's come to the same conclusion,
comparing the p-value and alpha.
Notice how the p-value is higher, greater than alpha.
And because the p-value is high,
the null must fly, or more firmly,
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 1% level of significance
from the sample data,
there is not sufficient evidence to conclude
that a student who graduates from college A
has taken more math classes on average
than a student who graduates from college B.
I hope you found this helpful.