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A welterweight boxing team has 4 boxers who weigh $68.9kg$, $70.7kg$, $72.5kg$ and $74.3kg$. Determine all possible random samples (with replacement) of $n=2$ and calculate the mean of these samples. Using this information, calculate the following:

a)The mean of the weight of the four boxers.

b) the standard deviation of the four boxers' weights

c) The mean of the sample means

d) The standard error of the sample means.

My workings:

Well a) is just $\frac{68.9\:+\:\:70.7\:+\:\:72.5\:+\:74.3}{4}=71.6$

b) I used my graphics calculator for this one. Here is the snippet:

enter image description here

c)Now the mean of the sample means. First I construct 6 combination of teams. As expected the mean of the sample means is $71.6$

d) Now the standard error of the sample means. I entered the sample data ($n=6$) into my calculator and got $1.27279$ as my standard error of the sample means. But according to my the answer provided:

enter image description here

***So how can I calculate the standard error of the sample means?

CountDOOKU
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1 Answers1

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So I haven't run the numbers to check the result, but this is how I interpret the question. "With replacement" means (in general) that we are allowed to draw the same element twice. Say we draw a sample of $n=2$ from a deck a cards with replacement. You draw one card and look at it, then you place it back into the deck (hence replacement) and mix, and then you draw another one. A more natural example is rolling a die twice.

So with that in mind, I think they 1) allow $(a,a)$ as a sample. And 2) want you to count $(a,b)$ and $(b,a)$ as different samples. Point 2) miiight be up for debate, but it would be very weird otherwise.

So if there were only $2$ boxers $a$ and $b$, we would get $4$ possible samples with equal probability: $$ (a,a), (a,b),(b,a),(b,b) $$

Milten
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