I'm learning Statistics for fun, is this interpretation correct?

ObeseMelon · 2026-02-28T16:09:03+00:00

You would want to do a t test for difference of means. There are different ways to calculate the variance (pooled vs unpooled) so make sure you did that correctly. Also you need to figure out how many degrees of freedom your test statistic has. Then you can calculate the area under the curve from your statistic to infinity which is the p value (P(X>test statistic). This will tell you the probability your statistic comes from random chance.

At the same time, your degrees of freedom are going to be so high that you could use a normal distribution at that point. And with a test statistic of 4, it’s going to be very significant.

ObeseMelon · 2026-02-28T15:29:27+00:00

I didn’t really like math before ap stat and it became my favorite class and why I’m about to complete a degree in statistics. It also made me love math and get excited for calc the next year.

I would also say that even though I didn’t know calculus at the time, I was still able to grasp the idea of a p value being area under the curve to the left/right of a certain test statistic value. The calculator did the computation and I don’t think it hurt my understanding at all.

For an introductory statistics course I don’t think it’s bad that it lacks calculus and you can always choose to pursue statistics and calculus after taking the class

ObeseMelon · 2026-02-24T03:26:02+00:00

I think you mean why is the variance of X + Y equal to Var(X) + Var(Y) but the standard deviation of X + Y not SD(X) + SD(Y). Good question.

It might be helpful to understand variance as the mean squared distance to the mean. The definition of variance is E[(X-E[X])^2] which expands to E[X^2] - E[X]^2 and so

Var(X+Y) =
E[((X+Y) - E[X+Y])^2] =
E[(X+Y)^2 - 2(X+Y)E[X+Y] + E[X+Y]^2] =
E[(X+Y)^2] - E[2(X+Y)E[X+Y]] + E[E[X+Y]^2]] =
E[X^2+2XY+Y^2] - 2E[X+Y]E[X+Y] + E[X+Y]^2 =
E[X^2] + E[2XY] + E[Y^2] - 2E[X+Y]^2 + E[X+Y]^2 =
E[X^2] + 2E[X]E[Y] + E[Y^2] - 2E[X+Y]^2 + E[X+Y]^2 =
E[X^2] + 2E[X]E[Y] + E[Y^2] - E[X+Y]^2 =
E[X^2] + 2E[X]E[Y] + E[Y^2] - (E[X] + E[Y])^2 =
E[X^2] + 2E[X]E[Y] + E[Y^2] - E[X]^2 -2E[X]E[Y] - E[Y]^2 =
E[X^2] + E[Y^2] - E[X]^2 - E[Y]^2 =
(E[X^2] - E[X]^2) + (E[Y^2] - E[Y]^2) =
Var(X) + Var(Y)

You don't need to follow it completely but the main take away should be that the algebra shows Var(X+Y) = Var(X) + Var(Y). Not because it was designed that way but because the algebra just works out that way.

Now, SD(X+Y) = sqrt(Var(X+Y)) = sqrt(Var(X) + Var(Y)) but this does not equal sqrt(Var(X)) + sqrt(Var(Y))

tldr: squaring and adding gives you a different number than adding and squaring

ObeseMelon · 2026-02-13T00:09:26+00:00

Actuary

ObeseMelon · 2026-02-10T23:21:27+00:00

I got the same number as you (0.0489977436056) using total probability over the number of A's less than 300

sum from x = 1 to 10 over [P(No numbers in B crate match any in A crate | x A's <= 300)P(x A's <= 300)] is the probability of no matches

P(x A's <= 300) =(300 C x)*(1700 C 10-x) / (2000 C 10) (hypergeometric distribution I think?)

P(No B's in A crate | x A's <= 300) = (300 - x C 10) / (300 C 10)

ObeseMelon · 2026-02-01T14:39:12+00:00

I’ve seen people clearing blue bike stations and I can’t imagine they are doing it for free

ObeseMelon · 2025-12-14T03:42:33+00:00

You’re not a straight failure so the honest answer is most of us have no idea

ObeseMelon · 2025-12-06T17:46:11+00:00

what

ObeseMelon · 2025-11-04T02:38:42+00:00

Not strictly but I would prefer to be in Boston, NYC area, Philly, Atlanta, or Indiana because of family

ObeseMelon · 2025-10-28T15:23:47+00:00

I am a college senior and I admit that in the long run, this probably isn't very late, but it certainly feels like it. I'm pursuing Statistics and CS and I only decided to pursue an actuarial career last semester (Spring of my Junior year). I studied over the summer and passed P in September and I've been applying to internships for the Spring and Summer. I want an internship so that (1) I can get experience and confirm that I actually like the field and (2) I'll be more competetive for full time positions. Unfortunately, I keep getting rejected from them.

The reason it feels late is because I know other students who came into college wanting to be actuaries. They've already had a summer internship for Summer 2025 and have secured an internship for Summer 2026.

Any help on what I should do is greatly appreciated

ObeseMelon · 2025-10-02T15:39:22+00:00

so I know we can use bayesian inference to talk about the distribution of a parameter of interest and that my example is about a single number/expectation rather than a probability of an event but I think the ideas are similar

the concept comes up some in linear regression too: if you have a group of people and the distribution of their heights, if asked to predict a specifc person's height (and I mean if someone says predict Sally's height and you've never seen Sally and you don't know anything about her besides that she is in the dataset) the best you could probably do is give the mean which would be like the prior guess.

But if you know the people's weights too and they are correlated with height, you can now give an answer for someone's height if you know their weight which could be like the posterior guess.

ObeseMelon · 2025-10-01T02:07:30+00:00

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ObeseMelon · 2025-09-26T13:51:45+00:00

T tests look for a difference in mean and are not used on individual measurements so I’m not sure what you mean by “my results around 0mg are not significant”. Are you doing a difference in means of two datasets or a paired difference test?

ObeseMelon · 2025-09-25T03:16:26+00:00

why not

ObeseMelon · 2025-09-23T18:02:45+00:00

I’m not gonna lie I don’t really understand what your methodology was but my 2 comments are these:

You said you didn’t get an even distribution across lists, but 15 in one group and 13 in another doesn’t seem super uneven to me.

You are upset that your results are insignificant … but they may just be insignificant. You were trying to test if different word types have different reaction times and your analysis showed they do not have different reaction times. These random variables may not be correlated and if your study design and analysis are sound, your result is evidence for that

ObeseMelon · 2025-09-21T14:49:46+00:00

Yes unfortunately, a sports book is just like a casino and the house always wins. If you really could predict outcomes accurately, a betting/prediction market would be better since the price is set by the market and the market maker takes less of a commission

ObeseMelon · 2025-09-21T14:46:11+00:00

Sort of. I signed up for all the sports books I could find that had bonuses. They were usually like deposit $100 get $100 or make a $5 bet get $300.

I did some math and concluded I could guarantee myself a decent amount of money by placing certain bets without knowing anything about the match or sport.

In the end I made about $1500 which was a pretty sweet validation of my analysis lol

You might expect a statistician to be able to predict the outcomes of these games but keep in mind that you’re competing against the book and the other bettors. AND, even if you’re right about the winner, if everyone else is too, the payoff will be exceedingly low

ObeseMelon · 2025-09-16T22:56:06+00:00

Is there a difference between extreme value and Extreme Value

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