In this video, I’ll be conducting paired or dependent samples t-tests.
It is also known as related samples t-test, matched-pairs t-test,
or correlated samples t-test.
It is used to determine if a significant difference exists between two sets of measurements
when those measurements are paired or are related.
The most common paired t-test design is the repeated measures t-test or what is
also called within subjects t-test.
The repeated measures t-test is commonly used to examine the effect of a treatment
or intervention on the same set of individuals…
where measurements are taken twice for every individual in that sample.
A common example is when measurements are taken before and after an intervention.
For example, it could be used to determine if there is a statistically significant difference
between starting salaries and current salaries of employees in a company.
A main advantage of the repeated measures t-test is that it requires a smaller sample
size compared to the independent t-test.
Another paired t-test design is the matched subjects design,
where each individual in one group is matched (based on common attributes of interest)
with an individual in the other group.
Note that the matched subjects design requires twice as many individuals as the corresponding
repeated measures t-test.
Let’s now examine a repeated measures example.
Here we have scores “before” and “after” some treatment.
We want to determine if a significant difference exists between the before and after scores.
This requires a paired-samples t-test because each pair of scores belongs to one subject.
We will be applying this test statistic.
d-bar here is the sample mean of the before and after differences.
Mu-d is the population mean of the differences and in the denominator, we have the standard
error of the mean of differences.
That is, the standard deviation of the differences divided by the square root of the number of
pairs or differences.
The test requires that the distribution of the differences be approximately normal.
We begin by taking the difference between each pair of scores.
I’m going to do “After – Before” for the differences.
Although, it does not really matter for 2-tailed tests whether we do After – Before or Before
– After.
However, in one-tailed tests, doing “After – Before” is recommended because it makes
the hypothesis setup easier to understand.
The sum of the differences here is 14.
And the number of differences is 6.
Therefore, the mean of the sample differences is 2.333.
Next, we can calculate the standard deviation using this formula.
We need the sum of squares of the differences, which adds up to 86
The standard deviation is thus 3.266.
We can also use this formula to calculate the standard deviation
by first calculating the deviations from the mean,
then summing the squared deviations, which results in the same standard deviation
as obtained before.
Now to test if there is a significant difference in the “before” and “after” scores,
the null hypothesis will be that there is no difference.
That is, the population mean difference is equal to zero.
And the alternative hypothesis will be that the mean difference is not equal to zero,
because testing for a difference suggests we are conducting a 2-tailed test.
Here again is the test statistic.
The denominator here is the standard error of the mean of the differences
which computes to 1.333.
Mu-d here is equal to zero as seen in the null hypothesis.
Thus, t is essentially equal to d-bar divided by the standard error, which gives 1.750.
Now, suppose we’re testing at the significance level alpha = 0.05,
we can find the critical or rejection region using software or from the t-table.
The degrees of freedom for the test is n_d – 1, which is 5.
For alpha = 0.05 in two tails, or 0.025 in one tail, the critical value is 2.571.
That is, we reject the null hypothesis if the observed t is less -2.571
or if it is greater than +2.571.
Since the observed test statistic is not in the rejection region,
we cannot reject the null hypothesis at alpha = 0.05.
Using software, we also find the P-value to be 0.141
which is greater than our significance level of 0.05.
As a result, we fail to reject the null hypothesis.
Consequently, we have insufficient evidence to conclude
that a significant difference exists between the Before and After Scores.
Now, suppose the observed t statistic is -2.750, which falls in the rejection region, we will
reject the null hypothesis and conclude that a significant difference
exists between the “before” and “after” scores.
To construct a 95% confidence interval for the mean difference,
we use this formula.
For 95% confidence, alpha is .05 and at df = 6-1 = 5,
the critical value is 2.571 as we saw earlier.
Now, substituting the values, we have a margin of error of 3.427,
a lower limit of -1.9 and upper limit of 5.76.
We can also use this interval to make a decision about rejecting the null hypothesis.
Since the confidence interval contains the hypothesized value, 0
we fail to reject the null hypothesis at alpha = 0.05.
If this -1.9 had been positive, then the interval would not contain zero,
and we would reject the null hypothesis.
To interpret, we say, we are 95% confident that the difference between the Before and
After scores lies between the lower and upper limits.
And that’s it.
Thanks for watching.