First, don’t ask why I’m doing stats on December 21. Long story that makes me want to tear my hair out.
Second, I suck at stats. So talk to me like I’m very slow. I keep reading and rereading my textbook and it could be in Greek for all I understand this. (In fact, an unfortunate percentage of it does appear to be in Greek.)
Third, I think I’ve isolated my confusion with this assignment.
How the hell do we know that? I plugged in 185 into this t distribution calculator, and sure enough, .5 pops up. But then I plugged in 24 (something from another problem I got stuck on), and it also came up as .5.
I just finished a fun stats filled semester. Have you looked at a t-test distribution chart? That might give you a better understanding of how the t-test works and how the df plays a large role in determining the t-test calculation. There are a number easily available online.
Yeah, my book has a t-distribution chart in it, but it seems to be dependent on the confidence interval? Confidence isn’t a factor in this question.
I am just baffled at how I’m supposed to get the p-value from my test statistic. I can’t find a formula at all, it’s some kind of conceptual thing that I’m not grasping.
There isn’t enough information in the OP to even guess what you’re asking, let alone try to formulate an answer. What’s the problem that you’re stuck on, and what does the bit you quoted have to do with it?
Are you sure that the confidence interval is explicitly what the book is identifying? Is it possibly 1 minus the p-value? For instance, if you have a p-value of 0.05 (quite common), then you look for the p-value including 1-.05, equals .95 or 95 % of values. It can be confusing when 95% is also the number usually used to establish a confidence interval.
Or perhaps it is phrased that you need that value to create a confidence interval of a certain range?
I agree with the previous poster that more information is needed to help you. I don’t want to add to the confusion.
In this case, with n=25, you were correct about 24 df.
Then, you have the p-value of .05.
So, all you have to do is look up in a chart, or calculate what the t-value.
I’m using a chart because it explains a little bit more what is going on. Your chart will probably have a list of degrees of freedom on the left hand side. Find 24.
Then at the top will probably be a listing of “Upper-tail probability.” Since this is a one sided hypothesis (ie- just > or< and not both), you want the entire p-value. Some charts might indicate the p-value using the alpha symbol. So look for the value at 1-p-value (1- .05), which is .95.
According to my old fashioned chart, that gives you a t-score of 1.711. That’s your test statistic.
Okay. I found this t-distribution chart, which is easier to read than the one in my book, because it specifically says that I’m looking at p-values. (The columns in my book say t.100, t.050, t.025, etc.) If I’m reading this right, if I have 24 degrees of freedom and my given p-value is .05, my t-score is 2.06. Am I right?
In my book, though, the 2.06 score appears in the column labelled t.025. This is a one-tail test v. two-tail test thing, right?
Thanks for your help, people. I swear, I have never EVER studied anything that baffled me as much as statistics.
That chart appears to be homemade by someone and includes that 2-sided values, listed under the total p-value. It’s confusing- use this one instead (http://www.statsoft.com/textbook/distribution-tables/). You’ll need to scroll down below the z-scores and you should get the same answer are my previous post.
Yes. The t distribution is symmetric around the point zero, so (for example) the area to the right of 1 is the same as the area to the left of -1. That allows you to do both kinds of tests given one table, but it means that you have to be very careful about making sure that you know which type of table you’re looking at.
