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andijo76

  • 10 months ago

three randomly selected households are surveyed. the numbers of people in the households are 1,3,8 assume that samples of size n =2 are randomly selelcted with replacement from the population of 1,3,8. Listed below are the nine different sampls 1,1 1,3 1,8 3,1 3,3 3,8 8,1 8,3 8,8 find the variance of each of the nine samples then summerize the sampling distribution of the variances in the format of a table representing the probability distribution of the distinct variance values s^2 Probability 3.3 1 6.3 49

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  1. andijo76
    • 10 months ago
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    can anyone help me ??????

  2. StClowers
    • 10 months ago
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    We have the following cases to consider (1,1), (1,3), (1,8), (3,1) ,(3.3) (3,8), (8,1) (8,3), (8.8). Well we know three of our samples will have a variance of 0. The variance for the case when we have 1 and 3 is 3.3; likewise, the variance for when we have 8 and 3 is 6.3 and the variance for when we have 1 and 8 is 49. The possible variances are 3.3, 1, 6.3, and 49. The probability of having a variance of 1 is 3/9. For a variance of 3.3 its 2/9, for 6.3 its 2/9 and lastly, for a variance of 49 its 2/9. As you can see, 3/9 + 2/9 + 2/9 + 2/9 = 1. I had this same question and this was the answer....see if you plug this in if its works.

  3. andijo76
    • 10 months ago
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    i hope it does to trying to do it on excel does not work so well thank you so much

  4. StClowers
    • 10 months ago
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    I think thats how you do it...I would plug in those fractions and see...tryin to help cuz I know not many other ppl are. lol

  5. andijo76
    • 10 months ago
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    someone tried to show me how to use excell but that was a nightmare

  6. StClowers
    • 10 months ago
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    Yeah, I dont like excel either.

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