DAREN STARNES: Hello, AP stat students,
and welcome to AP Daily practice sessions.
I'm Mr. Starnes coming to you from beautiful Hilton
Head, South Carolina.
And in this session, we're going to be practicing questions
of free response inference.
Now, if you haven't already downloaded the questions
and tried them, like working out your own solutions,
please pause the video now and do that.
You can find the link at the beginning of the video.
So that'll get you the most out of this session
if you've actually tried them and seen
where you had challenges, and then you
can turn the video back on and watch
me go through the solutions.
So I'll trust that you've done that, and here we go.
First question, a survey conducted by a National
Research Center asked a random sample of 920 teenagers
in the United States, how often they use a video streaming
service?
From the sample, 59% answered that they use a video streaming
service every day.
You'll notice I'm continuing to annotate questions
by underlining and circling some of the key phrases or numbers.
That's important to do on the AP Statistics exam
where there's a lot of wording in our questions,
and you don't want to miss an important word.
So part A says, construct and interpret
a 95% confidence interval for the proportion of all teenagers
in the United States who would respond
that they use a video streaming service every day.
So I'm sure by now you've constructed and interpreted
plenty of confidence intervals.
The first thing you want to do is define any parameters
and identify the appropriate inference method.
So in this case, the parameter is about a population
proportion, and the inference method
should be related to that.
I'm going to try to do it all in one efficient step.
So the parameter is p, the population proportion
of all teenagers in the United States
who would respond that they use a video streaming
service every day.
And the name of the inference method
is a one sample, z-interval for a proportion.
So I did it all in one, trying to show you efficient ways
to get your response down, so that you
can use your time to produce a high quality response.
What's next?
If you said check conditions, that's right.
So there are three conditions that we
need to check here for doing this one sample
z-interval for a proportion.
The conditions' purposes are first of all
to check for independence in the data collection.
And second, to make sure that the sampling distribution
of the sample proportion is approximately normally
distributed.
So first thing we check is did we have random data production.
What says in the stem of the question
that we had a random sample of 920 teenagers,
so we're good on that.
And the other thing we need for independence
is we need to check the 10% condition because we're
sampling without replacement from this finite population
of teenagers, so we need to make sure that 920 is less than 10%
of all US teenagers.
And there are a lot more than 9,200 US teenagers,
so we're definitely good there.
So we have the independence check done.
Now we need to be sure about the sampling distribution of p
being approximately normal.
That's our large counts condition
where we look at n times p and n times 1 minus p.
In this case, we don't know the population proportion, that's
what we're trying to estimate.
So we use the sample size of 920 times the sample proportion
of 0.59 that I circled up here.
And if you do that on your technology,
you don't get a whole number.
They rounded that 59% in the stem of the question.
So it's really 542.8 but it was 543 teenagers.
And that's at least 10, which is the criterion.
And then n times 1 minus p.
That in this case is in 920 times 1 minus 0.59.
Again, not a whole number if you punch it in, but it's
around 377.
That is also at least 10.
So because both of those are at least 10,
we meet the large counts condition,
sampling distribution of p hat is approximately normal.
What do you want to do now?
If you said, do the calculations, that's right.
On your formula sheet, there's a general form
for the confidence interval.
It's shown on the first page.
And then you can look for the specific standard error
that you need on the second page.
So for a sample proportion, here's
that formula standard error.
Now the critical value, you either
know it from memory for 95% confidence for a one proportion
z-interval.
It's 1.96.
Or you can look on the table that comes with the exam
or you can punch it into technology
and get the critical value that way.
So I know that's going to be 1.96.
The standard error is going to take a little bit
more work with the calculator.
So here we go, statistic is 0.59, plus or minus,
there's that critical value 1.96.
And the standard error is the square root
of 0.59 times 0.41 all over 920.
You use your calculator to simplify,
0.59 plus or minus 0.032.
That's a margin of error of 3.2 percentage m
and the final interval is 0.558 to 0.622.
If you prefer to use technology either as your method
or as a check on your calculation,
you can go to the one prop z menu in the calculator,
plug-in your sample number of successes and your sample size.
Make sure your confidence level is 0.95.
And you'll get the same answer to three decimal places
that we got with the formula.
Now again on the exam, all that you need to do
is produce the correct interval.
But it's an all or nothing proposition
if you use the calculator.
There's no work for the reader to look at.
So there's a little bit of a risk in using the calculator.
But there's also the potential benefit
that if you try to use the formula,
you could make an arithmetic error
and get the wrong interval that way.
So for some students, the technology is a big advantage.
And for others, it's less of an advantage.
It's certainly a good check.
What's the last step?
We're going to interpret the interval.
Went to all this trouble to get it, so let's interpret it.
So we say we are 95% confident that the proportion
of all teenagers in the United States,
who would respond that they use a streaming service every day
is between 0.558 and 0.622.
Notice that we are just parroting the words
in the stem of the question, the proportion
of all teenagers in the United States, et cetera.
That way, we don't unintentionally
change the mean of the population parameter
when we're writing our conclusion.
So I guess what I'm saying is use a template or a sentence
frame there, don't freelance or you might accidentally
change a word that's important.
Let's go to the second question.
Part B, based on the confidence interval in part A,
do the sample data provide convincing statistical evidence
that the proportion of all teenagers in the United States,
who would respond that they use a video streaming
service every day is not 0.5?
Justify your answer.
Well as a reminder, the interval in part A was 0.558 to 0.622.
Does that interval provide convincing evidence
that the true population proportion is not 0.5?
It does.
Why?
Because 0.5 is not in there.
So here might be a model response.
Yes, from part A the interval of plausible values
for the population proportion based on the sample data
is 0.558 to 0.622.
And because the value 0.5 is not contained in the interval,
the sample data provide convincing statistical evidence
that the proportion of all teenagers in the United States
who would use a streaming service every day is not 0.5.
Notice I'm parroting the stem of the question
again to make sure I get the wording just right.
So that's a follow up to part A. And that
completes the first free response
about confidence intervals.
Next up is one about significance tests.
This one's got a little bit of a long stem,
so I better get to it.
Patients experiencing symptoms of a heart attack
are routinely transported to a hospital in an ambulance.
In a study of a new treatment thought to reduce damage
to the heart, patients experiencing
symptoms of a heart attack were randomly
assigned to one of two groups.
During transportation to the hospital, patients in one group
received standard care and patients in the other group
received the new treatment consisting
of standard care and the application of a blood pressure
cuff.
So those are our two groups in the experiment.
The response variable measured for each patient
was a number between 0 and 1.
So it's a quantitative variable, referred to
as the myocardial salvage index, MSI.
A higher MSI value indicates a more positive outcome
for the patient.
Summary statistics for the MSI responses of the two groups
are shown in the table here.
Here's the question, do the data provide
convincing statistical evidence that the new treatment
results in a higher mean MSI value
than does the standard care among people
similar to the patients in the study?
This is a significance test type of prompt.
Do the data provide convincing evidence?
We're going to have to go through the process step
by step.
And our process starts with stating hypotheses, defining
parameters, and identifying the appropriate inference method.
So similar to confidence interval,
but we have to state hypotheses as part of this.
So hypotheses here, I'm going to use some notation that
might connect you to the treatment, new treatment
versus standard care.
So my null hypothesis is no difference in the effectiveness
of the treatments.
So the mean MSI value for n minus the mean MSI value for s
is equal to zero.
And the alternative is a greater than because of higher mean.
So mu sub n is greater than mu sub s, if you will.
And I'm defining my parameters here.
Mu n is the mean MSI response among all patients
like those in the study who receive the new treatment.
And mu sub s is similar in MSI response
among patients like those in the study who
receive the standard treatment.
Now you could state the hypotheses
in that other format.
It's mathematically equivalent, might even
help if you're going to use the calculator
to do your computations.
And the inference method here is called a two-sample t-test
for a difference in population means.
I used the symbolic two-sample t-test
for mu sub n minus mu sub s.
So be sure to identify that inference method
in some kind of name format or using a formula, which
is coming up shortly.
Next up, you know it, right?
Check conditions.
So there is the hypotheses.
We need to check for independence in the data
collection and the sampling distribution
in this case of the difference in sample means
is approximately normal.
So first up, we had random data protection,
because the patients were randomly
assigned into those standard care or new treatment groups.
We should not check the 10% condition.
Why?
There was no sampling without replacement
from a larger population.
These were recruited patients.
They were just randomly assigned.
So we should not check the 10% condition here.
Now for the sampling distribution
to be approximately normal, we need
to either know that the population distributions are
approximately normally distributed,
which we don't know or we need large sample sizes.
Well, we're in luck there because both sample
sizes are 56, and all we need is sample sizes of at least 30.
So we have large sample sizes here.
And that gives us the sampling distribution
of x bar n minus x bar s being approximately normal.
We can now proceed to calculations,
our favorite part.
Remember those hypotheses, because we
have a one-tailed alternative hypothesis or greater than.
Where standardized test statistic
is given on the formula sheet-- statistic minus parameter
over standard error of the statistic.
There's that formula for the standard error
of a difference in sample means, which
is on the second page of the formula sheet.
And so you can kind of model the formula from there.
The statistic is the difference in the sample means.
0.69 minus 0.57 from the table minus the null value, which
I'm getting from here is zero, no difference,
divided by the square root of the s1 squared over n1
plus s2 squared over n2.
Or we could use the calculator to get that statistic of 2.40
by going to our two samp t-test menu.
If we plug-in all of the values in the menu there.
And then run the 2 samp t-test, you can see we get a t
statistic of 2.40, if we round it two places.
The big difference is that we also
get the p value and the degrees of freedom.
That's important for a two-sample t procedure
because the degrees of freedom is a decimal.
There are alternatives, a conservative method
from the past where you could use the smaller of the two
degrees of freedom from the individual samples, that'd
be 55.
Don't recommend that.
It is acceptable on the exam, but it's a lot easier
to do what I just did and get the correct p
value and the correct degrees of freedom this way.
So I'm going to use the correct degrees of freedom
from technology--
0.0091, degrees of freedom 109.8.
Final step is to make some conclusion in context
about the hypotheses.
So here we go.
The p value, remember, is 0.0091.
And we say because the p value of 0.0091 is less than alpha
equals 0.05, which you should use as the default,
if you're not given a significance
level in the question, we reject the null hypothesis.
We still need a sentence about whether it does or doesn't
provide convincing statistical evidence,
so we say the data provide convincing statistical evidence
that the new treatment results in a higher mean MSI value
than does the standard care among people
similar to the patients in the study.
Notice I'm parroting the stem of the question again.
I know I've said it a couple of times on this video.
But that's an important strategy to help
you get the wording correct and not have to worry about that.
We got through two full free response questions--
one confidence interval, one significance test.
Up next, we will be looking at the investigative task that
is the one on our exam that gives students
a lot of trouble, because it has many different unexpected
parts.
And so I hope you'll be able to tune in for that.
See you next session.