A population consists of $4$ numbers $\{0, 2, 4, 6\}$. Consider drawing a random sample of size $n = 2$ with replacement.
(a) What is the sampling distribution of $\bar x$?
Is this a normal distribution ? Since $\bar x $~ $N\left(\mu, \dfrac{\sigma^2}{n}\right)$?
(b) Calculate the mean & standard deviation of the sampling distribution of $\bar x$.
I got the answer of mean $\mu$ by $\frac{0+2+4+6}{4} = 3$
Thereafter, I proceed to calculate $\sigma$
$\sigma = \frac{(0 – 3)^2 + (2 – 3)^2 + (4 – 3)^2 + (6 – 3)^2}{4} = 5$
Substituting it back into the sample distribution gives:
$\bar x $~ $N\left(3, \dfrac{5^2}{4}\right)$
Thus, I derive the standard deviation to be:
$\sqrt{\frac{\sigma^2}{n}} = \dfrac{5}{2}$.
However, the answer given was $\dfrac{\sqrt{5}}{\sqrt{2}}$.
Can someone explain why is this so? I'm really quite confused with the whole concept of sampling distribution..
Thanks a lot!
Best Answer
You may have confused variance and standard deviation
With $n=2$, the possible values of $\bar{x}$ are $\{0,1,2,3,4,5,6\}$ with respective probabilities $\frac1{16},\frac2{16},\frac3{16},\frac4{16},\frac3{16},\frac2{16},\frac1{16}$
This is not Gaussian. If anything, you might describe it as a discrete triangular distribution
Your $3$ and $5$ are the expectation and variance of a sample with $n=1$
For a sample with $n=2$ the sum would have expectation $6$ and variance $10$, while the mean of the sample would have expectation $3$ and variance $\frac{5}{2}$ and so standard deviation $\frac{\sqrt 5}{\sqrt 2}$