Using a table to estimate P-value from t statistic | AP Statistics | Khan Academy

Katarina was testing her null hypothesis that the true population mean of some data set is equal to zero versus her alternative hypothesis that it's not equal to zero. Then she takes a sample of six observations, and using that sample, her test statistic, I can never say that her test statistic was t is equal to 2.75. Assume that the conditions for inference were met. What is the approximate p-value for Katarina's test?

So, like always, pause this video and see if you can figure it out. Well, I just always like to remind ourselves what's going on here. So there's some population here. She has a null hypothesis that the mean is equal to zero, but the alternative is that it's not equal to zero. She wants to test her null hypothesis, so she takes a sample size, or she takes a sample of size 6.

From that, since we care about the population parameter, we care about the population mean. She would calculate the sample mean in order to estimate that and the sample standard deviation. Then from that, we can calculate this t-value. The t-value is going to be equal to the difference between her sample mean and the assumed population mean from the null hypothesis. That's what this little sub-zero means; it means it's the assumed mean from the null hypothesis divided by our estimate of the standard deviation of the sampling distribution.

I say estimate because unlike when we're dealing with proportions, with proportions we can actually calculate the assumed based on the null hypothesis sampling distribution standard deviation, but here we have to estimate it. So it's going to be our sample standard deviation divided by the square root of n.

Now in this example, they calculated all of this for us. They said, "Hey, this is going to be equal to 2.75," and so we can just use that to figure out our p-value. But let's just think about what that is asking us to do. So the null hypothesis is that the mean is zero; the alternative is that it is not equal to zero.

So this is a situation where, if we're looking at the t-distribution right over here, it's my quick drawing of a t-distribution. If this is the mean of our t-distribution, what we care about is things that are at least 2.75 above the mean and at least 2.75 below the mean because we care about things that are different from the mean, not just things that are greater than the mean or less than the mean.

So we would look at—we would say, "Well, what's the probability of getting a t value that is more than 2.75 or more above the mean?" And similarly, what's the probability of getting a t value that is 2.75 or less below or 2.75 or more below the mean? So this is negative 2.75 right over there.

What we have here is a t-table, and a t-table is a little bit different than a z-table because there are several things going on. First of all, you have your degrees of freedom; that's just going to be your sample size minus one. So in this example, our sample size is six, so six minus one is five.

We are going to be in this row right over here, and then what you want to do is look up your t value. This is the t-distribution critical value. So we want to look up 2.75 on this row, and we see 2.75. It's a little bit less than that, but that's the closest value. It's a good bit more than this right over here, but it's a little bit closer to this value than this value, and so our tail probability—remember this is only giving us this probability right over here—our tail probability is going to be between 0.025 and 0.02.

It's going to be closer to this one; it's going to be approximately this. It'll actually be a little bit greater because we're going to go a little bit in that direction because we are less than 2.757. So we could say this is approximately 0.02. Well, that's 0.02 approximately.

The t-distribution is symmetric; this is going to be approximately 0.02. So our p-value, which is going to be the probability of getting a t value that is at least 2.75 above the mean and/or 2.75 below the mean, the p-value is going to be approximately the sum of these areas, which is 0.04.

Then, of course, Katarina would want to compare that to her significance level that she set ahead of time. If this is lower than that, then she would reject the null hypothesis, and that would suggest the alternative. If this is not lower than her significance level, well then she would not be able to reject her null hypothesis.