Intro Stats help (accuracy not guaranteed)

Background:

If it is normally distributed, imagine a bell curve. The center is at 1000. It is pretty fat too, because the standard deviation is high. The x-axis is your score. The number of students who score within, say, 1200 and 1300 is the area of the curve between 1200 and 1300 on the x-axis. You need the score at which 5% of the total area of the curve is to the right of the score.

(Aside: This can't actually be really normally distributed, as the SAT exam has a max and min score...but for the purposes of the problem we might assume this.)

You could technically calculate this with the equation for Normal distributions but we won't. Instead, we shift and shrink this distribution into the standard normal, Z, and back again. This equation is

mytransformedscore = (myscore - meanscore)/SD

Why do we use the Z distribution? Because Z has an area of 1, so 5% = .05. Also it's easier to have just one Z table than a table for every Normal distribution.

Solution:

1. Go to your Z table, find the Z score that leaves .05 area to the right of it (area of right tail). This is your 'mytransformedscore'

2. Fill in meanscore and SD of previous equation. Now use some algebra and solve for myscore

For the percentage: I assume you were given data whose mean is 23.90% The nationwide average is likely a population average - the average of every person in the nation. Your data, most likely, is a sample - the average of only some (hopefully randomly selected) people in the nation. Sample means will vary, but the population mean will not.

As for your second question, consider the formula

(Average y Per x) = (Total y)/(Total x).

Therefore,

(Average Points Per Paper) = (Total Points)/(Total Papers).

Right now, if we substitute the numbers you have into the above equation,

70 = (Total Points)/19

You can solve what your total points are. What you want, however, is

71 = (New Total Points)/(Possibly New Total Papers)

I say possibly new total papers because I don't know if you're supposed to raise the average by adding more papers or just increasing the score of one paper. In any case, know that

New Total Points = Total Points + (Number of Added Points)

New Total Papers = Total Papers (which was 19) + (Number of Added Papers)

An SRS of 20 recent birth records at the local hospital were selected. In the sample, the average birth weight was 121.4 ounces and the standard deviation was 7.5 ounces. Assume that in the population of all babies born in this hospital, the birth weights follow a Normal distribution, with some mean μ.

The standard error of the mean is A. 27.1. B. 6.1. C. 1.7. D.0.4.

Background:

Knowing what any distribution looks like allows you to figure out things like what percentage of the population is bigger than this number or is in between values a and b. The Normal distribution is super amazing because, among other things, if you are looking at a distribution of an average (say, the sample mean), by Central Limit Theorem, that distribution will always become Normal as n increases regardless of the population distribution. This is great because that means it doesn't matter if you have some crazy population distribution or don't even know what it is, you can still make inferences on its sample mean with sufficient sample size.

Now "sufficient sample size" is very important-this sample size is bigger when your population distribution is more skewed, and smaller if it's fairly symmetric/Normal shaped already. If you know it is already Normal (as in here), then the sample mean is definitely normally distributed (actually don't even need CLT for that!).

Standard error of the mean is the same thing as the standard deviation of the sample mean, but it's just sort of nice to differentiate that from the parameter 'standard deviation' (there's other reasons, but I chose this one).

SE = (standard deviation of population)/(square root of sample size). This should be true for any iid sample, and can be gotten from variance rules and the fact that sample mean = (sum of iid variables)/n

Solution:

The underlying population is Normal, thus small sample size (20) is not an issue.

We know the true standard deviation of the population (7.5), so do not need t distribution.

Our sample mean Xbar is Normally distributed with the same mean as the original population, with standard deviation (ie, SE) of (pop SD)/(sqrt(sample size))

Take, for example the Lisa and Bart example on the wikipedia page, where Lisa improves 0/3, 5/7 papers, and Bart improves 1/7, 3/3 papers. If we look at only proportion of papers improved per day, Bart wins both times, but combining the numbers (5/10, 4/10) respectively, we see that Lisa is overall better.

The problem here is that Lisa's failure the first day (0/3) is given the same weight as Bart's number (1/7) that day, even though Bart failed to improve a lot more papers. With bigger numbers, it'd be like seeing that Lisa failed to improve the single paper she was given, but Bart managed to improve 1 out of the 100 papers he was given, so Bart is better. I think the great confusion is that people tend to think of proportions as equally important-and if the sample sizes in all the groups were the same (say, they each graded 5 papers both days), then such a paradox shouldn't happen (of course, you should still consider possible confounders).

People, do anyone know/study about statistics?

I have note idea what means my homework… :

” X~U(-PI,PI) and Y = C tg(x), determine f_y(y) “

Nevertheless, I’m guessing that your X is uniformly distributed from -pi to pi, so what is f(x)? Using f(x), substitute it as your x into your equation of y. You can probably use the transformation of variables formula in your book, though I personally like to start with the cdf P(Y < y) = P(h(X) < y) = P(X < h^(-1)(y)) = F(h^(-1)(y)), then use chain rule (though really it’s the same thing, just more intuitive IMO)

It was c*tan(x), c constant.

What is cdf and IMO?

cdf is the cumulative distribution function. It is the integral of the density (ie, the probability density function, or pdf). If X has a cdf F(X), then the value of F at X = k is simply P(X < k). The cdf is also called the distribution function (it's pretty confusing since some people will misspeak and refer to the pdf as the probability distribution function, so watch out!)

thestripedshirtgirl:

Seriously does anyone know anything about hypothesis testing I REALLY need help D:

If you’d still like help, what are you confused on?

thestripedshirtgirl:

Seriously does anyone know anything about hypothesis testing I REALLY need help D:

If you’d still like help, what are you confused on?

Oh my god, I can’t believe a blog exists for helping with stat stuff, I wish I knew about you earlier!

Anyway, yeah I still need help, specifically I’m having trouble undrsranding how to establish where the critical values are on a distribution so I can compare them to my t obt., and I have no understanding if what degrees of freedom are whatsoever or if theyre related to the other issue or what.

Background:

Presumably, you have some sample, and you are interested in its sample mean. Assuming you've shown that you can use a t-test here (or the book tells you to do it), you now want to test the hypothesis that the true mu = 5, ie, H_0: mu = 5. I'm also going to assume your significance level is .05

If your alternative is right tailed (H_1: mu > 5), you want to find sample mean M* such that P(M > M*) = .05 if the true mu = 5. In this case, you transform both sides of that inequality into the relevant t-distribution:

P( (M - 5)/SE > (M* - 5)/SE ) = .05

Because (M - 5)/SE is t-distributed with n-1 degrees of freedom (usually, I'll talk more on this later),

P( (M - 5)/SE > tstat ) = 1 - pt(tstat)

where pt(tstat) = P (T < tstat). Here, you look up tstat in a t-table with the relevant degrees of freedom, and then find the associated probability. Note that some books look at the right tail and some the left, so be careful! Anyway, this is what you want to do:

1. Find a tstat such that P(T > tstat) = .05. (do this using a computer or a t-table)

2. Solve for M* in  (M* - 5)/SE = tstat

  Now, for a two-tail alternative, it's very similar except you split that .05 region on either side, so on one side, that P(T > tstat) = .025 now:

1. Find a tstat such that P(T > tstat) = .025. (do this using a computer or a t-table)

2. Solve for M* in  (M* - 5)/SE = tstat

3. Your critical value is |M*|, and you compare if |M| > |M*|

  Left-tail is similar to right-tail, except switch the inequality.

And of course, if your null hypothesis wasn't 5 but some other number, replace the 5 with that number.

  A final note: When you talk about critical values, you are usually talking about the critical sample mean M*, so if you get a sample mean M that's beyond that critical M*, you reject, otherwise you fail to reject. However, if you've already got your t statistic, ie, you've done (M - mu_0)/SE, then you only have to compare to that 'critical' tstat, which can be found by looking it up in a t-table/on a computer.

  So, degrees of freedom:

Recall that t-distribution is used in cases where you know what you're estimating has a Normal distribution, but you don't know the true variance sigma^2. Instead of sigma^2, you approximate using the sample variance, but this introduces some bias (deviation from true variance, say) that luckily the t-distribution corrects for. However, you'd also expect that, with increasing sample size, your bias gets smaller. This is why as n increases, the t-distribution converges to the Z-distribution. Therefore, unlike the unique Z distribution, you actually get a separate t-distribution for every different degrees of freedom!

Degrees of freedom for a single i.i.d. sample of size n is n-1. It's used in a lot of other contexts, but I'm guessing you're interested in if you need it to calculate a critical value, and the answer is yes, because it determines the t-distribution you're using.

People, do anyone know/study about statistics?

I have note idea what means my homework… :

” X~U(-PI,PI) and Y = C tg(x), determine f_y(y) “

Nevertheless, I'm guessing that your X is uniformly distributed from -pi to pi, so what is f(x)? Using f(x), substitute it as your x into your equation of y. You can probably use the transformation of variables formula in your book, though I personally like to start with the cdf P(Y < y) = P(h(X) < y) = P(X < h^(-1)(y)) = F(h^(-1)(y)), then use chain rule (though really it's the same thing, just more intuitive IMO)