Instructor: Hello students,
welcome to Biostatistics ER, Example 2 Chapter 7.
In this example,
we will learn how to perform hypothesis testing
for two sample matched continuous outcome data.
The problem is from our textbook, Problem 18.
The problem states,
"Suppose the results of the analysis
in problem 15 through Problem 17
are reported and criticized because the participants
were not randomized to different diets,
and that there may be other factors
associated with changes in cholesterol.
A third study is run to estimate the effect
of the low-carbohydrate diet on cholesterol levels.
In the third study,
participants' cholesterol levels
are measured before starting the program,
and then again after 6 months on the program.
The data are shown here.
Is there a significant increase in cholesterol
after 6 months on the low-carbohydrate diet?
Run the appropriate test at a 5% level of significance."
So here, the samples are matched
because the cholesterol levels of the same participants
were measured before and after 6 months on the program.
As we know, cholesterol level is also measured
on a continuous scale,
and here we will use t-test
because our sample size is less than 30.
With that, we will move to Step 1,
which is set up hypothesis
and determine level of significance.
The null hypothesis here is mu d equal to 0
and the alternative hypothesis is mu d greater than 0,
and the level of significance is 0.05
as stated in the problem.
Here, I have provided you
with the appropriate test statistics,
which is a t-test.
In Step 3, we will set up the decision rule
and we will reject.
Now if our t calculated is greater than or equal to 2.015,
and this value was from the t-table
based on degrees of freedom being 5.
Now we are going to calculate the test statistics.
So here, I strongly encourage you
to either use Excel or SPSS
to calculate the average
and the standard deviation,
and I have used Excel to calculate the average
and the standard deviation.
So once those values are calculated,
we have to insert them into our formula.
and the numerator here is going to be 5,
which is the average,
and the denominator is going to be the standard deviation,
11.8 divided by the square root of sample size,
which is 6.
And once we perform the algebra,
our t calculated is 1.04.
So now in Step 5,
and we are going to reach our conclusion.
And here we will fail to reject null
because 1.04 is less than 2.015.
So we do not have statistically significant evidence
at the alpha of 0.05 to show
that there is a significant difference in cholesterol
after 6 months on the diet.
Thank you for your time and attention,
and I'll see you in the next video.