If a random sample of size nequals=6 is taken from a​ population, what is required in order to say that the sampling distribution of x overb

Question

If a random sample of size nequals=6 is taken from a​ population, what is required in order to say that the sampling distribution of x overbarx is approximately​ normal?

in progress 0
Parker 1 week 2021-11-21T10:57:38+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-11-21T10:59:30+00:00

    Answer:

     \bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

    In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can’t apply the central limit theorem.

    Step-by-step explanation:

    For this case we have that the sample size is n =6

    The sample man is defined as :

     \bar X = \frac{\sum_{i=1}^n X_i}{n}

    And we want a normal distribution for the sample mean

     \bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

    In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can’t apply the central limit theorem.

    So for this case we need to satisfy the following condition:

     X_i \sim N(\mu , \sigma), i=1,2,...,n

    Because if we find the parameters we got:

     E(\bar X) =\frac{1}{n} \sum_{i=1}^n E(X_i) = \frac{n\mu}{n}=\mu

     Var(\bar X)= \frac{1}{n^2} \sum_{i=1}^n Var(X_i) = \frac{n\sigma^2}{n^2}= \frac{\sigma^2}{n}

    And the deviation would be:

     Sd (\bar X) = \frac{\sigma}{\sqrt{n}}

    And we satisfy the condition:

     \bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )