Prediction error in a random sample

[Charlotte87]

I have an exercise that I do not understand how to solve (statistics and probability is really my weaker part...). The exercise goes as follow:

In a certain population, the random variable Y has variance equal to 490. Two independent random samples, each of size 20, are drawn. The first sample is used as the predictor of the second sample mean.

a) Calculate the expectation, expected square end variance of the prediction error.
b) Approximate the probability that the prediction error is less than 14 in absolute value.

any clues how I can start?

 

[Simon Bridge]

You need to look at how sample means are distributed.

 

[Charlotte87]

But how do I know how it is distributed? There is no information about that in the exercise.

 

[Simon Bridge]

The information is presented in your course notes - the part where it talks about how to combine two samples perhaps? You will also see that the variance calculation is different for a sample than for a population.

The distribution of the means of successive samples is related to the distribution of the population.

 

[blue_raver22]

As a scientist with expertise in statistics and probability, I would be happy to provide some guidance on how to approach this exercise.

Firstly, it is important to understand the concept of prediction error in a random sample. In simple terms, prediction error refers to the difference between the predicted value and the actual value. In this exercise, we are looking at the prediction error of the second sample mean using the first sample as a predictor.

To start solving this exercise, we need to use the given information to calculate the expectation, expected square end variance of the prediction error. The expectation of a random variable is its average value, and it is denoted by E(X). In this case, the random variable Y has a variance of 490, so the expectation of Y would be the square root of 490, which is approximately 22.1. This means that on average, the predicted value of Y would be 22.1.

Next, we need to calculate the expected square end variance of the prediction error. This can be done by first finding the variance of the first sample mean, which is equal to the variance of Y divided by the sample size (20 in this case). So the variance of the first sample mean would be 490/20 = 24.5. Then, we can find the variance of the prediction error by adding the variances of the first sample mean and the second sample mean, which would be 24.5 + 24.5 = 49. This means that the expected square end variance of the prediction error is 49.

Moving on to part b, we are asked to approximate the probability that the prediction error is less than 14 in absolute value. To do this, we can use the Central Limit Theorem, which states that the distribution of sample means from a population with a finite variance will be approximately normal, regardless of the shape of the original population. In this case, we can treat the prediction error as a normal distribution with a mean of 0 and a standard deviation of √49 = 7.

To approximate the probability, we can use the Z-score formula. The Z-score is calculated by subtracting the mean from the value of interest (in this case, 14) and then dividing by the standard deviation. So the Z-score would be (14-0)/7 = 2. This means that the probability of the prediction error being less than 14 in absolute value is approximately 97.