In Exercises 7–10, use the same population of {4, 5, 9} that w...

Question

In Exercises 7–10, use the same population of {4, 5, 9} that was used in Examples 2 and 5. As in Examples 2 and 5, assume that samples of size n = 2 are randomly selected with replacement.

c. Find the mean of the sampling distribution of the sample variance.

Accepted Answer

All right, hello, everyone. So this question is asking us to consider the population of 26, and 14. Calculate the mean of the sampling distribution of the sample variants if samples of size N equals 2 are randomly selected with replacement. Option A says 5.0, B says 6.1, C says 24.9, and D says 26.2. So, first and foremost, we have to understand. The list of all possible samples of size and equals 2, with replacement from the given population. So, because sampling is with replacement, each draw is independent, which leads a total of 9 possible samples. So the list of all possible samples are as follows. We can have 2 and 2. 2 and 6 2 and 14. 6 and 2 6 and 6. 6 and 14. And 14 and 2. 14 and 6. And 14 and 14. So now let's calculate the sample variants for each sample. Now These sample variants or as squared. For a sample size of 2. Is equal to X1. Subtracted by the mean squared. Added to X2 subtracted by the mean squared. So, because the sample mean, or X bar in this case, is equal to the sum of both values in the sample. Divided by 2, we can plug this into our formula of sample variance and obtain the following modified expression. The modified version of sample variants is equal to. X1 subtracted by X2 squared divided by 2. And so using the derived formula, you can calculate the sample variants for each possible sample, and then finish your calculations by calculating the mean. Of the sampling distribution of the sample variants. So here, I'm going to list out the sample variances for each possible sample. For 2 and 2, that's 0. And in fact, for every single sample where the two numbers are repeated. That's going to be true as well. For 2 and 6, that's going to be 8. And for 6 and 2, that's also going to be 8. For 2 and 14, that's 72. And the same is true for 14 and 2. For 6 and 14, that's going to be 32. And for 14 and 6, that's also going to be 32. So now, our final step to reach our final answer is to calculate the mean of all sample variances. So the mean Is equal to the sum of all sample variances calculated earlier. Divided by the total number of sample variances, which is 9. And this approximately equals 24.9. Which corresponds to option C in the multiple choice. And there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

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In Exercises 7–10, use the same population of {4, 5, 9} that was used in Examples 2 and 5. As in Examples 2 and 5, assume that samples of size n = 2 are randomly selected with replacement.

c. Find the mean of the sampling distribution of the sample variance.

Key Concepts

Sampling Distribution

Sample Variance

Mean of the Sampling Distribution

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In Exercises 7–10, use the same population of {4, 5, 9} that was used in Examples 2 and 5. As in Examples 2 and 5, assume that samples of size n = 2 are randomly selected with replacement.c. Find the mean of the sampling distribution of the sample variance.