Some user claims CG@Harvard is a plagiarist

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Some user claims CG@Harvard is a plagiarist

Lavender

Fisher's exact test is about comparing the probability parameters of two independent Binomials, and the test statistic is obtained by conditioning the random variable giving the proportion of successes in one sample (S1) on the random variable capturing the total number of successes across the two samples (S=S1+S2). S1 is neither the sample mean, nor the test statistic. The test statistic, call it T, is derived as P(T=t) = P(S1=s1 | S=s), using the simple definition of conditional probability and doing some simple algebra to show it follows a Hypergeometric distribution with known parameters. In no way does it follow from this example that the sample mean can be a test statistic.

Good lord. Yes, S1 is a test statistic there. That's why C&B literally refer to it as a "test statistic." No one claimed S1 is a sample mean. This example was pointed to because it illustrates a general principle that test statistics can be compared to distributions that are conditioned on observed statistics, which numerous posters claimed was not allowed. The specific example where the sample mean is used as a test statistic comes later, but uses the same principle. This has been explained roughly 8,000 times at this point, but they'll never, ever get it.

1 year ago # QUOTE 0 YEA 6 NAY !
Media

Fisher's exact test is about comparing the probability parameters of two independent Binomials, and the test statistic is obtained by conditioning the random variable giving the proportion of successes in one sample (S1) on the random variable capturing the total number of successes across the two samples (S=S1+S2). S1 is neither the sample mean, nor the test statistic. The test statistic, call it T, is derived as P(T=t) = P(S1=s1 | S=s), using the simple definition of conditional probability and doing some simple algebra to show it follows a Hypergeometric distribution with known parameters. In no way does it follow from this example that the sample mean can be a test statistic.

Good lord. Yes, S1 is a test statistic there. That's why C&B literally refer to it as a "test statistic." No one claimed S1 is a sample mean. This example was pointed to because it illustrates a general principle that test statistics can be compared to distributions that are conditioned on observed statistics, which numerous posters claimed was not allowed. The specific example where the sample mean is used as a test statistic comes later, but uses the same principle. This has been explained roughly 8,000 times at this point, but they'll never, ever get it.

S1 is definitely NOT the test statistic. Here's another way to see why you're wrong: C&B say that the test statistic they construct by conditioning S1 on S follows a Hypergeometric(n1+n2,n1,s) distribution. S1 is Binomial (n1,p1). Not Hypergeometric. Ergo it can't be the test statistic. The test statistic is a new random variable which I denoted T above when I showed you how the derivation process works. Again, it's derived as P(T=t) = P(S1=s1 | S=s).

1 year ago # QUOTE 6 YEA 1 NAY !
Lavender

S1 is definitely NOT the test statistic. Here's another way to see why you're wrong: C&B say that the test statistic they construct by conditioning S1 on S follows a Hypergeometric(n1+n2,n1,s) distribution. S1 is Binomial (n1,p1). Not Hypergeometric. Ergo it can't be the test statistic. The test statistic is a new random variable which I denoted T above when I showed you how the derivation process works. Again, it's derived as P(T=t) = P(S1=s1 | S=s).

C&B: "Given the value of S=s, it is reasonable to use S1 as a test statistic and reject H0 in favor of H1 for large values of S1..." Gee, real hard to tell if they use S1 as a test statistic, huh?

You also seem not to understand how conditioning works or what a test statistic is. Try reading the beginning of that chapter by C&B for some basics. Your last two sentences are complete gibberish -- there's no new variable, no T, and that's not how test statistics work. Other than that, real close!

1 year ago # QUOTE 0 YEA 5 NAY !
Media

S1 is definitely NOT the test statistic. Here's another way to see why you're wrong: C&B say that the test statistic they construct by conditioning S1 on S follows a Hypergeometric(n1+n2,n1,s) distribution. S1 is Binomial (n1,p1). Not Hypergeometric. Ergo it can't be the test statistic. The test statistic is a new random variable which I denoted T above when I showed you how the derivation process works. Again, it's derived as P(T=t) = P(S1=s1 | S=s).

C&B: "Given the value of S=s, it is reasonable to use S1 as a test statistic and reject H0 in favor of H1 for large values of S1..." Gee, real hard to tell if they use S1 as a test statistic, huh?
You also seem not to understand how conditioning works or what a test statistic is. Try reading the beginning of that chapter by C&B for some basics. Your last two sentences are complete gibberish -- there's no new variable, no T, and that's not how test statistics work. Other than that, real close!

You can't read. What they're saying is that they're using the *conditional* distribution of S1 given S=s as the test statistic. But in so doing they are implicitly creating a new random variable, different from S1, that acts as the test statistic, and has a Hypergeometric distribution (unlike S1, which is a Binomial).

1 year ago # QUOTE 5 YEA 1 NAY !
Lavender

S1 is definitely NOT the test statistic. Here's another way to see why you're wrong: C&B say that the test statistic they construct by conditioning S1 on S follows a Hypergeometric(n1+n2,n1,s) distribution. S1 is Binomial (n1,p1). Not Hypergeometric. Ergo it can't be the test statistic. The test statistic is a new random variable which I denoted T above when I showed you how the derivation process works. Again, it's derived as P(T=t) = P(S1=s1 | S=s).
C&B: "Given the value of S=s, it is reasonable to use S1 as a test statistic and reject H0 in favor of H1 for large values of S1..." Gee, real hard to tell if they use S1 as a test statistic, huh?
You also seem not to understand how conditioning works or what a test statistic is. Try reading the beginning of that chapter by C&B for some basics. Your last two sentences are complete gibberish -- there's no new variable, no T, and that's not how test statistics work. Other than that, real close!

You can't read. What they're saying is that they're using the *conditional* distribution of S1 given S=s as the test statistic. But in so doing they are implicitly creating a new random variable, different from S1, that acts as the test statistic, and has a Hypergeometric distribution (unlike S1, which is a Binomial).

C&B: "Given the value of S=s, it is reasonable to use S1 as a test statistic and reject H0 in favor of H1 for large values of S1..." Yeah, see where they talk about new variables there and how it's not S1 that's the test statistic. Oh, wait...

Seriously, though, you seem not to have even a basic understanding of how this works. A test statistic involves an observed value that you compare to a distribution. You can't condition an observation itself on another variable. You condition the *distribution* on the other variable. So yes, S1 is the test statistic, just as C&B clearly state. The distribution is conditioned on S.

1 year ago # QUOTE 0 YEA 6 NAY !
Lavender

Stats bros: No, see, you can't use S1 as a test statist--

C&B: "Given the value of S=s, it is reasonable to use S1 as a test statistic and..."

Stats bros: No! No! There must be a different variable called T that's the test stati--

C&B: "it is reasonable to use S1 as a test statistic and..."

Stats bros: No! My brain is melting! Help me!!

C&B: "S1 as a test statistic..."

Stats bros: It can't be... (gasp) ... (gurgle)

1 year ago # QUOTE 0 YEA 6 NAY !
Ryu

HOW THE F*^K ARE YOU PEOPLE STILL TALKING ABOUT THIS?????????????

Its been weeks. Let it go!

1 year ago # QUOTE 3 YEA 2 NAY !
Media

S1 is definitely NOT the test statistic. Here's another way to see why you're wrong: C&B say that the test statistic they construct by conditioning S1 on S follows a Hypergeometric(n1+n2,n1,s) distribution. S1 is Binomial (n1,p1). Not Hypergeometric. Ergo it can't be the test statistic. The test statistic is a new random variable which I denoted T above when I showed you how the derivation process works. Again, it's derived as P(T=t) = P(S1=s1 | S=s).
C&B: "Given the value of S=s, it is reasonable to use S1 as a test statistic and reject H0 in favor of H1 for large values of S1..." Gee, real hard to tell if they use S1 as a test statistic, huh?
You also seem not to understand how conditioning works or what a test statistic is. Try reading the beginning of that chapter by C&B for some basics. Your last two sentences are complete gibberish -- there's no new variable, no T, and that's not how test statistics work. Other than that, real close!
You can't read. What they're saying is that they're using the *conditional* distribution of S1 given S=s as the test statistic. But in so doing they are implicitly creating a new random variable, different from S1, that acts as the test statistic, and has a Hypergeometric distribution (unlike S1, which is a Binomial).

C&B: "Given the value of S=s, it is reasonable to use S1 as a test statistic and reject H0 in favor of H1 for large values of S1..." Yeah, see where they talk about new variables there and how it's not S1 that's the test statistic. Oh, wait...
Seriously, though, you seem not to have even a basic understanding of how this works. A test statistic involves an observed value that you compare to a distribution. You can't condition an observation itself on another variable. You condition the *distribution* on the other variable. So yes, S1 is the test statistic, just as C&B clearly state. The distribution is conditioned on S.

If you read further, they say "The conditional distribution S1 given S=s is hypergeometric (n1+n2,n1,s)." It's not possible for a random variable to have two distributions. Once you have conditioned a random variable on another random variable, you have created a new random variable. But it's a moot point anyway because this is a test of equality of two Binomial parameters, and is completely irrelevant to whether one can use the sample mean as a test statistic.

1 year ago # QUOTE 6 YEA 1 NAY !
Jeffery

Lavender:

If Xbar is the random variable that you claim to use as a test statistic in 8.37.c, explain to us:

1. Which other random variable are you conditioning Xbar on to get the conditional distribution that you’re taking about?

2. What distribution does Xbar given this other random variable follow? Either explain how you would derive its density (and what that resulting density is), or if it’s a named distribution state its names and parameters.

3. Which command and which arguments would you use (in R, or Stata) to get the relevant critical values?

No theatrics, no lamentations, no insulting other posters. Just straight answers to the questions above. Thanks.

1 year ago # QUOTE 4 YEA 1 NAY !
Lavender

If you read further, they say "The conditional distribution S1 given S=s is hypergeometric (n1+n2,n1,s)." It's not possible for a random variable to have two distributions. Once you have conditioned a random variable on another random variable, you have created a new random variable. But it's a moot point anyway because this is a test of equality of two Binomial parameters, and is completely irrelevant to whether one can use the sample mean as a test statistic.

"It's not possible for a random variable to have two distributions." Hahahahaha, it's not? How about if one distribution is conditional on something and the other is not? Ok, done.

Your inability to grasp the most elementary points is incredible. Yes, this example is relevant. For the millionth time, the example shows that you can use test statistics where the distribution is conditioned on an observed statistic, which is the only coherent objection anyone ever made to the sample mean or coefficient being used as a test statistic. This shows that that is actually fine. I'll just keep repeating this until you get it. How much time should I set aside? Is 10,000 years enough?

1 year ago # QUOTE 1 YEA 3 NAY !
Lavender

Lavender:
If Xbar is the random variable that you claim to use as a test statistic in 8.37.c, explain to us:
1. Which other random variable are you conditioning Xbar on to get the conditional distribution that you’re taking about?
2. What distribution does Xbar given this other random variable follow? Either explain how you would derive its density (and what that resulting density is), or if it’s a named distribution state its names and parameters.
3. Which command and which arguments would you use (in R, or Stata) to get the relevant critical values?
No theatrics, no lamentations, no insulting other posters. Just straight answers to the questions above. Thanks.

1. Cool, this has only been said 183,439 times. The estimated standard error. Exactly as C&B do in the example.

2. Also has been said maybe 18,345,101 times. It's a scaled t distribution. Thus, you compare the sample mean to the standard t critical values times the standard error. Exactly as C&B do in the example.

3. You calculate a critical t value and multiply it by the standard error. Very difficult code, but maybe you can figure it out.

I just love the pretense at the end that I haven't been crystal clear in my answers and argument from the very beginning. Hilarious stuff. I literally wrote down explicit equations (amazingly, the *exact* one you find in C&B) and gave consistent straight answers literally dozens of times. Doesn't matter to the diehard stats bros, though, who will never get it or deal with clear answers honestly.

1 year ago # QUOTE 1 YEA 3 NAY !
Lavender

HOW THE F*^K ARE YOU PEOPLE STILL TALKING ABOUT THIS?????????????
Its been weeks. Let it go!

I actually agree with you at this point. I'm beyond bored of this. I proved my point, then found iron-clad support in a stats textbook. Most posters begrudgingly admitted I was right. Now I'm just getting the same questions that have already been answered over and over again. If some people are just incapable of admitting they were wrong, great, fine with me.

1 year ago # QUOTE 1 YEA 5 NAY !

Reply

Post <blockquote>Everyone disagrees with you, including Casella and Berger. The so-called "kill shot" is Exercise 8.37.c from Casella & Berger, which you misinterpreted as an example showing that the sample mean can be a test statistic. The official solutions, written by the authors themselves, revealed that the correct approach is to construct a test statistic T_{n-1} which follows a Student t distribution, which you can compare to the given critical value t_{n-1, \alpha}. Therefore, the sample mean cannot be used as a test statistic. Despite being proved wrong with official evidence, you keep insisting that you're right and everyone agrees with you. Just because you keep telling that lie in the face of evidence from the authors themselves proving you wrong will never make your claims true. How can you stand to look at yourself in the mirror knowing you're such a compulsive liar? This is getting just cringe-worthy. Basically every stats bro troll has now admitted I'm right (the clear C&B evidence was a real killer for them), but we got a couple real diehards who are impervious to the most basic definitions and clear evidence. Admitting they were wrong is just too painful so no amount of logic or facts will ever matter. This complete idiot is still going with the laughable argument that how you prove something to be true negates it being true. That's his actual argument! He actually believes y=3-x is not an equation since you can turn it into y+x=3 and that's an equation. Did you know only one thing can be an equation? That's what this guy's selling! It's just incredible stuff. C&B literally set up a problem showing that you can compare the sample mean to a distribution under the null. They call it a "test." They show it can test your null hypothesis. This is the exact definition of a test statistic. But does it count as one to the stats bros? No, because you compute something else to *prove* the sample mean works as a test statistic. So that like eliminates it working by the made-up Conservation of Test Statistics or something. Do these people have actual brain damage?</blockquote>

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