Instructor: Hello students.
Welcome to Biostats ER, example one for chapter seven.
In this example,
we will learn how to use
a one-sample continuous outcome formula.
The problem here is from our textbook, problem 20.
The problem states,
"The mean lifetime for cardiac stents is 8.9 years.
"A medical device company has implemented some improvements
"in the manufacturing process,
"and hypothesizes that the lifetime is now longer.
"A study of 14 new devices reveals
"a mean lifetime of 9.7 years
"with a standard deviation of 3.4 years.
"Is there statistical evidence
"of a prolonged lifetime of the stents?"
"Run the test at a 5% level of significance."
Now, before I move forward with solving the problem,
again, I would like to let you know
that for this Biostats ER example,
the solution is typed,
because again, as I mentioned previously,
once I heard from a student,
that all of my Biostats ER examples are solved by hand.
Hence, I have again typed the solution here,
but I'm going to go through it step-by-step.
So the first step here
is for us to set up the hypothesis,
and determine the level of significance.
So here, the null hypothesis states,
that the mean lifetime for cardiac stents
is equal to 8.9 years,
and the alternative states
that it is greater than 8.9 years,
and the significance level is 0.05 as stated in the problem.
Now we have to select the appropriate test statistics.
Now, given our sample size is 40,
we can select a Z test.
But then the question is,
which type of Z test are we going to use?
Now here we have to look at the problem again,
and as we can see that this is
a one-sample continuous outcome,
because we are comparing our mean
from a single sample to an expected value.
And our outcome is continuous as we are measuring years.
Hence, we will select the correct Z test,
which is provided here.
Then we have to set up the decision rule,
and given that alpha is 0.05,
we will reject null if our Z calculated
is greater than or equal to 1.645.
Now, in step four we have to compute the test statistics.
The X bar here is 9.7,
and the mu naught is 8.9.
The standard deviation is 3.4,
and it is divided by the
square root of the sample size, which is 40.
So after we do all the algebra,
our Z calculated is 1.49.
Now, we have to draw the conclusion.
So the conclusion here,
we realize that we fail to reject the null
because 1.49 is less than 1.645.
So, we do not have statistically significant evidence
at alpha of 0.05 to show
that the lifetime is longer than 8.9 years.
I hope this was helpful,
and if you have any questions,
please let me know,
and I'll see you in the next video.
Thank you.