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### Course: AP®︎/College Statistics > Unit 11

Lesson 3: Carrying out a test for a population mean- When to use z or t statistics in significance tests
- Example calculating t statistic for a test about a mean
- Calculating the test statistic in a t test for a mean
- Using TI calculator for P-value from t statistic
- Using a table to estimate P-value from t statistic
- Calculating the P-value in a t test for a mean
- Comparing P-value from t statistic to significance level
- Making conclusions in a t test for a mean
- Free response example: Significance test for a mean

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# Free response example: Significance test for a mean

AP Statistics free response on significance test for a mean.

## Want to join the conversation?

- Is it possible to calculate the P-Value without the use of technology?(9 votes)
- I tried this question on my own prior to watching the rest of the video, and came to the same conclusion using an estimate from this t-table: https://www.stat.tamu.edu/~lzhou/stat302/T-Table.pdf(4 votes)

- I do believe the H0 should be < 128 as the problem mentioned the status que is at least 128 - what do you think ?(4 votes)
- "At least 128" means 128
*or more*(e.g. 129, 130...)(2 votes)

- If they fail and add more milk, that wouldn’t be coinsidered not with the regulations? meaning that Ha = mu != 128? Double tail?(4 votes)
- how can we calculate the P-value from the t-statistic manually without a TI-83.(2 votes)
- By using a t table. There is a video on KA that shows how to do that.(4 votes)

- Would this video also help with preparation for the 2021 AP stats exam?

I assume yes but want to check(3 votes) - In the question, it states that the mean number of fluid ounces of milk in the containers is at least 128, but why the null hypothesis is mu=128? And why we can conclude that the alternative hypothesis is that mu is smaller than 128?(2 votes)
- The sample mean is below the minimum requirement of 128 fl.oz., which makes us suspect that the population mean is also below the minimum requirement (i.e., 𝜇 < 128), and that is what we're testing for.(2 votes)

- Is it possible to calculate the P-Value without the use of technology?(2 votes)
- Is it not P(t < t-value | (given that) mu = 128)?

Also, why is it assumed that alpha is .05?(2 votes) - Good accompanying article to this unit: Statistics for people in a hurry https://towardsdatascience.com/statistics-for-people-in-a-hurry-a9613c0ed0b(2 votes)
- Is it also a valid answer to construct a confidence interval and see if the mean falls in that interval.(2 votes)
- It is, but from what I understand, when no specific method is mentioned in the question, the AP exam expects you to use a z- or t-test(1 vote)

## Video transcript

- [Instructor] Regulations
require that product labels on containers of food that
are available for sale to the public accurately
state the amount of food in those containers. Specifically, if milk
containers are labeled to have 128 fluid ounces and
the mean number of fluid ounces of milk in the containers is at least 128, the milk processor is considered to be in compliance with the regulations. The filling machines can be
set to the labeled amount. Variability in the filling process causes the actual contents
of milk containers to be normally distributed. A random sample of 12
containers of milk was drawn from the milk processing line in a plant, and the amount of milk in
each container was recorded. The sample mean and standard
deviation of this sample of 12 containers of milk were 127.2 ounces and 2.1 ounces, respectively. Is there sufficient evidence to conclude that the packaging plant is not in compliance with the regulations? Provide statistical
justification for your answer. So pause this video, and see
if you can have a go at it. All right, now let's do this together. So first, let's say what
we're talking about. So let me define mu. And this is going to be the mean, mean amount, amount of milk in population, population of containers, containers at the plant that we care about. And so then we can set up our hypotheses. Our null hypothesis over here
is that we are in compliance. We could say that the mean for our population of
containers is actually 128. That's our minimum we
need to be in compliance. And then our alternative hypothesis is that we are not in compliance. So that's that our mean, the true population mean, is less than 128 fluid ounces. And so this is a situation
where we are not in compliance, not in compliance, compliance in the alternative hypothesis. Now, if you're going to
do a significance test, you need to set a significance level. So let's do that over here, significance level. And if you haven't noticed, I'm trying to do, in this video, what would be expected of you on a test. This is an actual
question from an AP exam. So our significance level here,
I'll just pick it to be 0.05 because, well, that's
a fairly typical one. And since they didn't give one to us, it's important to set one ahead of time. And now we want to check our
conditions for inference. So let me do that over here, conditions, conditions for inference. And this is to feel good that
the sample that we're using to make our inference, to
do our significance test, that it's a reasonable one
to make inferences from. And so the first one is,
are the random condition. And do we need that? Well, they tell us here
it's a random sample of 12 containers of milk. If I was doing this on the AP exam, I would write it out here. So I would say, in the
passage or in the question, in the question, they say, they say a random, a random sample of 12, and then they go
on to say more things. And so I would say that meets condition, meets condition. Now the next one we want to care about is our normal condition,
and this is to feel good that our sampling distribution
is roughly normal. Now, there's a couple of
ways that we could do that. One is if our sample
size is greater than 30 or greater than or equal
to 30, then we'd say, okay, our sampling distribution's
going to be roughly normal. But in this situation, our
sample size N, so sample size, sample size is less than 30, but, but there's another way to
meet the normal condition. And that's if the underlying parent data is normally distributed. And they actually say it right over here. Variability in the filling process causes the actual contents of milk to be normally distributed. So we could say in passage, in passage says, and let's see, I could quote part of this. So actual contents, actual contents and dot, dot, dot, normally distributed, normally distributed. So that meets condition, meets condition. And then the last condition we want to think about is the
independence condition, independence. And this is to feel good
that the observations, that the individual
observations in our sample can be considered to
be roughly independent. Now one way is if they were
sampling with replacement, which they're not doing here. Looks like they took all
12 containers at once. But another way is if
this is less than 10% of the overall population, then you could say, okay, they're gonna, you could view them as
roughly independent. And so you can say didn't, didn't sample with replacement, with replacement, but, but assume, assume that 12 is less than 10% of the population. And in that case, you
would meet condition, meet this condition as well. So it looks like we are, we've met these three conditions that we needed to make for inference, or we can assume we've done it. They haven't given us any
information to the contrary. And so now, what we can do
is calculate a t-statistic and then, from that,
calculate our P-value, compare our P-value to
our significance level, and see what kind of
conclusions we can make. And so our t-statistic right
over here, and, once again, if at any point you're
inspired and if you haven't done so already, try to do it on your own. Our t-statistic is going
to be our sample mean minus the assumed mean
from the null hypothesis. And let me, since I'm
introducing this notation, this little sub zero, I'll
say that's the assumed, assumed mean from my null hypothesis. So I'll do that, and then I'll divide, ideally, if I was doing a z-statistic, I would divide by the standard deviation of the sampling distribution
of the sample mean, which is often known as the
standard error of the mean. But the whole reason why I'm
doing a t-statistic is, well, I don't know exactly what
that is, but I could estimate the standard deviation of
the sampling distribution of the sample mean using the
sample standard deviation divided by the square root of n. And once again, it's always good, if you're doing this on a
test, to explain what n is or what some of these things are. If you're using standard notation, people might assume what they are, but, if you have time on these tests,
you can always explain more of what these actual variables are. But in this case, this
is going to be 127.2, that is our sample mean, minus our assumed mean from
our null hypothesis, minus 128, all of that over our sample
standard deviation is 2.1, divided by the square root of 12. And so this is going to
be approximately equal to, get a calculator out here, and so we have, let's see,
the numerator, we have 127.2 minus 128. And then we're gonna divide that by, I'll do another parentheses, 2.1 divided by the square root of 12. And then let me close my parentheses. Did I type that in correctly? Yeah, that looks right. Click enter, and so this is negative, I'll say it's approximately negative 1.32. So negative 1.32. And now we can figure out our P-value, our P-value, which is the
same thing as the probability of getting a t-statistic
this low or lower. So we could say t is less than
or equal to negative 1.32, is equal to, so I'll get
my calculator back out. And so here, what I would
use is I would use the cumulative distribution
function for t-statistic. So that's that right over there. And so I do care about the left tail. So I care about the area under the curve from negative infinity up to
and including negative 1.32. So let's do negative, negative 1.32. And then my degrees of freedom, well, it's gonna be my
sample size minus one. My sample size was 12,
so that minus one is 11. And then I do paste. And so I have this tcdf from negative E 99 to negative 1.32 comma 11. And actually, you'd
want to write this down on your exam if you were doing it, just so they know where you got that from. And so this is, this is equal to 0.107. So let me write it. This is approximately 0.107. And it's important to say
how you calculated this, so used, used tcdf. And we went from negative one times 10 to the 99th power. And we went up to negative 1.32, and then we had 11 degrees of freedom to get this result right over here. And it also might be good practice to draw your t-distribution
right over here. So that's our t-distribution. That's the mean of our t-distribution. So we say that this is the
area that we care about. So that is that right over there, just to make sure people know
what we're talking about. And so here, now we're
ready to make a conclusion. We can compare this to
our significance level. And so we can say since, since 0.107 is greater than, our significance level
is greater than 0.05, which is alpha, we fail, we fail to reject, reject the null hypothesis. And so let's just make sure
we read their question right. Is there sufficient evidence to conclude that the packaging plant
is not in compliance with the regulations? And so another way of
saying this is there is not, there is not sufficient, sufficient, I'm gonna have
to scroll down a little bit. I'm trying to squeeze it on the page, but I'm gonna have to go down. There is not sufficient evidence, sufficient evidence to conclude, to conclude that the plant is not in compliance with regulations. And then we are done.