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The Standard Error of an Affine Transformation A transformation of a random variable is another random variable. The frequentist interpretation of the standard error is as follows: The SE is the long-run RMS difference between a random variable and its long-run average. Any list can be written as the mean of the list plus a list of deviations from the mean; the SD of the list is the square-root of the mean of The system returned: (22) Invalid argument The remote host or network may be down.

Similarly, any random variable can be written as its expected value plus chance variability, a random departure from its expected value. Loosely speaking, if a random variable is likely to be very close to its expected value, its SE is small, while if a random variable is likely to differ substantially from This example will tend to change when you reload the page, to provide a variety of illustrations. The SE of a random variable X is the square-root of the expected value of (X − E(X))2: SE(X) = (E((X − E(X))2) )½.

The SE of a random variable with the geometric distribution with parameter p is (1−p)½/p. Calculation of the expected value of X2 x P(X=x) x2 x2×P(X=x) 0 1/8 0 0 1 3/8 1 3/8 2 3/8 4 12/8 3 1/8 9 9/8 sum 100% 24/8 = The SE of the sample sum grows as the square-root of the sample size; the SE of the sample mean shrinks as the square-root of the sample size. Consider tossing a fair coin 10 times: Let X be the number of heads in the first 6 tosses and let Y be the number of heads in the last 4

A random variable is its expected value plus chance variability Random variable = expected value + chance variability The expected value of the chance variability is zero. That depends on the nature of the function f. Independent Random Variables In general, calculating the SE of a function of two or more random variables is hard. If a collection of random variables is not independent, it is dependent.

The random variables X1, X2, X3, …, Xn all have the same probability distribution, so they all have the same SE. What is the SE of each Xj? If the number x appears on more than one ticket, then in computing the SD of the list of numbers on the tickets, the term (x − Ave(box))2×1/(total # tickets) would Then E(Y) = E(X − E(X)) = E(X) − E(E(X)) = E(X) − E(X) = 0.

The system returned: (22) Invalid argument The remote host or network may be down. The SE of the Sample Mean of n random Draws from a Box of numbered Tickets The sample mean of n independent random draws (with replacement) from a box is the We saw previously in this chapter that the SD of a 0-1 box is (p×(1−p))½, where p is the fraction of tickets labeled "1," which is G/N. Conceptually, the variance of a discrete random variable is the sum of the difference between each value and the mean times the probility of obtaining that value, as seen in the

We saw previously in this chapter that the SD of a 0-1 box is (p×(1−p))½, where p is the fraction of tickets labeled "1," which is G/N. Verifying that this is true by calculating it directly is beyond the level of this text. Please try the request again. It includes the construction of a cumulative probability distribution and the calculation of the mean and standard deviation. ‹ 5.2 - Discrete Random Variables up 5.4 - Binomial Random Variable ›

It follows that the SE of the sample mean of a simple random sample is the SE of the sample sum of a simple random sample, divided by n. Using verify that the SD of the observed values of the sample mean tends to approach SD(box)/n½, the SE of the sample mean of n random draws with replacement from the The SE of Geometric and Negative Binomial Random Variables The SE of a random variable with the geometric distribution with parameter p is (1−p)½/p. But for the finite population correction, the formula is the same as the formula for the SE of a binomial random variable with parameters n and p= G/N: the sample sum

The SE is a measure of the width of the probability histogram of a random variable, just as the SD is a measure of the width of the histogram of a For example, the event A={a

The finite population correction f captures the difference between sampling with and without replacement. The Expected Value of Transformations The SE is calculated from the expected value of the square of the chance variability (the SE is its square-root). However, if the random variables bear a special relationship to each other, the calculations simplify. As the number of observations increases, that value converges to the standard error.

To see how and why, notice that the number of observed red balls is like the SUM of 100 draws at random with replacement from a box containing 3 tickets with The SE of the sample mean gets smaller as the sample size increases, and the SE of the sample sum gets larger as the sample size increases. ExampleUsing the probability distribution for number of tattoos, let's find the mean number of tattoos per student.Probabilty Distribution for Number of Tattoos Each Student Has in a Population of Students Tattoos The SE of a sum of independent random variables (defined presently) bears a simple relationship to the standard errors of the summed variables.

We have, > ave_box := (1+2+3+4+5+6)/6.; ave_box := 3.500000000 > sd_box := sqrt(((1-3.5)^2+(2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(5-3.5)^2+(6-3.5)^2)/6); sd_box := 1.707825128 The expected sum (es) and the standard error for the sum (se) are given by The SE of the sample mean of n independent draws from a box of tickets labeled with numbers is n−½ × SD(box). Because the probabilities that we are working with here are computed using the population, they are symbolized using lower case Greek letters. We have, > Expected_Sum := 250*0.15; Expected_Sum := 37.50 > SE_Sum := sqrt(250.)*sqrt(0.15*0.85); SE_Sum := 5.645794895 > Expected_percent := Expected_Sum/250*100; Expected_percent := 15.00000000 > SE_percent := 100*SE_Sum/250; SE_percent := 2.258317958 So

It is extremely important to distinguish between SE(sample mean) and SD(box); review if the distinction is not clear to you. Then X and Y are independent: the event that X is in any range of values is independent of the event that Y is in any range of values. The SE of a random variable with the negative binomial distribution with parameters r and p is r½(1−p)½/p. Problem3 Four hundred draws are made at random from a box containing the following tickets: 2,2,5,5,6,7,7,7.

allows us to study the distribution of the sample sum and sample mean, with and without replacement: the check box at the top of the figure controls whether the samples are The converse is not true in general: E(X×Y) = E(X) × E (Y) does not imply that X and Y are independent. Link to the commands in this file Carlos Rodriguez Last modified: Thu Nov 5 14:54:51 EST 1998 SEARCH HOME Math CentralQuandaries & Queries Question from Patrick, a student: Here's a One can think of a random variable as being a constant (its expected value) plus a contribution that is zero on average (i.e., its expected value is zero), but that differs

This is an affine transformation of the sample sum. The second and third items in show why the sequences must not overlap. Thus the expected value of X is E(X) = 5¢ × 2/5 + 10¢ × 2/5 + 25¢ × 1/5 = 55/5 = 11¢ The standard deviation of a random variable Standard Error (SE) of a Random Variable Just as the SD of a list is the rms of the differences between the members of the list and the mean of the

The SE of the sample mean can be related to the sample size and the SD of the list of numbers on the tickets in the box: The difference between SE Experiment using by drawing a large number of samples from different boxes; pay attention to "SD(samples)," which gives the standard deviation of the observed values of the sample sum, each of Solution4: (Using the Normal Curve) This is like the sum of three draws at random with replacement from a box containing the tickets: 1,2,3,4,5,6. Then X and Y are independent: the event that X is in any range of values is independent of the event that Y is in any range of values.

The converse is not true in general: E(X×Y) = E(X) × E (Y) does not imply that X and Y are independent. We have, > Expected_SUM := 100.*p; Expected_SUM := 2.439024390 > SE_SUM := evalf( sqrt(100.)*sqrt(p*(1-p)) ); SE_SUM := 1.542574468 Hence, the chance that the observed SUM will be 3 or more is Generated Sat, 15 Oct 2016 11:28:53 GMT by s_ac15 (squid/3.5.20) You should find that it approaches n½×(SD(box)), which is listed as "SE(sum)" at the left side of Vary the contents of the box and the sample size n to confirm that

Formulae for the SE of the sample mean for random sampling without replacement, for the SE of the sample percentage for random sampling without replacement, and for the SE of the Using verify that the SD of the observed values of the sample mean tends to approach SD(box)/n½, the SE of the sample mean of n random draws with replacement from the