Welcome to an example
of a hypothesis test of two population means
with unknown population standard deviations given data.
A professor at a large community college wants
to determine whether there is a difference in the means
of final exam scores between students
who took his stats course online
and the students who took it in-person.
He believes that the mean of the final exam scores
for the online class is lower than the in-person class.
Is the professor correct?
The randomly selected 30 final exams scores from each group
are listed in the tables below.
On the left we have the online class final exam scores.
On the right we have the in-person final exam scores.
So again is the mean of the final exam scores
of the online class lower than the mean
of the final exam scores of the in-person class?
Test at a 5% significance level.
So if we let Mu sub one equal the mean
of the final exam scores for the online class,
and Mu sub two equal the mean for the final exam scores
for the in-person class.
We want to test whether Mu sub one is less than Mu sub two,
which is the alternative hypothesis.
And if Mu sub one is not less than Mu sub two,
we assume the means are equal,
which gives us Mu sub of one equals Mu sub two
for the null hypothesis.
The null hypothesis may also be stated as Mu sub one
is greater than or equal to Mu sub two.
And because the level of significance is 5%,
alpha is equal to 0.05.
And we have a left tail test because we're testing
to see whether Mu sub one is less than Mu sub two.
Because we have unknown population standard deviations,
we will perform a two-sample t-test on the TI-84.
So the test statistic is a t-score,
and then we'll also find the p-value.
So going the calculator,
I've already entered the data to save time.
To enter data, press Stat, Enter.
Again I have the online final exam scores in L1
and I have the in-person
or face-to-face class final exam scores in L2.
Next we press Stat, right arrow to test,
and we select option four; two sample t-test.
I've already entered the key information,
but let's review it.
The input is data, down.
List one is L1, down.
List two is L2, down.
We are not using a frequency table,
and therefore we enter one for frequency one
and one for frequency two.
We're testing whether Mu sub one is less than Mu sub two.
We select less than Mu sub two here in the middle.
For pooled, we select no, arrow down to calculate,
and press Enter.
We have a t-score of approximately negative 3.2286
and p-value of approximately 0.0011.
Let's go ahead and record these values.
Before we compare the p-value and alpha,
let's take a closer look at the t-score of negative 3.2286.
If we were doing this test by hand,
we would take the alpha of 0.05,
and knowing we have a left tail test,
and find the corresponding z-score,
which I've already done here on the left.
Notice how the z-score is negative 1.645.
Next we calculate the t-score using the formula shown here.
Notice how this requires us to find the sample means
and the sample standard deviations.
We also assume that Mu sub one minus Mu sub two
is equal to zero.
So to find the sample means and sample standard deviations,
again we can use the TI-84
by determining the 1-Var statistics.
To do this, press Stat, right arrow to calculate,
select 1-Var Stats.
First select L1 which I've already done.
Go down to calculate, Enter.
X bar is the sample mean.
Sx is the sample standard deviation.
We repeat the process for L2.
Stat, Calc,
option one.
Now we select list two by pressing second two.
Go down to calculate, press Enter.
And again we have the sample mean
and the sample standard deviation.
Performing the calculations,
notice we get a t-score that's slightly different
than the t-score from the two sample t-test
because I am using rendered values
for the sample standard deviations.
Either way notice how the t-score of negative 3.2288
is in the rejection region.
And therefore, we are going to reject the null hypothesis.
Let's also draw the same conclusion
by comparing the p-value and alpha.
Actually before we compare the p-value and alpha,
let's take a look at this graph here,
showing the p-value of 0.0011.
If the sample means are equal, then the difference is zero,
which is why we have zero
at the center of this distribution.
And then the difference of the sample means
is negative 9.98,
which is shown here on the left of zero
on the horizontal axis.
So the p-value is equal to the shaded area here,
which is also equal to the probability
that the difference of the means
is less than or equal to negative 9.98.
And now comparing the p-value and alpha,
notice how the p-value is small or lower than alpha.
And if the p-value is low, the null must go.
Or more formally,
because the p-value is less or equal to alpha,
or because alpha is greater than or equal to the p-value,
we reject the null hypothesis.
Which in conclusion means the professor is correct.
The evidence shows that the mean
of the final exam scores for the online class
is lower than that of the in-person class.
At a 5% level of significance from the sample data,
there is sufficient evidence to conclude
that the mean of the final exam scores for the online class
is less than the mean of the final exam scores
of the in-person class.
I hope you found this helpful.