Consider the random experiment of drawing n goods at random...

Consider the random experiment of drawing n goods at random from a block of d defective goods and g qualified ones, where $d, g > 0$ and $n < d + g$. Problem 1. On population and sample: • Identify the population random variable $X$. • Identify the population probability distribution. Problem 2. Scenario I of sample: Drawing with replacement • Is the sample $X_1, \dots, X_n$ a SRS? Why? • Calculate the probability to get a defective one in the 2nd trial, i.e., $P(X_2 = 1)$. • Identify the probability distribution of the sample count $N_1$ of defective goods within the sample. • Evaluate $E[N_1]$ and $Var[N_1]$. Problem 3. Scenario II of sample: Drawing without replacement • Is the sample $X_1, \dots, X_n$ a SRS? Why? • Calculate the probability to get a defective one in the 2nd trial, i.e., $P(X_2 = 1)$. • Identify the probability distribution of the sample count $N_2$ of defective goods within the sample. • Evaluate $E[N_2]$. Problem 4. Whether $E[N_2] = E[N_1]$? Why?

Transcript

00:01 So for this problem, we know that we have that x is distributed as a binomial random variable, where the number of trials, n, is equal to 100, and the probability of success on a single trial is 0 .75.

00:16 We have that the expected value of x would be equal to n times p, so the expected value of x would be not 0 .0 anything, pardon me.

00:27 The expected value of x would be 75, and we'd have that the value of x would be 75, and we'd have that the variance of x is equal to n times p times one minus p so that's going to be 100 times 0 .75 times 0 .25 which gives a result of, one second here gives us a result of 18 .75.

00:55 Now if we have that p hat, our sample proportion, is equal to x over n, then that tells us that the expected value of, of p hat or the mean value of p hat is going to be equal to 75 over 100, e of x over n.

01:16 So that gives us expected value of p hat is 0 .75.

01:20 And we'd have that the variance of p hat is going to be equal to 18 .75 divided by n again.

01:36 So we'd be dividing by 100.

01:39 Or actually, pardon me, i need to back up one second here.

01:43 So if we're talking about x as an individual sort of random variable, then it's true that the variance of x would be 18 .75.

01:54 But if we're dealing with samples, so x is an individual sample mean value, or yeah, individual sample value, sample proportion, we'd actually have that the variance is going to be, oh, actually, i'm just, i would end up overcomplicating this here.

02:11 I'll go the little bit simpler route here.

02:13 So we'd have that the variance of p -hat is going to be given by our variance for the original variable, 18 .75, divide by 100.

02:26 But we have the catch that when we're dealing with...

02:31 Okay, actually, i'll put it this way.

02:32 So we have variance of p -had is going to be 18 .75 over 100, or 0 .1875.

02:39 But when we're applying the central limit theorem, we have that the mean value of the sampling distribution, of p hat values, it's going to be the same thing as the expected value of p hat, but the standard deviation of those p hat values is going to be equal to the square root of the variance, so square root of 0 .1875, over the square root of the sample size, over 100.

03:08 Or pardon me, over square root of 100...

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