Answer to Question #218758 in Statistics and Probability for Dinly

Question #218758

A population consists of the four numbers 1,2,8 and 9. List all the possible sample of size n=2 which can be drawn without replacement from the population find the following:


a. Population mean

b. Population variance

c. Population standard deviation

d. Mean of the sampling distribution of sample means

e. Variance of the sampling distribution of sample means

f. Standard deviation of the sampling distribution of sample means



1
Expert's answer
2021-07-20T06:12:55-0400

We have population values 1,2,8,91,2,8,9 population size N=4,N=4, and sample size n=2.n=2. Thus, the number of possible samples which can be drawn without replacement is



(42)=6\dbinom{4}{2}=6


a.


mean=μ=1+2+8+94=5mean=\mu=\dfrac{1+2+8+9}{4}=5

b.

Variance=σ2Variance=\sigma^2

=14((15)2+(25)2+(85)2=\dfrac{1}{4}((1-5)^2+(2-5)^2+(8-5)^2

+(95)2)=12.5+(9-5)^2)=12.5

c.


σ=σ2=12.5=2.523.535534\sigma=\sqrt{\sigma^2}=\sqrt{12.5}=2.5\sqrt{2}\approx3.535534

d.

NoSampleMean1(1,2)1.52(1,8)4.53(1,9)54(2,8)55(2,9)5.56(8,9)8.5\def\arraystretch{1.5} \begin{array}{c:c:c} No & Sample & Mean \\ \hline 1 & (1,2) & 1.5 \\ \hdashline 2 & (1,8) & 4.5 \\ \hdashline 3 & (1,9) & 5 \\ \hdashline 4 & (2,8) & 5 \\ \hdashline 5 & (2,9) & 5.5 \\ \hdashline 6 & (8,9) & 8.5 \\ \hdashline \end{array}


The sampling distribution of the sample mean xˉ\bar{x} and its mean and standard deviation are:


xˉff(xˉ)xˉf(xˉ)xˉ2f(xˉ)1.511/63/129/244.511/69/1281/24522/620/12200/245.511/611/12121/248.511/617/12289/24Total615175/6\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:} \bar{x} & f & f(\bar{x})& \bar{x}f(\bar{x})& \bar{x}^2f(\bar{x}) \\ \hline 1.5 & 1 & 1/6 & 3/12 & 9/24 \\ \hdashline 4.5 & 1 & 1/6 & 9/12 & 81/24 \\ \hdashline 5 & 2 & 2/6 & 20/12 & 200/24 \\ \hdashline 5.5 & 1 & 1/6 & 11/12 & 121/24 \\ \hdashline 8.5 & 1 & 1/6 & 17/12 & 289/24 \\ \hdashline Total & 6 & 1 & 5 &175 /6 \\ \hdashline \end{array}

E(Xˉ)=xˉf(xˉ)=5E(\bar{X})=\sum\bar{x}f(\bar{x})=5

e.

Var(Xˉ)=σXˉ2=xˉ2f(xˉ)(xˉf(xˉ))2Var(\bar{X})=\sigma_{\bar{X}}^2=\sum\bar{x}^2f(\bar{x})-(\sum\bar{x}f(\bar{x}))^2

=175652=256=\dfrac{175}{6}-5^2=\dfrac{25}{6}

f.


σXˉ=σXˉ2=256=5662.041241\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{\dfrac{25}{6}}=\dfrac{5\sqrt{6}}{6}\approx2.041241

Check

E(Xˉ)=5=μE(\bar{X})=5=\mu

Var(Xˉ)=256=σ2n(NnN1)=12.52(4241)Var(\bar{X})=\dfrac{25}{6}=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{12.5}{2}(\dfrac{4-2}{4-1})




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment