Statistics and probability
- Writing hypotheses for a significance test about a mean
- Writing hypotheses for a test about a mean
- Conditions for a t test about a mean
- Reference: Conditions for inference on a mean
- Conditions for a t test about a 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
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?(6 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(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?(3 votes)
- how can we calculate the P-value from the t-statistic manually without a TI-83.(1 vote)
- I do believe the H0 should be < 128 as the problem mentioned the status que is at least 128 - what do you think ?(2 votes)
- Would this video also help with preparation for the 2021 AP stats exam?
I assume yes but want to check(2 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?(1 vote)
- 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.(1 vote)
- Good accompanying article to this unit: Statistics for people in a hurry https://towardsdatascience.com/statistics-for-people-in-a-hurry-a9613c0ed0b(1 vote)
- Is it also a valid answer to construct a confidence interval and see if the mean falls in that interval.(1 vote)
- [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.